CHEM 122 - GENERAL CHEMISTRY II Precipitation calculations Question Bank Set 8

CHEM 122 - GENERAL CHEMISTRY

II - Precipitation calculations

Question Bank - Set 8

Liberty University

Question 1

Question

Calculate the concentration of barium sulfate (BaSO4) in a solution if 50.0 mL

of a 0.0250 M barium nitrate (Ba(NO3)2) solution is mixed with 25.0 mL of

a 0.0200 M sodium sulfate (Na2SO4) solution. The reaction between barium

nitrate and sodium sulfate forms barium sulfate precipitate according to the

following balanced chemical equation:

Ba(NO3)2+ Na2SO4→BaSO4↓+2NaNO3

Solution

Step 1: Calculate the moles of Ba2+ and SO2−

4ions in each solution.

Moles of Ba2+ from barium nitrate solution:

Moles of Ba2+ = Volume×Concentration = 50.0 mL×0.0250 mol/L = 0.00125 mol

Moles of SO2−

4from sodium sulfate solution:

Moles of SO2−

4= Volume×Concentration = 25.0 mL×0.0200 mol/L = 0.0005 mol

Step 2: Determine the limiting reactant in the reaction by comparing the

moles of Ba2+ and SO2−

4ions.

Since the balanced chemical equation indicates a 1:1 ratio of Ba(NO3)2and

Na2SO4needed to form BaSO4, the limiting reactant will be the one that pro-

duces the least amount of BaSO4. In this case, SO2−

4from sodium sulfate is the

limiting reactant because it produces 0.0005 mol of BaSO4compared to 0.00125

mol of Ba2+ from barium nitrate.

Step 3: Calculate the concentration of BaSO4ions in the final solution.

The volume of the final solution is 50.0 mL + 25.0 mL = 75.0 mL = 0.0750

L. Since 0.0005 mol of BaSO4is produced in this volume, the concentration of

BaSO4is:

Concentration of BaSO4=0.0005 mol

0.0750 L = 0.00667 M

Therefore, the concentration of barium sulfate in the final solution is 0.00667

Question 2

Question

A solution contains 0.3 M calcium nitrate and 0.6 M sodium sulfate. Will a

precipitate form when these two solutions are combined? If so, what mass of

the precipitate is formed?

Solution

Step 1: Write the balanced equation for the precipitation reaction between

calcium nitrate and sodium sulfate:

Ca(N O3)2+Na2SO4→CaSO4+ 2NaNO3

Step 2: Determine the possible products of the reaction. Calcium sulfate is

insoluble and will precipitate out of solution, while sodium nitrate will stay in

solution.

Step 3: Determine the net ionic equation for the reaction:

Ca2+ +SO2−

4→CaSO4

Step 4: Calculate the concentrations of the ions in the combined solution.

Since both solutions are dissolved in water, assume that the volumes are addi-

tive.

For calcium ions: [Ca2+]=0.3M

For sulfate ions: [SO2−

4]=0.6M

Step 5: Use the concentrations of the ions to determine the reaction quotient,

Q, for the precipitation reaction.

Q= [Ca2+][SO2−

4] = (0.3)(0.6) = 0.18

Step 6: Compare the reaction quotient to the solubility product constant,

Ksp, for calcium sulfate. If Q > Ksp, a precipitate will form.

The Ksp for calcium sulfate is 2.4×10−5. Since Q>Ksp (0.18 ¿ 2.4×10−5),

a precipitate of calcium sulfate will form.

Step 7: Calculate the mass of the precipitate formed using the stoichiometry

of the balanced equation. Since 1 mole of calcium sulfate is formed for every 1

mole of calcium nitrate reacted,

Moles of CaSO4= Molarity of Ca(NO3)2×Volume of Ca(NO3)2

= 0.3 mol/L ×volume of calcium nitrate solution in L

Then, use the molar mass of calcium sulfate to calculate the mass of the

precipitate formed.

Question 3

Question

Calculate the concentration of sulfate ions in a solution that results from mixing

250.0 mL of 0.200 M barium chloride and 300.0 mL of 0.150 M sodium sulfate.

Assume complete reaction and that the final volume of the solution is 500.0 mL.

Solution

Step 1: Write the balanced chemical equation for the reaction between barium

chloride (BaCl2) and sodium sulfate (Na2SO4) to determine the mole ratio.

BaCl2+ Na2SO4→BaSO4↓+2NaCl

Step 2: Calculate the moles of barium chloride and sodium sulfate, respec-

tively, using the given concentrations and volumes.

Moles of BaCl2= 0.200 M ×0.2500 L = 0.0500 mol

Moles of Na2SO4= 0.150 M ×0.3000 L = 0.0450 mol

Step 3: Determine the limiting reactant by comparing the mole ratios from

the balanced chemical equation. Since the mole ratio of BaCl2to Na2SO4is

1:1, BaCl2is the limiting reactant.

Step 4: Use the limiting reactant to calculate the moles of barium sulfate

formed.

Moles of BaSO4= 0.0500 mol

Step 5: Calculate the concentration of sulfate ions in the final solution. Since

1 mole of barium sulfate produces 1 mole of sulfate ions, the moles of sulfate

ions in the final solution is also 0.0500 mol. Now, we need to find the final

volume of the solution, which is 500.0 mL.

Step 6: Calculate the final concentration of sulfate ions in the solution.

Concentration of sulfate ions = 0.0500 mol

0.5000 L = 0.100 M

Therefore, the concentration of sulfate ions in the final solution is 0.100 M.

Question 4

Question

A solution contains 0.15 M BaCl2and 0.20 M Na2SO4. Will a precipitation

reaction occur when these solutions are mixed? If so, what mass of BaSO4will

be formed? (Given: Ksp = 1.1×10−10 for BaSO4)

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction:

BaCl2(aq) + Na2SO4(aq) −−→ BaSO4(s) + 2 NaCl (aq)

Step 2: Determine the ions that will be present in solution after mixing: -

Ba2+ from BaCl2- SO42−from Na2SO4

Step 3: Calculate the ion product (Qsp) to determine if precipitation will

occur:

Qsp = [Ba2+][SO42−] = (0.15)(0.20) = 0.03

Step 4: Compare Qsp to Ksp to determine if precipitation will occur: Since

Qsp = 0.03 > Ksp = 1.1×10−10, a precipitation reaction will occur.

Step 5: Calculate the mass of BaSO4formed: - Let’s assume the volume of

the resulting solution is 1 L for easier calculations. - Convert moles of BaSO4

to mass using its molar mass (233.4 g/mol):

1 mol of BaSO4= 233.4 g

0.15 M BaSO4×233.4 g/mol ×1 L = 34.31 g

Therefore, 34.31 g of BaSO4will be formed during the precipitation reaction.

Question 5

Question

At a certain location, the monthly average precipitation for January is 3.6 inches,

while the monthly average precipitation for July is 4.8 inches. If the annual av-

erage precipitation is 49.2 inches, determine the average precipitation for each of

the remaining months (February to December) to maintain a constant monthly

average precipitation for the year.

Solution

Let xrepresent the average precipitation for each of the remaining months

(February to December).

Step 1: Determine the total precipitation for the year.

The total precipitation for the year can be calculated as:

Total precipitation = January precipitation+July precipitation+Remaining months precipitation

Total precipitation = 3.6+4.8 + 11x

Step 2: Set up the equation for the total precipitation.

Since the total annual precipitation is given as 49.2 inches, we have:

3.6+4.8 + 11x= 49.2

Step 3: Solve for x.

Combine the known values and solve for x:

8.4 + 11x= 49.2

11x= 40.8

x=40.8

x= 3.72 inches

Step 4: Answer

Therefore, the average precipitation for each of the remaining months (Febru-

ary to December) should be 3.72 inches to maintain a constant monthly average

precipitation for the year.

Question 6

Question

A solution contains 0.2 M of silver nitrate (AgNO3) and 0.1 M of sodium chloride

(NaCl). If these two solutions are mixed together, what is the maximum mass

of silver chloride (AgCl) that can precipitate out? (Assume 100

Given: The molar mass of silver chloride (AgCl) is 143.32 g/mol.

Solution

Step 1: Write the balanced chemical equation for the reaction between silver

nitrate and sodium chloride to form silver chloride. Identify the limiting reagent.

The balanced equation is:

AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)

From the balanced chemical equation, we can see that silver nitrate and

sodium chloride react in a 1:1 mole ratio.

Step 2: Calculate the moles of silver nitrate and sodium chloride in the

solution.

Given: [AgNO3] = 0.2 M [NaCl] = 0.1 M

Volume of solution does not affect the amount of precipitate formed; hence,

1 L of solution is assumed. Therefore, for 1 L of solution: Moles of AgNO3= 0.2

Moles of NaCl = 0.1

Step 3: Identify the limiting reagent. Since the reaction occurs in a 1:1 mole

ratio, the limiting reactant will be the one that is present in lesser moles. In

this case, sodium chloride is the limiting reagent.

Step 4: Calculate the maximum mass of silver chloride that can be formed.

Given: Molar mass of AgCl = 143.32 g/mol

Since the limiting reagent is sodium chloride, all of the 0.1 moles of NaCl

will react to form AgCl.

Number of moles of AgCl = 0.1 moles Mass of AgCl = Molar mass ×

Number of moles Mass of AgCl = 143.32 g/mol×0.1 mol Mass of AgCl = 14.332

Therefore, the maximum mass of silver chloride that can precipitate out is

14.332 g.

Question 7

Question

A chemist mixes 100.0 mL of 0.200 M silver nitrate (AgNO3) solution with

150.0 mL of 0.100 M sodium chloride (NaCl) solution. Determine the mass of

silver chloride (AgCl) precipitate that will form. Assume the reaction goes to

completion.

Solution

Step 1: Write the balanced chemical equation for the reaction between silver

nitrate and sodium chloride:

AgNO3+ NaCl →AgCl + NaNO3

Step 2: Determine the limiting reactant.

Calculate the moles of silver nitrate:

moles of AgNO3= M ×V=0.200 M ×0.100 L = 0.020 mol

Calculate the moles of sodium chloride:

moles of NaCl = M ×V=0.100 M ×0.150 L = 0.015 mol

Since NaCl is the limiting reactant (less moles), we will base our calculations

on it.

Step 3: Calculate the mass of silver chloride produced.

From the balanced chemical equation, we see that 1 mol of NaCl produces

1 mol of AgCl.

Calculate the moles of AgCl formed:

moles of AgCl = moles of NaCl = 0.015 mol

Calculate the mass of AgCl formed:

mass of AgCl = moles of AgCl ×molar mass of AgCl

= 0.015 mol ×(107.87 g/mol)

= 1.62 g

Therefore, the mass of silver chloride precipitate that will form is 1.62 g.

Question 8

Question

A chemist wants to prepare a saturated solution of silver chloride by mixing

100.0 mL of 0.200 M silver nitrate solution with excess calcium chloride. If the

solubility product constant of silver chloride is 1.77 ×10−10, what mass of silver

chloride can be precipitated? (Assume complete precipitation)

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction

between silver nitrate and calcium chloride to form silver chloride precipitation.

AgNO3(aq) + CaCl2(aq)→AgCl ↑+Ca(NO3)2(aq)

Step 2: Determine the limiting reactant in the reaction. Using the molarity

of silver nitrate solution, calculate the moles of silver nitrate:

Moles of AgNO3= Volume ×Molarity = 100.0 mL ×0.200 mol/L = 0.0200 mol

Step 3: Use stoichiometry to find the moles of silver chloride that can be

precipitated. From the balanced equation, 1 mol of silver nitrate produces 1

mol of silver chloride. Therefore, moles of silver chloride precipitated = 0.0200

mol.

Step 4: Calculate the mass of silver chloride precipitated. The molar mass

of silver chloride (AgCl) is 143.32 g/mol. Therefore, mass of silver chloride =

moles of AgCl ×molar mass of AgCl

Mass of AgCl = 0.0200 mol ×143.32 g/mol = 2.87 g

Therefore, the mass of silver chloride that can be precipitated is

2.87 g.

Question 9

Question

A solution of silver nitrate (AgNO3) with a concentration of 0.150 M is mixed

with a solution of sodium chloride (NaCl) with a concentration of 0.200 M. If the

solubility product constant of silver chloride (AgCl) is 1.8×10−10, determine if

a precipitation reaction will occur when the two solutions are mixed.

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction. The

balanced chemical equation for the reaction between silver nitrate and sodium

chloride to form silver chloride precipitate is: AgNO3+ NaCl →AgCl + NaNO3

Step 2: Write the expression for the solubility product constant. The

solubility product constant expression for the formation of silver chloride is:

Ksp = [Ag+][Cl−]

Step 3: Calculate the initial concentrations of Ag+and Cl−ions. Initial con-

centration of Ag+([Ag+]initial) from silver nitrate: 0.150 M Initial concentration

of Cl−([Cl−]initial) from sodium chloride: 0.200 M

Step 4: Determine the maximum amount of AgCl that can dissolve. Use

the solubility product constant to calculate the maximum amount of AgCl ions

that can dissolve: Ksp = [Ag+]initial[Cl−]initial 1.8×10−10 = (0.150)(0.200)

Step 5: Compare the actual and maximum amount of AgCl that can dis-

solve. The calculated value in step 4 is much greater than the solubility product

constant Ksp, indicating that a precipitation reaction will occur when the silver

nitrate and sodium chloride solutions are mixed.

Therefore, a white precipitate of silver chloride will form when the two so-

lutions are mixed.

Question 10

Question

Determine the mass of lead(II) chloride (P bCl2) that can be formed when a

solution containing 3.50 grams of lead(II) nitrate (P b(NO3)2) is mixed with a

solution containing 2.80 grams of sodium chloride (NaCl). Assume that the

reactions proceed to completion.

Solution

Step 1: Write the balanced chemical equation for the reaction between lead(II)

nitrate and sodium chloride to form lead(II) chloride and sodium nitrate.

Pb(NO3)2+ 2NaCl →PbCl2+ 2NaNO3

Step 2: Calculate the molar mass of each compound. The molar mass of

P b(NO3)2is 207.2 + 2(14 + 3(16)) = 331.2 g/mol. The molar mass of NaCl is

23 + 35.5 = 58.5 g/mol. The molar mass of P bCl2is 207.2 + 2(35.5) = 278.2

g/mol.

Step 3: Determine the number of moles of each reactant.



Moles of P b(NO3)2:3.50 g

331.2 g/mol = 0.0106 mol



Moles of NaCl:2.80 g

58.5 g/mol = 0.0478 mol

Step 4: Determine the limiting reactant. In this case, P b(N O3)2is the

limiting reactant because it produces the least amount of product (P bCl2).

Step 5: Calculate the mass of P bCl2formed using the limiting reactant.

Using the stoichiometry of the reaction, 1 mol of P b(NO3)2produces 1 mol of

P bCl2. Therefore, 0.0106 mol of P b(N O3)2will produce 0.0106 mol of P bCl2.

Finally,

Mass of P bCl2= 0.0106 mol ×278.2 g/mol = 2.95 g

Therefore, 2.95 grams of lead(II) chloride can be formed.

Question 11

Question

Calculate the precipitation in inches when 1 inch of rain falls on a 5-acre plot

of land.

Solution

Step 1: Convert 5 acres to square feet.

1 acre = 43,560 square feet

So, 5 acres = 5 ×43,560 square feet = 217,800 square feet.

Step 2: Convert 1 inch of rain to cubic feet.

1 foot = 12 inches

So, 1 inch of rain = 1

12 feet = 1

12 cubic feet.

Step 3: Calculate the volume of rainfall on the plot of land. The volume of

rainfall is given by the formula:

Volume = Area ×Height

where Area = 217,800 square feet and Height = 1

12 cubic feet. Plugging in the

values, we get:

Volume = 217,800 ×1

Volume = 18,150 cubic feet

Step 4: Convert the volume to inches.

1 cubic foot = 1728 cubic inches

So, 18,150 cubic feet = 18,150 ×1728 cubic inches = 31,401,600 cubic inches.

Step 5: Convert the volume in cubic inches to inches of precipitation. Since

31,401,600 cubic inches of rain falls on the 5-acre plot of land, the precipitation

depth in inches is

31,401,600 cubic inches

217,800 square feet = 144.17 inches

Therefore, the precipitation depth is 144.17 inches.

Question 12

Question

Calculate the concentration of a precipitate formed when 100.0 mL of 0.200

M lead(II) nitrate, Pb(NO3)2, is mixed with 200.0 mL of 0.500 M potassium

iodide, KI. Assume the reaction goes to completion and that lead(II) iodide,

PbI2, is the only precipitate formed.

Solution

Step 1: Write the balanced chemical equation for the reaction between lead(II)

nitrate and potassium iodide.

Pb(NO3)2+ 2 KI →PbI2+ 2 KNO3

Step 2: Determine the limiting reactant. First, calculate the moles of lead(II)

nitrate and potassium iodide: For lead(II) nitrate, moles = volume×molarity =

0.100 L ×0.200 mol/L = 0.0200 mol For potassium iodide, moles = volume ×

molarity = 0.200 L ×0.500 mol/L = 0.100 mol

Since lead(II) nitrate is limiting (0.0200 moles ¡ 0.100 moles), we will calcu-

late the amount of precipitate formed based on lead(II) nitrate.

Step 3: Use the balanced equation to determine the moles of lead(II) iodide

formed. From the balanced equation, we see that 1 mole of lead(II) nitrate forms

1 mole of lead(II) iodide. Therefore, moles of lead(II) iodide = 0.0200 mol

Step 4: Calculate the concentration of the lead(II) iodide precipitate. Vol-

ume of lead(II) iodide = 0.100 L + 0.200 L = 0.300 L Concentration of lead(II)

iodide = moles of PbI2

volume =0.0200 mol

0.300 L = 0.0667 mol/L

Therefore, the concentration of lead(II) iodide precipitate formed is 0.0667

Question 13

Question

A chemist wants to determine the concentration of chloride ions in a water

sample. To do so, the chemist adds an excess of silver nitrate to a 100 mL

sample of the water. The resulting precipitate is filtered, dried, and found to

weigh 0.345 g. Given that the molar mass of silver chloride is 143.32 g/mol,

determine the concentration of chloride ions in the water sample in parts per

million (ppm). Assume that the only source of chloride ions in the water sample

is from sodium chloride.

Solution

Step 1: Calculate the moles of silver chloride precipitated. Given: - Mass of

silver chloride precipitate = 0.345 g - Molar mass of silver chloride = 143.32

g/mol

We can use the formula:

moles of substance = mass of substance

molar mass

Substitute the given values:

moles of silver chloride = 0.345 g

143.32 g/mol = 0.0024051 mol

Step 2: Determine the moles of chloride ions in the water sample. Given

that each formula unit of silver chloride contains 1 chloride ion. Therefore, the

moles of chloride ions = moles of silver chloride precipitated Hence, moles of

chloride ions = 0.0024051 mol

Step 3: Calculate the concentration of chloride ions in the water sample. -

Volume of water sample = 100 mL = 0.1 L - Concentration of chloride ions in

ppm is given by the formula:

ppm = mass of solute (in mg)

volume of solution (in L) ×106

Given that 1 ppm = 1 mg/L, we need to convert moles of chloride ions to

mass in mg before calculating the concentration in ppm. Using the molar mass

of chloride ions (35.45 g/mol):

mass of chloride ions = 0.0024051 mol ×35.45 g/mol = 0.0851861 g = 85.2 mg

Now, calculate the concentration of chloride ions in the water sample:

ppm = 85.2 mg

0.1 L ×106= 852,000 ppm

Therefore, the concentration of chloride ions in the water sample is 852,000

ppm.

Question 14

Question

Calculate the mass of lead iodide (PbI2) that will precipitate when 50.0 mL of a

0.200 M lead(II) nitrate (Pb(NO3)2) solution is mixed with 50.0 mL of a 0.250

M potassium iodide (KI) solution. Assume that lead(II) iodide (PbI2) is the

only precipitate formed and that it has a 1:2 stoichiometry with lead(II) nitrate.

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction

between lead(II) nitrate and potassium iodide.

Pb(NO3)2+ 2KI →PbI2+ 2KNO3

Step 2: Determine the limiting reactant by calculating the moles of each

reactant. The limiting reactant is the one that produces the least amount of

product. - Moles of lead(II) nitrate:

Moles = Molarity ×Volume (L)

Moles of Pb(NO3)2= 0.200 mol/L ×0.0500 L = 0.0100 mol

- Moles of potassium iodide:

Moles of KI = 0.250 mol/L ×0.0500 L = 0.0125 mol

Since lead(II) nitrate produces the least amount of lead(II) iodide, it is the

limiting reactant.

Step 3: Calculate the mass of lead iodide precipitated using the stoichiometry

of the balanced chemical equation. - Moles of lead iodide precipitated: From

the balanced equation, 1 mole of lead(II) nitrate produces 1 mole of lead(II)

iodide. Therefore, 0.0100 moles of lead(II) nitrate will produce 0.0100 moles of

lead(II) iodide.

Moles of PbI2= 0.0100 mol

- Mass of lead iodide precipitated:

Molar mass of PbI2= Pb:207.2 g/mol + I:126.9 g/mol ×2 = 459.0 g/mol

Mass of PbI2= Moles ×Molar mass = 0.0100 mol ×459.0 g/mol = 4.59 g

Therefore, 4.59 grams of lead iodide will precipitate in the reaction.

Question 15

Question

A solution is prepared by mixing 50.0 mL of 0.200 M barium chloride (BaCl2)

with 75.0 mL of 0.150 M sodium sulfate (Na2SO4). Calculate the concentra-

tion of each ion remaining in the solution after precipitation of barium sul-

fate (BaSO4) is complete. The solubility product constant of barium sulfate is

1.1×10−10.

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction

between barium chloride and sodium sulfate:

BaCl2(aq) + Na2SO4(aq)→BaSO4(s) + 2NaCl(aq)

Step 2: Determine the limiting reactant to find out how much barium sulfate

precipitates.

First, calculate the moles of each reactant: For barium chloride: 0.0500 L ×

0.200 mol/L = 0.010 mol For sodium sulfate: 0.0750 L×0.150 mol/L = 0.0113 mol

Since there is more sodium sulfate, it is the limiting reactant. Therefore, all

of the barium sulfate will precipitate.

Step 3: Calculate the moles of sulfate ions from sodium sulfate used in the

reaction: 0.0113 mol ×1=0.0113 mol SO2−

Step 4: Calculate the concentrations of each ion in the final solution.

For barium ions: Initial moles of barium ions = 0.010 mol Moles of bar-

ium ions consumed in reaction = 0.010 mol Moles of barium ions remaining =

0.010 mol −0.010 mol = 0 mol Concentration of barium ions = 0 mol

0.125 L = 0 M

For chloride ions: Initial moles of chloride ions = 0.010 mol Moles of chlo-

ride ions consumed in reaction = 0.010 mol Moles of chloride ions remaining =

0.010 mol −0.010 mol = 0 mol Concentration of chloride ions = 0 mol

0.125 L = 0 M

For sulfate ions: Initial moles of sulfate ions = 0.0113 mol Moles of sul-

fate ions consumed in reaction = 0.0113 mol Moles of sulfate ions remaining =

0.0113 mol −0.0113 mol = 0 mol Concentration of sulfate ions = 0 mol

0.125 L = 0 M

Therefore, the concentrations of barium ions, chloride ions, and sulfate ions

remaining in the solution after precipitation is complete are 0 M.

Question 16

Question

In a precipitation reaction, 50.0 mL of a 0.200 M solution of lead(II) nitrate,

Pb(NO3)2, is mixed with 75.0 mL of a 0.150 M solution of sodium iodide, NaI.

Determine the mass of lead(II) iodide, PbI2, that will precipitate. (Assume the

reaction goes to completion and there are no volume changes.)

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction.

Pb(NO3)2+ 2NaI →PbI2+ 2NaNO3

Step 2: Determine the limiting reagent in the reaction. To find the limiting

reagent, we need to calculate the moles of each reactant. We can use the formula:

moles = molarity ×volume (L)

For lead(II) nitrate, Pb(NO3)2:

molesPb(NO3)2= 0.200 M ×50.0×10−3L

1= 0.010 mol

For sodium iodide, NaI:

molesNaI = 0.150 M ×75.0×10−3L

1= 0.01125 mol

Since the stoichiometry of the reaction is 1:2 between lead(II) nitrate and

sodium iodide, lead(II) nitrate is the limiting reagent.

Step 3: Calculate the mass of lead(II) iodide, PbI2, that will precipitate.

Using the stoichiometry of the reaction, we can see that 1 mol of lead(II) nitrate

produces 1 mol of lead(II) iodide. The molar mass of PbI2is 461.01 g/mol.

molesPbI2= molesPb(NO3)2= 0.010 mol

massPbI2= 0.010 mol ×461.01 g/mol = 4.61 g

Therefore, 4.61 g of lead(II) iodide, PbI2, will precipitate.

Question 17

Question

A chemist wants to calculate the amount of precipitate formed when 50.0 mL of

0.200 M silver nitrate (AgNO3) solution is mixed with excess sodium chloride

(NaCl) solution. If the balanced chemical equation is:

AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)

and the molar mass of AgCl is 143.32 g/mol, what mass of precipitate will

be formed?

Solution

Step 1: Calculate moles of silver nitrate (AgNO3) from the volume and molarity

given.

Moles of AgNO3= Volume ×Molarity = 0.0500 L ×0.200 mol/L = 0.0100 mol

Step 2: Use the balanced chemical equation to determine the stoichiometry

of the reaction. From the equation, 1 mole of AgNO3reacts with 1 mole of

NaCl to form 1 mole of AgCl. Therefore, the moles of AgCl formed will also be

0.0100 mol.

Step 3: Calculate the mass of precipitate formed.

Mass of AgCl = Moles ×Molar mass = 0.0100 mol ×143.32 g/mol = 1.43 g

Therefore, the mass of precipitate formed when 50.0 mL of 0.200 M silver

nitrate solution is mixed with excess sodium chloride solution is 1.43 g.

Question 18

Question

Calculate the concentration of chloride ions in a solution after mixing 100 mL

of 0.2 M sodium chloride with 200 mL of 0.1 M silver nitrate. Assume complete

precipitation of silver chloride and no volume change upon mixing.

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction

occurring.

NaCl(aq) + AgNO3(aq)→NaNO3(aq) + AgCl(s)

Step 2: Determine which reactant will limit the formation of the precipitate,

silver chloride.

Step 3: Calculate the amount of silver chloride precipitated.



Moles of NaCl = 0.2 mol/L ×0.1 L = 0.02 mol



Moles of AgNO3= 0.1 mol/L ×0.2 L = 0.02 mol



Since the moles of both reactants are equal, both are in stoichiometric

quantities. Therefore, the limiting reagent is NaCl.



The amount of AgCl precipitated will also be 0.02 mol.

Step 4: Calculate the concentration of chloride ions after precipitation.



Initial moles of chloride ions = 0.1 mol/L ×0.1 L = 0.01 mol



Moles of chloride ions remaining = 0.01 mol - 0.02 mol = -0.01 mol (neg-

ative because all chloride ions have precipitated)



Volume of solution after mixing = 100 mL + 200 mL = 300 mL = 0.3 L



Concentration of chloride ions after precipitation = −0.01 mol

0.3 L =−0.033 M

Step 5: Interpretation of the negative concentration: Since concentrations

cannot be negative, the negative sign in this case indicates that all the chloride

ions have precipitated as silver chloride, and there are no chloride ions left in

the solution.

Question 19

Question

A certain stream has a flow rate of 10 m3/s. If the stream receives 2 cm of rain

over an area of 500 km2, how long will it take for the stream to rise by 1 meter

assuming all the rain contributes to the stream flow?

Solution

Step 1: Convert the area of land receiving rain to square meters. Given that

the area is 500 km2and 1 km2= 106m2, we have:

500 ×106= 5 ×108m2

Step 2: Calculate the volume of rain that fell over the area. Given that 2

cm of rain fell over the area, we first convert it to meters:

2 cm = 0.02 m

Then, the volume of rain that fell over the area is:

0.02 m ×5×108m2= 1 ×107m3

Step 3: Calculate the time it takes for the stream to rise by 1 meter. Given

the stream flow rate is 10 m3/s, we can determine the time it takes for the

stream to rise by 1 meter using the formula:

time = volume of rain

flow rate

Substitute the values:

time = 1×107m3

10 m3/s= 1 ×106s

Therefore, it will take 1,000,000 seconds for the stream to rise by 1 meter.

Question 20

Question

A water sample contains 0.05 M MgCl2, 0.02 M Na2CO3, and 0.03 M Ca(NO3)2.

If you mix 50.0 mL of this solution with another 50.0 mL of a 0.04 M Na2SO4

solution, will a precipitate form? If so, calculate the mass of the precipitate

formed.

Given: - Ksp of MgCO3= 6.82 ×10−6-Ksp of CaCO3= 3.95 ×10−9

Solution

Step 1: Write the balanced chemical equations for the possible precipitates.

For MgCl2and Na2CO3:

Mg2+ (aq) + CO32−(aq) −−→ MgCO3(s)

For Ca(NO3)2and Na2CO3:

Ca2+ (aq) + CO32−(aq) −−→ CaCO3(s)

Step 2: Calculate the initial concentrations and find the ions’ concentrations

after dilution.

Given initial concentrations: - [Mg2+] = 0.05 M - [CO32−] = 0.02 M -

[Ca2+] = 0.03 M - [CO32−] = 0.02 M

After dilution with the Na2SO4 solution: - [Na+]=0.02 M - [SO42−]=0.04

Step 3: Determine the ion product, Q, for each possible precipitation reac-

tion.

For MgCO3:

Q= [Mg2+]×[CO32−]=0.05 ×0.02 = 1.0×10−3

For CaCO3:

Q= [Ca2+]×[CO32−]=0.03 ×0.02 = 6.0×10−4

Step 4: Compare Qto the corresponding Ksp for each precipitate.

For MgCO3:

Q < Ksp (No precipitation)

For CaCO3:

Q < Ksp (No precipitation)

Conclusion: No precipitate will form when the two solutions are mixed.

Question 21

Question

A solution is prepared by dissolving 15.0 g of barium chloride in enough water

to make 250 mL of solution. This solution is then mixed with 100.0 mL of a

0.150 M solution of sodium sulfate. Will a precipitate form? If so, what mass

of precipitate will form?

(Note: The molar mass of barium chloride is 208.23 g/mol and the molar

mass of sodium sulfate is 142.04 g/mol. The solubility product constant (Ksp)

for barium sulfate is 1.08 ×10−10 at 25

C.)

Solution

Step 1: Calculate the moles of barium chloride (BaCl2) present in the solution.

Given: - Mass of BaCl2= 15.0 g - Molar mass of BaCl2(MBaCl2) = 208.23

g/mol

The number of moles of BaCl2can be calculated using the formula:

Moles of BaCl2=Mass of BaCl2

MBaCl2

Moles of BaCl2=15.0 g

208.23 g/mol

Moles of BaCl2≈0.072 mol

Step 2: Calculate the moles of sodium sulfate (N a2SO4) added to the solu-

tion.

Given: - Volume of Na2SO4solution added = 100.0 mL = 0.100 L - Molarity

of Na2SO4solution = 0.150 M

The number of moles of Na2SO4can be calculated using the formula:

Moles of Na2SO4= Volume ×Molarity

Moles of Na2SO4= 0.100 L ×0.150 mol/L

Moles of Na2SO4= 0.015 mol

Step 3: Determine the limiting reactant to predict if a precipitate will form.

From the balanced chemical equation for the reaction between BaCl2and

Na2SO4:

BaCl2(aq) + N a2SO4(aq)→BaSO4(s)+2NaCl(aq)

1 mole of BaCl2reacts with 1 mole of Na2SO4to form 1 mole of BaSO4.

Since we have 0.072 moles of BaCl2and 0.015 moles of Na2SO4,N a2SO4

is the limiting reactant.

Step 4: Calculate the mass of precipitate formed (barium sulfate, BaSO4).

Given: - Molar mass of BaSO4(MBaSO4) = 233.39 g/mol

The mass of BaSO4formed can be calculated using the formula:

Mass of BaSO4= Moles of Na2SO4×MBaSO4

Mass of BaSO4= 0.015 mol ×233.39 g/mol

Mass of BaSO4≈3.50 g

Therefore, a precipitate of barium sulfate will form, and the mass of the

precipitate will be approximately 3.50 g.

Question 22

Question

In a laboratory experiment, a student mixes 50.0 mL of 0.200 M lead(II) nitrate

with 75.0 mL of 0.150 M potassium iodide solution. The balanced chemical

equation for the reaction is:

P b(NO3)2(aq)+2KI(aq)→P bI2(s)+2KNO3(aq)

Calculate the mass of lead(II) iodide, PbI2, that can be formed from this

reaction.

Solution

Step 1: Calculate the number of moles of lead(II) nitrate and potassium iodide

used in the reaction.

Moles of Pb(NO3)2= Volume (L) ×Concentration (M)

Moles of Pb(NO3)2= 0.0500 L ×0.200 M

Moles of Pb(NO3)2= 0.010 mol

Moles of KI = Volume (L) ×Concentration (M)

Moles of KI = 0.0750 L ×0.150 M

Moles of KI = 0.01125 mol

Step 2: Determine the limiting reactant.

Using the balanced chemical equation, we can see that 1 mole of lead(II)

nitrate reacts with 2 moles of potassium iodide to form 1 mole of lead(II) iodide.

Since the stoichiometry is 1:2 for Pb(NO3)2:KI, the limiting reactant is

Pb(NO3)2.

Step 3: Calculate the theoretical yield of lead(II) iodide.

Moles of PbI2=0.010 mol Pb(NO3)2

1×1 mol PbI2

1 mol Pb(NO3)2

Moles of PbI2= 0.010 mol ×1

Moles of PbI2= 0.010 mol

Step 4: Calculate the mass of lead(II) iodide.

Mass of PbI2= Moles of PbI2×Molar mass of PbI2

Mass of PbI2= 0.010 mol ×461.01 g/mol

Mass of PbI2≈4.61 g

Therefore, the mass of lead(II) iodide that can be formed from this reaction

is approximately 4.61 grams.

Question 23

Question

A solution is prepared by dissolving 20.0 g of potassium iodide (KI) in 100.0 g

of water at 25

C. Determine if a precipitate will form when 35.0 mL of 0.150 M

lead(II) nitrate (Pb(NO3)2) solution is added to the potassium iodide solution.

The solubility products constants are Ksp = 7.1×10−9for lead(II) iodide (PbI2)

and Ksp = 2.5×10−7for lead(II) bromide (PbBr2). Assume the volume of the

solution does not change upon mixing.

Solution

Step 1: Write the balanced equation for the precipitation reaction between

lead(II) nitrate and potassium iodide:

Pb(NO3)2+ 2KI →PbI2↓+2KNO3

Step 2: Calculate the moles of potassium iodide (KI) and lead(II) nitrate

(Pb(NO3)2) using their respective masses and molar masses: - Moles of KI:

Molar mass of KI = 39.10 g/mol + 126.90 g/mol = 166.00 g/mol

Moles of KI = 20.0 g

166.00 g/mol = 0.1205 mol

- Moles of Pb(NO3)2:

Molar mass of Pb(NO3)2= 207.2 g/mol+3(16.00 g/mol+14.01 g/mol) = 331.2 g/mol

Moles of Pb(NO3)2= M ×V=0.150 mol/L ×0.0350 L = 0.00525 mol

Step 3: Determine the limiting reactant: The molar ratio between Pb(NO3)2

and KI is 1:2, meaning that 1 mole of Pb(NO3)2will react with 2 moles of KI

to form PbI2. Since there is less moles of Pb(NO3)2than required, Pb(NO3)2

is the limiting reactant.

Step 4: Calculate the concentration of lead(II) ions in the solution:

Volume of solution = 100.0 g water + 35.0 mL Pb(NO3)2

Volume of solution (L) = 100.0 g

1.0 g/mL + 0.0350 L = 0.1350 L

[Pb2+] = moles of Pb(NO3)2

volume of solution in L =0.00525 mol

0.1350 L = 0.0389 M

Step 5: Calculate the ion product Qand compare with the solubility product

constant Ksp:

Q= [Pb2+][I−]2

Question 24

Question

A chemical reaction takes place in a solution resulting in the formation of a

precipitate. If 50.0 mL of a 0.200 M silver nitrate solution is mixed with 75.0 mL

of a 0.150 M sodium chloride solution, what mass of silver chloride precipitate

is formed? (Assume the reaction goes to completion)

Solution

Step 1: Write the balanced chemical equation for the reaction between silver

nitrate (AgNO3) and sodium chloride (NaCl):

AgNO3+ NaCl →AgCl + NaNO3

Step 2: Determine the limiting reactant to find out which reactant will run

out first. The limiting reactant is the one that produces the least amount of

product. We can do this by calculating the moles of each reactant:

moles of AgNO3= Molarity ×Volume = 0.200 mol/L ×0.0500 L = 0.0100 mol

moles of NaCl = Molarity ×Volume = 0.150 mol/L ×0.0750 L = 0.01125 mol

Since AgNO3has fewer moles (0.0100 mol) compared to NaCl (0.01125 mol),

AgNO3is the limiting reactant.

Step 3: Use the stoichiometry of the balanced chemical equation to find the

moles of AgCl formed. From the equation, 1 mole of AgNO3reacts with 1 mole

of NaCl to produce 1 mole of AgCl:

moles of AgCl = 0.0100 mol

Step 4: Calculate the mass of silver chloride precipitate formed using the

molar mass of AgCl, which is 143.32 g/mol:

mass of AgCl = 0.0100 mol ×143.32 g/mol = 1.43 g

Therefore, the mass of silver chloride precipitate formed is 1.43 g.

Question 25

Question

A student is performing a series of precipitation reactions in the laboratory.

They start with a solution containing 0.300 M calcium chloride (CaCl2) and

then add an excess of 0.150 M sodium sulfate (Na2SO4) solution.

Given that the Ksp of calcium sulfate (CaSO4) is 2.4×10−5, calculate the

mass of calcium sulfate that will precipitate.

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction

between CaCl2and Na2SO4:

CaCl2+ Na2SO4→CaSO4+ 2NaCl

Step 2: Determine the ions present in solution: - Initial concentrations: -

[Ca2+]initial = 0.300 M - [SO2−

4]initial = 0.150 M

- Once the reaction occurs, the Ca2+ ions will fully react with the SO2−

ions to form CaSO4. The limiting reactant will be CaCl2because the initial

concentration of Ca2+ ions is higher than the SO2−

4ions.

Step 3: Calculate the concentration of CaSO4formed: - The reaction con-

sumes all of the Ca2+ ions to form CaSO4, which means the final concentra-

tion of Ca2+ ions will be zero. - The concentration of Ca2+ ions that reacted

is 0.150 M. - Therefore, the concentration of Ca2+ ions that reacted to form

CaSO4is 0.150 M.

Step 4: Calculate the mass of CaSO4that precipitates: - The number of

moles of CaSO4formed is equal to the number of moles of Ca2+ ions that re-

acted. - Moles of CaSO4= [Ca2+]reacted = 0.150 mol - The molar mass of CaSO4

is 40.08 g/mol + 32.06 g/mol + 4(16.00 g/mol) = 136.16 g/mol. - Therefore, the

mass of CaSO4that will precipitate is:

Mass = Moles ×Molar mass = 0.150 mol ×136.16 g/mol = 20.42 g

Question 26

Question

A certain metal hydroxide, M(OH)2, has a solubility product constant Ksp of

1.6×10−15. Calculate the molar solubility of this metal hydroxide in a solution

with a pH of 9.2.

Solution

Step 1: Write the chemical equation for the dissolution of M(OH)2. The disso-

lution of M(OH)2can be represented by the equation:

M(OH)2⇌M2+ + 2OH−

Step 2: Write the equilibrium expression for the dissolution of M(OH)2. The

solubility product constant expression for M(OH)2is:

Ksp = [M2+][OH−]2

Step 3: Calculate the concentration of hydroxide ions ([OH]−) from the given

pH. We know that [H+] = 10−pH, so for a solution with pH 9.2:

[H+] = 10−9.2= 6.31 ×10−10

Since the solution is neutral, [H+] = [OH−]=6.31 ×10−10.

Step 4: Substitute the values into the solubility product constant expression.

Substitute [M2+] = xand [OH−]=6.31 ×10−10 into the Ksp expression:

1.6×10−15 = (x)(6.31 ×10−10)2

Step 5: Solve for the molar solubility of M(OH)2. Solve for x:

x=1.6×10−15

(6.31 ×10−10)2

x= 4.00 ×10−6M

Therefore, the molar solubility of the metal hydroxide in a solution with a

pH of 9.2 is 4.00 ×10−6M.

Question 27

Question

A solution is prepared by mixing 50.0 mL of 0.20 M silver nitrate with 75.0 mL

of 0.15 M sodium chloride. Determine if a precipitate will form. (For reference,

Ksp for silver chloride is 1.6×10−10.)

Solution

Step 1: Write the balanced chemical equation for the reaction of silver nitrate

and sodium chloride:

AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)

Step 2: Determine the ions present in solution after mixing the two solutions.

From the 50.0 mL of 0.20 M silver nitrate, we have: moles of Ag+= 0.050 L×

0.20 mol/L = 0.010 mol

From the 75.0 mL of 0.15 M sodium chloride, we have: moles of Cl−=

0.075 L×0.15 mol/L = 0.01125 mol

Step 3: Determine the molar solubility of AgCl. Let’s assume x mol/L

of AgCl is formed. The concentration of Ag+will be 0.010 mol +x. The

concentration of Cl−will be 0.01125 mol +x. At equilibrium, Ksp = [Ag+]×

[Cl−] = x×xThus, Ksp =x2

Therefore, 1.6×10−10 =x2Solving for x, we get x=√1.6×10−10 =

1.26 ×10−5

Step 4: Check if a precipitate will form. The molar solubility of AgCl is

1.26 ×10−5mol/L. Since the initial concentration of Ag+is 0.010 mol/L and

Cl−is 0.01125 mol/L, which are both greater than the molar solubility of AgCl,

a precipitate of AgCl will form.

Question 28

Question

A chemist has 500 mL of a solution containing 0.2 M lead nitrate (Pb(NO3)2).

The chemist wants to precipitate all the lead as lead sulfate (PbSO4) using

sulfuric acid (H2SO4). The balanced equation for the reaction is:

Pb(NO3)2(aq)+H2SO4(aq)→PbSO4(s) + 2HNO3(aq)

Assuming that lead dissolves completely as Pb2+ ions, calculate the volume

of 2 M sulfuric acid required to precipitate all the lead as lead sulfate. (Hint:

PbSO4remains insoluble in an excess of sulfuric acid.)

Solution

Step 1: Write the balanced chemical equation for the reaction between lead

nitrate and sulfuric acid to form lead sulfate and nitric acid:

Pb(NO3)2(aq)+H2SO4(aq)→PbSO4(s) + 2HNO3(aq)

Step 2: Calculate the number of moles of lead nitrate in the solution. Given:

- Concentration of lead nitrate solution = 0.2 M - Volume of lead nitrate solution

= 500 mL = 0.5 L

Number of moles of Pb(NO3)2= Concentration×Volume = 0.2 mol/L×0.5 L = 0.1 mol

Step 3: Use the stoichiometry of the balanced chemical equation to determine

the amount of sulfuric acid needed to precipitate all the lead as lead sulfate.

From the balanced equation, 1 mole of lead nitrate reacts with 1 mole of sulfuric

acid to produce 1 mole of lead sulfate.

Since the molar ratio is 1:1, the number of moles of sulfuric acid required

is equal to the number of moles of lead nitrate. Thus, we need 0.1 moles of

sulfuric acid.

Step 4: Calculate the volume of 2 M sulfuric acid required to supply 0.1

moles of sulfuric acid. Given: - Concentration of sulfuric acid = 2 M

Volume of sulfuric acid = Number of moles of H2SO4

Concentration =0.1 mol

2 mol/L = 0.05 L = 50 mL

Therefore, 50 mL of 2 M sulfuric acid is required to precipitate all the lead

as lead sulfate.

Question 29

Question

Calculate the concentration of sulfate ions in a solution that forms when 150.0

mL of a 0.20 M solution of barium chloride is mixed with 300.0 mL of a 0.10 M

solution of sodium sulfate.

Solution

Step 1: Write the balanced chemical equation for the reaction between barium

chloride and sodium sulfate:

BaCl2+ Na2SO4→BaSO4+ 2NaCl

Step 2: Determine the limiting reagent to find the maximum amount of

precipitate (barium sulfate) that can form. Since barium sulfate (BaSO4) is

insoluble in water, it will precipitate out of the solution. From the equation,

the stoichiometry shows that 1 mole of barium chloride will react with 1 mole

of sodium sulfate to form 1 mole of barium sulfate.

Calculate the moles of barium chloride:

moles of BaCl2= Molarity ×Volume (L) = 0.20 mol/L ×0.150 L = 0.030 mol

and the moles of sodium sulfate:

moles of Na2SO4= Molarity ×Volume (L) = 0.10 mol/L ×0.300 L = 0.030 mol

Since the moles of barium chloride and sodium sulfate are equal, there is

a 1:1 ratio of reactants. Therefore, both reactants will be fully consumed and

there will be no excess reagent.

Step 3: Calculate the concentration of sulfate ions in the final solution after

the precipitation reaction. The moles of sulfate ions in the final solution come

from the sodium sulfate:

moles of sulfate ions = moles of Na2SO4= 0.030 mol

The total volume of the final solution is the sum of the volumes of the two

initial solutions:

Total volume = 150 mL + 300 mL = 450 mL = 0.450 L

Therefore, the concentration of sulfate ions in the final solution is:

Concentration of sulfate ions = moles of sulfate ions

total volume (L) =0.030 mol

0.450 L = 0.067 M

Question 30

Question

A chemistry student is performing a precipitation reaction by mixing 100.0 mL

of 0.150 M silver nitrate (AgNO3) with 150.0 mL of 0.200 M sodium chloride

(NaCl). If the Ksp of silver chloride (AgCl) is 1.8×10−10, will a precipitate

of silver chloride form? If so, calculate the mass of silver chloride that will

precipitate.

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction

between silver nitrate and sodium chloride:

AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)

Step 2: Determine the ions present in solution after mixing the two solu-

tions. Calculate the concentrations of Ag+and Cl−ions in the mixture: - The

initial concentration of Ag+ions = 0.100 L ×0.150 M = 0.0150 mol - The initial

concentration of Cl−ions = 0.150 L ×0.200 M = 0.0300 mol

Step 3: Determine the reaction quotient (Qsp) by multiplying the concen-

trations of the products raised to their stoichiometric coefficients:

Qsp = [Ag+]×[Cl−] = (0.0150)(0.0300) = 0.00045

Step 4: Compare Qsp to Ksp to determine if a precipitate will form: - Since

Qsp < Ksp, a precipitate will not form.

Step 5: The mass of silver chloride that could potentially precipitate can

be calculated using the stoichiometry of the balanced chemical equation: - The

molar mass of AgCl is 107.87 g/mol. - The moles of AgCl that could potentially

precipitate is equal to the minimum of the moles of Ag+and Cl−ions, which

is 0.0150 mol. - Therefore, the mass of silver chloride that could potentially

precipitate is 0.0150 mol ×107.87 g/mol = 1.62 g.

Question 31

Question

A solution contains 20 grams of a compound AB2in 100 mL of water. If the

solubility product constant (Ksp) for AB2is 4.0×10−8, calculate the concen-

tration of AB2ions in the solution, its molar solubility, and the maximum mass

of AB2that can dissolve in 1 L of water.

Solution

Step 1: Calculate the concentration of AB2ions in the solution.

Molar mass of AB2= 2(molar mass of A) + 1(molar mass of B)

= 2(40.0 g/mol) + 1(16.0 g/mol)

= 112.0 g/mol

Number of moles of AB2in 20 g:

Number of moles = Mass in grams

Molar mass

=20 g

112.0 g/mol

≈0.179 mol

The volume of the solution is 100 mL = 0.1 L.

Concentration of AB2ions:

Concentration = Number of moles

Volume in L

=0.179 mol

0.1 L

= 1.79 M

Step 2: Calculate the molar solubility of AB2. Let the molar solubility of

AB2be x.

The solubility-product constant equation is given by:

Ksp = [A][B]2=x(2x)2= 4.0×10−8

Solve for x:

4x3= 4.0×10−8

x3= 1.0×10−8

x= 1.0×10−2M=0.01 M

So, the molar solubility of AB2is 0.01 M.

Step 3: Calculate the maximum mass of AB2that can dissolve in 1 L of

water. Molar mass of AB2is 112.0 g/mol.

Maximum mass that can dissolve in 1 L = molar solubility ×molar mass of

AB2:

Max mass = 0.01 mol/L ×112.0 g/mol = 1.12 g/L

Therefore, the maximum mass of AB2that can dissolve in 1 L of water is

1.12 grams.

Question 32

Question

A certain region receives an average annual precipitation of 850 mm. If the area

of the region is 1200 km2, determine the total volume of water that falls on the

region annually in cubic meters.

Solution

Step 1: Convert the area from km2to m2. Step 2: Convert the precipitation

from mm to m. Step 3: Calculate the total volume of water that falls on the

region annually.

Step 1: To convert the area from km2to m2, we use the conversion factor

1 km2= 1,000,000 m2. So, the area in m2is:

1200 km2×1,000,000 m2/km2= 1,200,000,000 m2

Step 2: To convert the precipitation from mm to m, we use the conversion

factor 1 mm = 0.001 m. So, the precipitation in m is:

850 mm ×0.001 m/mm = 0.85 m

Step 3: To calculate the total volume of water that falls on the region

annually, we multiply the area by the precipitation depth. Therefore, the total

volume of water in cubic meters is:

1,200,000,000 m2×0.85 m = 1,020,000,000 m3

Hence, the total volume of water that falls on the region annually is 1,020,000,000

cubic meters.

Question 33

Question

A solution contains 0.15 M barium chloride. If solid silver nitrate is added to the

solution, what concentration of silver nitrate is needed to completely precipitate

all the chloride ions as silver chloride? The solubility product constant Ksp for

silver chloride is 1.77 ×10−10.

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction:

BaCl2+ 2AgNO3→Ba(NO3)2+ 2AgCl

Step 2: Determine the moles of chloride ions present in the barium chloride

solution:

Moles of Cl−= 0.15 M ×volume of solution

Step 3: Calculate the moles of silver ions needed to precipitate all the chloride

ions: Moles of Ag+=Moles of Cl−

Step 4: Use the mole ratio from the balanced equation to find the moles of

silver nitrate required: Moles of AgNO3= 2 ×Moles of Ag+

Step 5: Calculate the concentration of silver nitrate needed based on the

volume of solution: Concentration of AgNO3=Moles of AgNO3

volume of solution

Step 6: Substitute the given values into the equation to find the concentra-

tion of silver nitrate required.

Question 34

Question

A solution contains 0.15 M barium chloride and 0.20 M silver nitrate. Calculate

the mass of barium sulfate that will precipitate in a volume of 500 mL. (Given:

Ksp of barium sulfate is 1.1×10−10)

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction

between barium chloride and silver nitrate to form barium sulfate and silver

chloride.

BaCl2+ AgNO3→BaSO4+ 2AgCl

Step 2: Determine the limiting reactant by calculating the moles of barium

sulfate that can be formed from each reactant. From the balanced chemical

equation, 1 mole of BaCl2produces 1 mole of BaSO4.

Moles of BaSO4= 0.15 M ×0.5 L = 0.075 mol

From the balanced chemical equation, 1 mole of AgNO3produces 1 mole of

BaSO4.

Moles of BaSO4= 0.20 M ×0.5 L = 0.1 mol

Since 0.075 mol of BaSO4can be formed from BaCl2and 0.1 mol of BaSO4

can be formed from AgNO3, BaCl2is the limiting reactant.

Step 3: Calculate the mass of barium sulfate that will precipitate by finding

the mass of 0.075 mol of BaSO4.

Molar mass of BaSO4= 137.3 g/mol+32.1 g/mol+(4×16.0 g/mol) = 233.3 g/mol

Mass of BaSO4= 0.075 mol ×233.3 g/mol = 17.5 g

Therefore, 17.5 grams of barium sulfate will precipitate in a volume of 500

mL.

Question 35

Question

A sample of water has a concentration of sulfate ions of 0.025 M. If a solution of

barium chloride is added to the water sample, what will be the concentration of

sulfate ions in the solution once the precipitation reaction reaches completion?

The balanced chemical equation for the precipitation reaction is:

BaCl2(aq) + SO2−

4(aq)→BaSO4(s) + 2Cl−(aq)

The solubility product constant Ksp of barium sulfate is 1.1×10−10.

Solution

Step 1: Write the expression for the solubility product constant Ksp. The

solubility product constant Ksp for the reaction is given by:

Ksp = [Ba2+][SO2−

Step 2: Let xbe the concentration of sulfate ions at equilibrium. Since

1 mole of barium sulfate produces 1 mole of sulfate ions in the reaction, the

concentration of sulfate ions ([SO2−

4]) at equilibrium will be equal to x.

Step 3: Set up the equilibrium constant expression and solve for x. Sub-

stitute the known values into the equilibrium constant expression and solve for

Ksp = [Ba2+][SO2−

1.1×10−10 = 2x×0.025

x=1.1×10−10

0.05

The volume of the final solution is 50.0 mL + 25.0 mL = 75.0 mL = 0.0750

L. Since 0.0005 mol of BaSO4is produced in this volume, the concentration of

BaSO4is:

Concentration of BaSO4=0.0005 mol

0.0750 L = 0.00667 M

Therefore, the concentration of barium sulfate in the final solution is 0.00667

Question 2

Question

A solution contains 0.3 M calcium nitrate and 0.6 M sodium sulfate. Will a

precipitate form when these two solutions are combined? If so, what mass of

the precipitate is formed?

Solution

Step 1: Write the balanced equation for the precipitation reaction between

calcium nitrate and sodium sulfate:

Ca(N O3)2+Na2SO4→CaSO4+ 2N aNO3

Step 2: Determine the possible products of the reaction. Calcium sulfate is

insoluble and will precipitate out of solution, while sodium nitrate will stay in

solution.

Step 3: Determine the net ionic equation for the reaction:

Ca2+ +SO2−

4→CaSO4

Step 4: Calculate the concentrations of the ions in the combined solution.

Since both solutions are dissolved in water, assume that the volumes are addi-

tive.

For calcium ions: [Ca2+]=0.3M

For sulfate ions: [SO2−

4]=0.6M

Step 5: Use the concentrations of the ions to determine the reaction quotient,

Q, for the precipitation reaction.

Q= [Ca2+][SO2−

4] = (0.3)(0.6) = 0.18

Step 6: Compare the reaction quotient to the solubility product constant,

Ksp, for calcium sulfate. If Q > Ksp, a precipitate will form.

The Ksp for calcium sulfate is 2.4×10−5. Since Q>Ksp (0.18 ¿ 2.4×10−5),

a precipitate of calcium sulfate will form.

Step 7: Calculate the mass of the precipitate formed using the stoichiometry

of the balanced equation. Since 1 mole of calcium sulfate is formed for every 1

mole of calcium nitrate reacted,

Moles of CaSO4= Molarity of Ca(NO3)2×Volume of Ca(NO3)2

= 0.3 mol/L ×volume of calcium nitrate solution in L

Then, use the molar mass of calcium sulfate to calculate the mass of the

precipitate formed.

Question 3

Question

Calculate the concentration of sulfate ions in a solution that results from mixing

250.0 mL of 0.200 M barium chloride and 300.0 mL of 0.150 M sodium sulfate.

Assume complete reaction and that the final volume of the solution is 500.0 mL.

Solution

Step 1: Write the balanced chemical equation for the reaction between barium

chloride (BaCl2) and sodium sulfate (Na2SO4) to determine the mole ratio.

BaCl2+ Na2SO4→BaSO4↓+2NaCl

Step 2: Calculate the moles of barium chloride and sodium sulfate, respec-

tively, using the given concentrations and volumes.

Moles of BaCl2= 0.200 M ×0.2500 L = 0.0500 mol

Moles of Na2SO4= 0.150 M ×0.3000 L = 0.0450 mol

Step 3: Determine the limiting reactant by comparing the mole ratios from

the balanced chemical equation. Since the mole ratio of BaCl2to Na2SO4is

1:1, BaCl2is the limiting reactant.

Step 4: Use the limiting reactant to calculate the moles of barium sulfate

formed.

Moles of BaSO4= 0.0500 mol

Step 5: Calculate the concentration of sulfate ions in the final solution. Since

1 mole of barium sulfate produces 1 mole of sulfate ions, the moles of sulfate

ions in the final solution is also 0.0500 mol. Now, we need to find the final

volume of the solution, which is 500.0 mL.

Step 6: Calculate the final concentration of sulfate ions in the solution.

Concentration of sulfate ions = 0.0500 mol

0.5000 L = 0.100 M

Therefore, the concentration of sulfate ions in the final solution is 0.100 M.

Question 4

Question

A solution contains 0.15 M BaCl2and 0.20 M Na2SO4. Will a precipitation

reaction occur when these solutions are mixed? If so, what mass of BaSO4will

be formed? (Given: Ksp = 1.1×10−10 for BaSO4)

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction:

BaCl2(aq) + Na2SO4(aq) −−→ BaSO4(s) + 2 NaCl (aq)

Step 2: Determine the ions that will be present in solution after mixing: -

Ba2+ from BaCl2- SO42−from Na2SO4

Step 3: Calculate the ion product (Qsp) to determine if precipitation will

occur:

Qsp = [Ba2+][SO42−] = (0.15)(0.20) = 0.03

Step 4: Compare Qsp to Ksp to determine if precipitation will occur: Since

Qsp = 0.03 > Ksp = 1.1×10−10, a precipitation reaction will occur.

Step 5: Calculate the mass of BaSO4formed: - Let’s assume the volume of

the resulting solution is 1 L for easier calculations. - Convert moles of BaSO4

to mass using its molar mass (233.4 g/mol):

1 mol of BaSO4= 233.4 g

0.15 M BaSO4×233.4 g/mol ×1 L = 34.31 g

Therefore, 34.31 g of BaSO4will be formed during the precipitation reaction.

Question 5

Question

At a certain location, the monthly average precipitation for January is 3.6 inches,

while the monthly average precipitation for July is 4.8 inches. If the annual av-

erage precipitation is 49.2 inches, determine the average precipitation for each of

the remaining months (February to December) to maintain a constant monthly

average precipitation for the year.

Solution

Let xrepresent the average precipitation for each of the remaining months

(February to December).

Step 1: Determine the total precipitation for the year.

The total precipitation for the year can be calculated as:

Total precipitation = January precipitation+July precipitation+Remaining months precipitation

Total precipitation = 3.6+4.8 + 11x

Step 2: Set up the equation for the total precipitation.

Since the total annual precipitation is given as 49.2 inches, we have:

3.6+4.8 + 11x= 49.2

Step 3: Solve for x.

Combine the known values and solve for x:

8.4 + 11x= 49.2

11x= 40.8

x=40.8

x= 3.72 inches

Step 4: Answer

Therefore, the average precipitation for each of the remaining months (Febru-

ary to December) should be 3.72 inches to maintain a constant monthly average

precipitation for the year.

Question 6

Question

A solution contains 0.2 M of silver nitrate (AgNO3) and 0.1 M of sodium chloride

(NaCl). If these two solutions are mixed together, what is the maximum mass

of silver chloride (AgCl) that can precipitate out? (Assume 100

Given: The molar mass of silver chloride (AgCl) is 143.32 g/mol.

Solution

Step 1: Write the balanced chemical equation for the reaction between silver

nitrate and sodium chloride to form silver chloride. Identify the limiting reagent.

The balanced equation is:

AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)

From the balanced chemical equation, we can see that silver nitrate and

sodium chloride react in a 1:1 mole ratio.

Step 2: Calculate the moles of silver nitrate and sodium chloride in the

solution.

Given: [AgNO3] = 0.2 M [NaCl] = 0.1 M

Volume of solution does not affect the amount of precipitate formed; hence,

1 L of solution is assumed. Therefore, for 1 L of solution: Moles of AgNO3= 0.2

Moles of NaCl = 0.1

Step 3: Identify the limiting reagent. Since the reaction occurs in a 1:1 mole

ratio, the limiting reactant will be the one that is present in lesser moles. In

this case, sodium chloride is the limiting reagent.

Step 4: Calculate the maximum mass of silver chloride that can be formed.

Given: Molar mass of AgCl = 143.32 g/mol

Since the limiting reagent is sodium chloride, all of the 0.1 moles of NaCl

will react to form AgCl.

Number of moles of AgCl = 0.1 moles Mass of AgCl = Molar mass ×

Number of moles Mass of AgCl = 143.32 g/mol×0.1 mol Mass of AgCl = 14.332

Therefore, the maximum mass of silver chloride that can precipitate out is

14.332 g.

Question 7

Question

A chemist mixes 100.0 mL of 0.200 M silver nitrate (AgNO3) solution with

150.0 mL of 0.100 M sodium chloride (NaCl) solution. Determine the mass of

silver chloride (AgCl) precipitate that will form. Assume the reaction goes to

completion.

Solution

Step 1: Write the balanced chemical equation for the reaction between silver

nitrate and sodium chloride:

AgNO3+ NaCl →AgCl + NaNO3

Step 2: Determine the limiting reactant.

Calculate the moles of silver nitrate:

moles of AgNO3= M ×V=0.200 M ×0.100 L = 0.020 mol

Calculate the moles of sodium chloride:

moles of NaCl = M ×V=0.100 M ×0.150 L = 0.015 mol

Since NaCl is the limiting reactant (less moles), we will base our calculations

on it.

Step 3: Calculate the mass of silver chloride produced.

From the balanced chemical equation, we see that 1 mol of NaCl produces

1 mol of AgCl.

Calculate the moles of AgCl formed:

moles of AgCl = moles of NaCl = 0.015 mol

Calculate the mass of AgCl formed:

mass of AgCl = moles of AgCl ×molar mass of AgCl

= 0.015 mol ×(107.87 g/mol)

= 1.62 g

Therefore, the mass of silver chloride precipitate that will form is 1.62 g.

Question 8

Question

A chemist wants to prepare a saturated solution of silver chloride by mixing

100.0 mL of 0.200 M silver nitrate solution with excess calcium chloride. If the

solubility product constant of silver chloride is 1.77 ×10−10, what mass of silver

chloride can be precipitated? (Assume complete precipitation)

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction

between silver nitrate and calcium chloride to form silver chloride precipitation.

AgNO3(aq) + CaCl2(aq)→AgCl ↑+Ca(NO3)2(aq)

Step 2: Determine the limiting reactant in the reaction. Using the molarity

of silver nitrate solution, calculate the moles of silver nitrate:

Moles of AgNO3= Volume ×Molarity = 100.0 mL ×0.200 mol/L = 0.0200 mol

Step 3: Use stoichiometry to find the moles of silver chloride that can be

precipitated. From the balanced equation, 1 mol of silver nitrate produces 1

mol of silver chloride. Therefore, moles of silver chloride precipitated = 0.0200

mol.

Step 4: Calculate the mass of silver chloride precipitated. The molar mass

of silver chloride (AgCl) is 143.32 g/mol. Therefore, mass of silver chloride =

moles of AgCl ×molar mass of AgCl

Mass of AgCl = 0.0200 mol ×143.32 g/mol = 2.87 g

Therefore, the mass of silver chloride that can be precipitated is

2.87 g.

Question 9

Question

A solution of silver nitrate (AgNO3) with a concentration of 0.150 M is mixed

with a solution of sodium chloride (NaCl) with a concentration of 0.200 M. If the

solubility product constant of silver chloride (AgCl) is 1.8×10−10, determine if

a precipitation reaction will occur when the two solutions are mixed.

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction. The

balanced chemical equation for the reaction between silver nitrate and sodium

chloride to form silver chloride precipitate is: AgNO3+ NaCl →AgCl + NaNO3

Step 2: Write the expression for the solubility product constant. The

solubility product constant expression for the formation of silver chloride is:

Ksp = [Ag+][Cl−]

Step 3: Calculate the initial concentrations of Ag+and Cl−ions. Initial con-

centration of Ag+([Ag+]initial) from silver nitrate: 0.150 M Initial concentration

of Cl−([Cl−]initial) from sodium chloride: 0.200 M

Step 4: Determine the maximum amount of AgCl that can dissolve. Use

the solubility product constant to calculate the maximum amount of AgCl ions

that can dissolve: Ksp = [Ag+]initial[Cl−]initial 1.8×10−10 = (0.150)(0.200)

Step 5: Compare the actual and maximum amount of AgCl that can dis-

solve. The calculated value in step 4 is much greater than the solubility product

constant Ksp, indicating that a precipitation reaction will occur when the silver

nitrate and sodium chloride solutions are mixed.

Therefore, a white precipitate of silver chloride will form when the two so-

lutions are mixed.

Question 10

Question

Determine the mass of lead(II) chloride (P bCl2) that can be formed when a

solution containing 3.50 grams of lead(II) nitrate (P b(NO3)2) is mixed with a

solution containing 2.80 grams of sodium chloride (NaCl). Assume that the

reactions proceed to completion.

Solution

Step 1: Write the balanced chemical equation for the reaction between lead(II)

nitrate and sodium chloride to form lead(II) chloride and sodium nitrate.

Pb(NO3)2+ 2NaCl →PbCl2+ 2NaNO3

Step 2: Calculate the molar mass of each compound. The molar mass of

P b(NO3)2is 207.2 + 2(14 + 3(16)) = 331.2 g/mol. The molar mass of NaCl is

23 + 35.5 = 58.5 g/mol. The molar mass of P bCl2is 207.2 + 2(35.5) = 278.2

g/mol.

Step 3: Determine the number of moles of each reactant.



Moles of P b(NO3)2:3.50 g

331.2 g/mol = 0.0106 mol



Moles of NaCl:2.80 g

58.5 g/mol = 0.0478 mol

Step 4: Determine the limiting reactant. In this case, P b(NO3)2is the

limiting reactant because it produces the least amount of product (P bCl2).

Step 5: Calculate the mass of P bCl2formed using the limiting reactant.

Using the stoichiometry of the reaction, 1 mol of P b(NO3)2produces 1 mol of

P bCl2. Therefore, 0.0106 mol of P b(N O3)2will produce 0.0106 mol of P bCl2.

Finally,

Mass of P bCl2= 0.0106 mol ×278.2 g/mol = 2.95 g

Therefore, 2.95 grams of lead(II) chloride can be formed.

Question 11

Question

Calculate the precipitation in inches when 1 inch of rain falls on a 5-acre plot

of land.

Solution

Step 1: Convert 5 acres to square feet.

1 acre = 43,560 square feet

So, 5 acres = 5 ×43,560 square feet = 217,800 square feet.

Step 2: Convert 1 inch of rain to cubic feet.

1 foot = 12 inches

So, 1 inch of rain = 1

12 feet = 1

12 cubic feet.

Step 3: Calculate the volume of rainfall on the plot of land. The volume of

rainfall is given by the formula:

Volume = Area ×Height

where Area = 217,800 square feet and Height = 1

12 cubic feet. Plugging in the

values, we get:

Volume = 217,800 ×1

Volume = 18,150 cubic feet

Step 4: Convert the volume to inches.

1 cubic foot = 1728 cubic inches

So, 18,150 cubic feet = 18,150 ×1728 cubic inches = 31,401,600 cubic inches.

Step 5: Convert the volume in cubic inches to inches of precipitation. Since

31,401,600 cubic inches of rain falls on the 5-acre plot of land, the precipitation

depth in inches is

31,401,600 cubic inches

217,800 square feet = 144.17 inches

Therefore, the precipitation depth is 144.17 inches.

Question 12

Question

Calculate the concentration of a precipitate formed when 100.0 mL of 0.200

M lead(II) nitrate, Pb(NO3)2, is mixed with 200.0 mL of 0.500 M potassium

iodide, KI. Assume the reaction goes to completion and that lead(II) iodide,

PbI2, is the only precipitate formed.

Solution

Step 1: Write the balanced chemical equation for the reaction between lead(II)

nitrate and potassium iodide.

Pb(NO3)2+ 2 KI →PbI2+ 2 KNO3

Step 2: Determine the limiting reactant. First, calculate the moles of lead(II)

nitrate and potassium iodide: For lead(II) nitrate, moles = volume×molarity =

0.100 L ×0.200 mol/L = 0.0200 mol For potassium iodide, moles = volume ×

molarity = 0.200 L ×0.500 mol/L = 0.100 mol

Since lead(II) nitrate is limiting (0.0200 moles ¡ 0.100 moles), we will calcu-

late the amount of precipitate formed based on lead(II) nitrate.

Step 3: Use the balanced equation to determine the moles of lead(II) iodide

formed. From the balanced equation, we see that 1 mole of lead(II) nitrate forms

1 mole of lead(II) iodide. Therefore, moles of lead(II) iodide = 0.0200 mol

Step 4: Calculate the concentration of the lead(II) iodide precipitate. Vol-

ume of lead(II) iodide = 0.100 L + 0.200 L = 0.300 L Concentration of lead(II)

iodide = moles of PbI2

volume =0.0200 mol

0.300 L = 0.0667 mol/L

Therefore, the concentration of lead(II) iodide precipitate formed is 0.0667

Question 13

Question

A chemist wants to determine the concentration of chloride ions in a water

sample. To do so, the chemist adds an excess of silver nitrate to a 100 mL

sample of the water. The resulting precipitate is filtered, dried, and found to

weigh 0.345 g. Given that the molar mass of silver chloride is 143.32 g/mol,

determine the concentration of chloride ions in the water sample in parts per

million (ppm). Assume that the only source of chloride ions in the water sample

is from sodium chloride.

Solution

Step 1: Calculate the moles of silver chloride precipitated. Given: - Mass of

silver chloride precipitate = 0.345 g - Molar mass of silver chloride = 143.32

g/mol

We can use the formula:

moles of substance = mass of substance

molar mass

Substitute the given values:

moles of silver chloride = 0.345 g

143.32 g/mol = 0.0024051 mol

Step 2: Determine the moles of chloride ions in the water sample. Given

that each formula unit of silver chloride contains 1 chloride ion. Therefore, the

moles of chloride ions = moles of silver chloride precipitated Hence, moles of

chloride ions = 0.0024051 mol

Step 3: Calculate the concentration of chloride ions in the water sample. -

Volume of water sample = 100 mL = 0.1 L - Concentration of chloride ions in

ppm is given by the formula:

ppm = mass of solute (in mg)

volume of solution (in L) ×106

Given that 1 ppm = 1 mg/L, we need to convert moles of chloride ions to

mass in mg before calculating the concentration in ppm. Using the molar mass

of chloride ions (35.45 g/mol):

mass of chloride ions = 0.0024051 mol ×35.45 g/mol = 0.0851861 g = 85.2 mg

Now, calculate the concentration of chloride ions in the water sample:

ppm = 85.2 mg

0.1 L ×106= 852,000 ppm

Therefore, the concentration of chloride ions in the water sample is 852,000

ppm.

Question 14

Question

Calculate the mass of lead iodide (PbI2) that will precipitate when 50.0 mL of a

0.200 M lead(II) nitrate (Pb(NO3)2) solution is mixed with 50.0 mL of a 0.250

M potassium iodide (KI) solution. Assume that lead(II) iodide (PbI2) is the

only precipitate formed and that it has a 1:2 stoichiometry with lead(II) nitrate.

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction

between lead(II) nitrate and potassium iodide.

Pb(NO3)2+ 2KI →PbI2+ 2KNO3

Step 2: Determine the limiting reactant by calculating the moles of each

reactant. The limiting reactant is the one that produces the least amount of

product. - Moles of lead(II) nitrate:

Moles = Molarity ×Volume (L)

Moles of Pb(NO3)2= 0.200 mol/L ×0.0500 L = 0.0100 mol

- Moles of potassium iodide:

Moles of KI = 0.250 mol/L ×0.0500 L = 0.0125 mol

Since lead(II) nitrate produces the least amount of lead(II) iodide, it is the

limiting reactant.

Step 3: Calculate the mass of lead iodide precipitated using the stoichiometry

of the balanced chemical equation. - Moles of lead iodide precipitated: From

the balanced equation, 1 mole of lead(II) nitrate produces 1 mole of lead(II)

iodide. Therefore, 0.0100 moles of lead(II) nitrate will produce 0.0100 moles of

lead(II) iodide.

Moles of PbI2= 0.0100 mol

- Mass of lead iodide precipitated:

Molar mass of PbI2= Pb:207.2 g/mol + I:126.9 g/mol ×2 = 459.0 g/mol

Mass of PbI2= Moles ×Molar mass = 0.0100 mol ×459.0 g/mol = 4.59 g

Therefore, 4.59 grams of lead iodide will precipitate in the reaction.

Question 15

Question

A solution is prepared by mixing 50.0 mL of 0.200 M barium chloride (BaCl2)

with 75.0 mL of 0.150 M sodium sulfate (Na2SO4). Calculate the concentra-

tion of each ion remaining in the solution after precipitation of barium sul-

fate (BaSO4) is complete. The solubility product constant of barium sulfate is

1.1×10−10.

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction

between barium chloride and sodium sulfate:

BaCl2(aq) + Na2SO4(aq)→BaSO4(s) + 2NaCl(aq)

Step 2: Determine the limiting reactant to find out how much barium sulfate

precipitates.

First, calculate the moles of each reactant: For barium chloride: 0.0500 L ×

0.200 mol/L = 0.010 mol For sodium sulfate: 0.0750 L×0.150 mol/L = 0.0113 mol

Since there is more sodium sulfate, it is the limiting reactant. Therefore, all

of the barium sulfate will precipitate.

Step 3: Calculate the moles of sulfate ions from sodium sulfate used in the

reaction: 0.0113 mol ×1=0.0113 mol SO2−

Step 4: Calculate the concentrations of each ion in the final solution.

For barium ions: Initial moles of barium ions = 0.010 mol Moles of bar-

ium ions consumed in reaction = 0.010 mol Moles of barium ions remaining =

0.010 mol −0.010 mol = 0 mol Concentration of barium ions = 0 mol

0.125 L = 0 M

For chloride ions: Initial moles of chloride ions = 0.010 mol Moles of chlo-

ride ions consumed in reaction = 0.010 mol Moles of chloride ions remaining =

0.010 mol −0.010 mol = 0 mol Concentration of chloride ions = 0 mol

0.125 L = 0 M

For sulfate ions: Initial moles of sulfate ions = 0.0113 mol Moles of sul-

fate ions consumed in reaction = 0.0113 mol Moles of sulfate ions remaining =

0.0113 mol −0.0113 mol = 0 mol Concentration of sulfate ions = 0 mol

0.125 L = 0 M

Therefore, the concentrations of barium ions, chloride ions, and sulfate ions

remaining in the solution after precipitation is complete are 0 M.

Question 16

Question

In a precipitation reaction, 50.0 mL of a 0.200 M solution of lead(II) nitrate,

Pb(NO3)2, is mixed with 75.0 mL of a 0.150 M solution of sodium iodide, NaI.

Determine the mass of lead(II) iodide, PbI2, that will precipitate. (Assume the

reaction goes to completion and there are no volume changes.)

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction.

Pb(NO3)2+ 2NaI →PbI2+ 2NaNO3

Step 2: Determine the limiting reagent in the reaction. To find the limiting

reagent, we need to calculate the moles of each reactant. We can use the formula:

moles = molarity ×volume (L)

For lead(II) nitrate, Pb(NO3)2:

molesPb(NO3)2= 0.200 M ×50.0×10−3L

1= 0.010 mol

For sodium iodide, NaI:

molesNaI = 0.150 M ×75.0×10−3L

1= 0.01125 mol

Since the stoichiometry of the reaction is 1:2 between lead(II) nitrate and

sodium iodide, lead(II) nitrate is the limiting reagent.

Step 3: Calculate the mass of lead(II) iodide, PbI2, that will precipitate.

Using the stoichiometry of the reaction, we can see that 1 mol of lead(II) nitrate

produces 1 mol of lead(II) iodide. The molar mass of PbI2is 461.01 g/mol.

molesPbI2= molesPb(NO3)2= 0.010 mol

massPbI2= 0.010 mol ×461.01 g/mol = 4.61 g

Therefore, 4.61 g of lead(II) iodide, PbI2, will precipitate.

Question 17

Question

A chemist wants to calculate the amount of precipitate formed when 50.0 mL of

0.200 M silver nitrate (AgNO3) solution is mixed with excess sodium chloride

(NaCl) solution. If the balanced chemical equation is:

AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)

and the molar mass of AgCl is 143.32 g/mol, what mass of precipitate will

be formed?

Solution

Step 1: Calculate moles of silver nitrate (AgNO3) from the volume and molarity

given.

Moles of AgNO3= Volume ×Molarity = 0.0500 L ×0.200 mol/L = 0.0100 mol

Step 2: Use the balanced chemical equation to determine the stoichiometry

of the reaction. From the equation, 1 mole of AgNO3reacts with 1 mole of

NaCl to form 1 mole of AgCl. Therefore, the moles of AgCl formed will also be

0.0100 mol.

Step 3: Calculate the mass of precipitate formed.

Mass of AgCl = Moles ×Molar mass = 0.0100 mol ×143.32 g/mol = 1.43 g

Therefore, the mass of precipitate formed when 50.0 mL of 0.200 M silver

nitrate solution is mixed with excess sodium chloride solution is 1.43 g.

Question 18

Question

Calculate the concentration of chloride ions in a solution after mixing 100 mL

of 0.2 M sodium chloride with 200 mL of 0.1 M silver nitrate. Assume complete

precipitation of silver chloride and no volume change upon mixing.

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction

occurring.

NaCl(aq) + AgNO3(aq)→NaNO3(aq) + AgCl(s)

Step 2: Determine which reactant will limit the formation of the precipitate,

silver chloride.

Step 3: Calculate the amount of silver chloride precipitated.



Moles of NaCl = 0.2 mol/L ×0.1 L = 0.02 mol



Moles of AgNO3= 0.1 mol/L ×0.2 L = 0.02 mol



Since the moles of both reactants are equal, both are in stoichiometric

quantities. Therefore, the limiting reagent is NaCl.



The amount of AgCl precipitated will also be 0.02 mol.

Step 4: Calculate the concentration of chloride ions after precipitation.



Initial moles of chloride ions = 0.1 mol/L ×0.1 L = 0.01 mol



Moles of chloride ions remaining = 0.01 mol - 0.02 mol = -0.01 mol (neg-

ative because all chloride ions have precipitated)



Volume of solution after mixing = 100 mL + 200 mL = 300 mL = 0.3 L



Concentration of chloride ions after precipitation = −0.01 mol

0.3 L =−0.033 M

Step 5: Interpretation of the negative concentration: Since concentrations

cannot be negative, the negative sign in this case indicates that all the chloride

ions have precipitated as silver chloride, and there are no chloride ions left in

the solution.

Question 19

Question

A certain stream has a flow rate of 10 m3/s. If the stream receives 2 cm of rain

over an area of 500 km2, how long will it take for the stream to rise by 1 meter

assuming all the rain contributes to the stream flow?

Solution

Step 1: Convert the area of land receiving rain to square meters. Given that

the area is 500 km2and 1 km2= 106m2, we have:

500 ×106= 5 ×108m2

Step 2: Calculate the volume of rain that fell over the area. Given that 2

cm of rain fell over the area, we first convert it to meters:

2 cm = 0.02 m

Then, the volume of rain that fell over the area is:

0.02 m ×5×108m2= 1 ×107m3

Step 3: Calculate the time it takes for the stream to rise by 1 meter. Given

the stream flow rate is 10 m3/s, we can determine the time it takes for the

stream to rise by 1 meter using the formula:

time = volume of rain

flow rate

Substitute the values:

time = 1×107m3

10 m3/s= 1 ×106s

Therefore, it will take 1,000,000 seconds for the stream to rise by 1 meter.

Question 20

Question

A water sample contains 0.05 M MgCl2, 0.02 M Na2CO3, and 0.03 M Ca(NO3)2.

If you mix 50.0 mL of this solution with another 50.0 mL of a 0.04 M Na2SO4

solution, will a precipitate form? If so, calculate the mass of the precipitate

formed.

Given: - Ksp of MgCO3= 6.82 ×10−6-Ksp of CaCO3= 3.95 ×10−9

Solution

Step 1: Write the balanced chemical equations for the possible precipitates.

For MgCl2and Na2CO3:

Mg2+ (aq) + CO32−(aq) −−→ MgCO3(s)

For Ca(NO3)2and Na2CO3:

Ca2+ (aq) + CO32−(aq) −−→ CaCO3(s)

Step 2: Calculate the initial concentrations and find the ions’ concentrations

after dilution.

Given initial concentrations: - [Mg2+] = 0.05 M - [CO32−] = 0.02 M -

[Ca2+] = 0.03 M - [CO32−] = 0.02 M

After dilution with the Na2SO4 solution: - [Na+]=0.02 M - [SO42−]=0.04

Step 3: Determine the ion product, Q, for each possible precipitation reac-

tion.

For MgCO3:

Q= [Mg2+]×[CO32−]=0.05 ×0.02 = 1.0×10−3

For CaCO3:

Q= [Ca2+]×[CO32−]=0.03 ×0.02 = 6.0×10−4

Step 4: Compare Qto the corresponding Ksp for each precipitate.

For MgCO3:

Q < Ksp (No precipitation)

For CaCO3:

Q < Ksp (No precipitation)

Conclusion: No precipitate will form when the two solutions are mixed.

Question 21

Question

A solution is prepared by dissolving 15.0 g of barium chloride in enough water

to make 250 mL of solution. This solution is then mixed with 100.0 mL of a

0.150 M solution of sodium sulfate. Will a precipitate form? If so, what mass

of precipitate will form?

(Note: The molar mass of barium chloride is 208.23 g/mol and the molar

mass of sodium sulfate is 142.04 g/mol. The solubility product constant (Ksp)

for barium sulfate is 1.08 ×10−10 at 25

C.)

Solution

Step 1: Calculate the moles of barium chloride (BaCl2) present in the solution.

Given: - Mass of BaCl2= 15.0 g - Molar mass of BaCl2(MBaCl2) = 208.23

g/mol

The number of moles of BaCl2can be calculated using the formula:

Moles of BaCl2=Mass of BaCl2

MBaCl2

Moles of BaCl2=15.0 g

208.23 g/mol

Moles of BaCl2≈0.072 mol

Step 2: Calculate the moles of sodium sulfate (N a2SO4) added to the solu-

tion.

Given: - Volume of Na2SO4solution added = 100.0 mL = 0.100 L - Molarity

of Na2SO4solution = 0.150 M

The number of moles of Na2SO4can be calculated using the formula:

Moles of Na2SO4= Volume ×Molarity

Moles of Na2SO4= 0.100 L ×0.150 mol/L

Moles of Na2SO4= 0.015 mol

Step 3: Determine the limiting reactant to predict if a precipitate will form.

From the balanced chemical equation for the reaction between BaCl2and

Na2SO4:

BaCl2(aq) + N a2SO4(aq)→BaSO4(s)+2NaCl(aq)

1 mole of BaCl2reacts with 1 mole of Na2SO4to form 1 mole of BaSO4.

Since we have 0.072 moles of BaCl2and 0.015 moles of Na2SO4,N a2SO4

is the limiting reactant.

Step 4: Calculate the mass of precipitate formed (barium sulfate, BaSO4).

Given: - Molar mass of BaSO4(MBaSO4) = 233.39 g/mol

The mass of BaSO4formed can be calculated using the formula:

Mass of BaSO4= Moles of Na2SO4×MBaSO4

Mass of BaSO4= 0.015 mol ×233.39 g/mol

Mass of BaSO4≈3.50 g

Therefore, a precipitate of barium sulfate will form, and the mass of the

precipitate will be approximately 3.50 g.

Question 22

Question

In a laboratory experiment, a student mixes 50.0 mL of 0.200 M lead(II) nitrate

with 75.0 mL of 0.150 M potassium iodide solution. The balanced chemical

equation for the reaction is:

P b(NO3)2(aq)+2KI(aq)→P bI2(s)+2KNO3(aq)

Calculate the mass of lead(II) iodide, PbI2, that can be formed from this

reaction.

Solution

Step 1: Calculate the number of moles of lead(II) nitrate and potassium iodide

used in the reaction.

Moles of Pb(NO3)2= Volume (L) ×Concentration (M)

Moles of Pb(NO3)2= 0.0500 L ×0.200 M

Moles of Pb(NO3)2= 0.010 mol

Moles of KI = Volume (L) ×Concentration (M)

Moles of KI = 0.0750 L ×0.150 M

Moles of KI = 0.01125 mol

Step 2: Determine the limiting reactant.

Using the balanced chemical equation, we can see that 1 mole of lead(II)

nitrate reacts with 2 moles of potassium iodide to form 1 mole of lead(II) iodide.

Since the stoichiometry is 1:2 for Pb(NO3)2:KI, the limiting reactant is

Pb(NO3)2.

Step 3: Calculate the theoretical yield of lead(II) iodide.

Moles of PbI2=0.010 mol Pb(NO3)2

1×1 mol PbI2

1 mol Pb(NO3)2

Moles of PbI2= 0.010 mol ×1

Moles of PbI2= 0.010 mol

Step 4: Calculate the mass of lead(II) iodide.

Mass of PbI2= Moles of PbI2×Molar mass of PbI2

Mass of PbI2= 0.010 mol ×461.01 g/mol

Mass of PbI2≈4.61 g

Therefore, the mass of lead(II) iodide that can be formed from this reaction

is approximately 4.61 grams.

Question 23

Question

A solution is prepared by dissolving 20.0 g of potassium iodide (KI) in 100.0 g

of water at 25

C. Determine if a precipitate will form when 35.0 mL of 0.150 M

lead(II) nitrate (Pb(NO3)2) solution is added to the potassium iodide solution.

The solubility products constants are Ksp = 7.1×10−9for lead(II) iodide (PbI2)

and Ksp = 2.5×10−7for lead(II) bromide (PbBr2). Assume the volume of the

solution does not change upon mixing.

Solution

Step 1: Write the balanced equation for the precipitation reaction between

lead(II) nitrate and potassium iodide:

Pb(NO3)2+ 2KI →PbI2↓+2KNO3

Step 2: Calculate the moles of potassium iodide (KI) and lead(II) nitrate

(Pb(NO3)2) using their respective masses and molar masses: - Moles of KI:

Molar mass of KI = 39.10 g/mol + 126.90 g/mol = 166.00 g/mol

Moles of KI = 20.0 g

166.00 g/mol = 0.1205 mol

- Moles of Pb(NO3)2:

Molar mass of Pb(NO3)2= 207.2 g/mol+3(16.00 g/mol+14.01 g/mol) = 331.2 g/mol

Moles of Pb(NO3)2= M ×V=0.150 mol/L ×0.0350 L = 0.00525 mol

Step 3: Determine the limiting reactant: The molar ratio between Pb(NO3)2

and KI is 1:2, meaning that 1 mole of Pb(NO3)2will react with 2 moles of KI

to form PbI2. Since there is less moles of Pb(NO3)2than required, Pb(NO3)2

is the limiting reactant.

Step 4: Calculate the concentration of lead(II) ions in the solution:

Volume of solution = 100.0 g water + 35.0 mL Pb(NO3)2

Volume of solution (L) = 100.0 g

1.0 g/mL + 0.0350 L = 0.1350 L

[Pb2+] = moles of Pb(NO3)2

volume of solution in L =0.00525 mol

0.1350 L = 0.0389 M

Step 5: Calculate the ion product Qand compare with the solubility product

constant Ksp:

Q= [Pb2+][I−]2

Question 24

Question

A chemical reaction takes place in a solution resulting in the formation of a

precipitate. If 50.0 mL of a 0.200 M silver nitrate solution is mixed with 75.0 mL

of a 0.150 M sodium chloride solution, what mass of silver chloride precipitate

is formed? (Assume the reaction goes to completion)

Solution

Step 1: Write the balanced chemical equation for the reaction between silver

nitrate (AgNO3) and sodium chloride (NaCl):

AgNO3+ NaCl →AgCl + NaNO3

Step 2: Determine the limiting reactant to find out which reactant will run

out first. The limiting reactant is the one that produces the least amount of

product. We can do this by calculating the moles of each reactant:

moles of AgNO3= Molarity ×Volume = 0.200 mol/L ×0.0500 L = 0.0100 mol

moles of NaCl = Molarity ×Volume = 0.150 mol/L ×0.0750 L = 0.01125 mol

Since AgNO3has fewer moles (0.0100 mol) compared to NaCl (0.01125 mol),

AgNO3is the limiting reactant.

Step 3: Use the stoichiometry of the balanced chemical equation to find the

moles of AgCl formed. From the equation, 1 mole of AgNO3reacts with 1 mole

of NaCl to produce 1 mole of AgCl:

moles of AgCl = 0.0100 mol

Step 4: Calculate the mass of silver chloride precipitate formed using the

molar mass of AgCl, which is 143.32 g/mol:

mass of AgCl = 0.0100 mol ×143.32 g/mol = 1.43 g

Therefore, the mass of silver chloride precipitate formed is 1.43 g.

Question 25

Question

A student is performing a series of precipitation reactions in the laboratory.

They start with a solution containing 0.300 M calcium chloride (CaCl2) and

then add an excess of 0.150 M sodium sulfate (Na2SO4) solution.

Given that the Ksp of calcium sulfate (CaSO4) is 2.4×10−5, calculate the

mass of calcium sulfate that will precipitate.

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction

between CaCl2and Na2SO4:

CaCl2+ Na2SO4→CaSO4+ 2NaCl

Step 2: Determine the ions present in solution: - Initial concentrations: -

[Ca2+]initial = 0.300 M - [SO2−

4]initial = 0.150 M

- Once the reaction occurs, the Ca2+ ions will fully react with the SO2−

ions to form CaSO4. The limiting reactant will be CaCl2because the initial

concentration of Ca2+ ions is higher than the SO2−

4ions.

Step 3: Calculate the concentration of CaSO4formed: - The reaction con-

sumes all of the Ca2+ ions to form CaSO4, which means the final concentra-

tion of Ca2+ ions will be zero. - The concentration of Ca2+ ions that reacted

is 0.150 M. - Therefore, the concentration of Ca2+ ions that reacted to form

CaSO4is 0.150 M.

Step 4: Calculate the mass of CaSO4that precipitates: - The number of

moles of CaSO4formed is equal to the number of moles of Ca2+ ions that re-

acted. - Moles of CaSO4= [Ca2+]reacted = 0.150 mol - The molar mass of CaSO4

is 40.08 g/mol + 32.06 g/mol + 4(16.00 g/mol) = 136.16 g/mol. - Therefore, the

mass of CaSO4that will precipitate is:

Mass = Moles ×Molar mass = 0.150 mol ×136.16 g/mol = 20.42 g

Question 26

Question

A certain metal hydroxide, M(OH)2, has a solubility product constant Ksp of

1.6×10−15. Calculate the molar solubility of this metal hydroxide in a solution

with a pH of 9.2.

Solution

Step 1: Write the chemical equation for the dissolution of M(OH)2. The disso-

lution of M(OH)2can be represented by the equation:

M(OH)2⇌M2+ + 2OH−

Step 2: Write the equilibrium expression for the dissolution of M(OH)2. The

solubility product constant expression for M(OH)2is:

Ksp = [M2+][OH−]2

Step 3: Calculate the concentration of hydroxide ions ([OH]−) from the given

pH. We know that [H+] = 10−pH, so for a solution with pH 9.2:

[H+] = 10−9.2= 6.31 ×10−10

Since the solution is neutral, [H+] = [OH−]=6.31 ×10−10.

Step 4: Substitute the values into the solubility product constant expression.

Substitute [M2+] = xand [OH−]=6.31 ×10−10 into the Ksp expression:

1.6×10−15 = (x)(6.31 ×10−10)2

Step 5: Solve for the molar solubility of M(OH)2. Solve for x:

x=1.6×10−15

(6.31 ×10−10)2

x= 4.00 ×10−6M

Therefore, the molar solubility of the metal hydroxide in a solution with a

pH of 9.2 is 4.00 ×10−6M.

Question 27

Question

A solution is prepared by mixing 50.0 mL of 0.20 M silver nitrate with 75.0 mL

of 0.15 M sodium chloride. Determine if a precipitate will form. (For reference,

Ksp for silver chloride is 1.6×10−10.)

Solution

Step 1: Write the balanced chemical equation for the reaction of silver nitrate

and sodium chloride:

AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)

Step 2: Determine the ions present in solution after mixing the two solutions.

From the 50.0 mL of 0.20 M silver nitrate, we have: moles of Ag+= 0.050 L×

0.20 mol/L = 0.010 mol

From the 75.0 mL of 0.15 M sodium chloride, we have: moles of Cl−=

0.075 L×0.15 mol/L = 0.01125 mol

Step 3: Determine the molar solubility of AgCl. Let’s assume x mol/L

of AgCl is formed. The concentration of Ag+will be 0.010 mol +x. The

concentration of Cl−will be 0.01125 mol +x. At equilibrium, Ksp = [Ag+]×

[Cl−] = x×xThus, Ksp =x2

Therefore, 1.6×10−10 =x2Solving for x, we get x=√1.6×10−10 =

1.26 ×10−5

Step 4: Check if a precipitate will form. The molar solubility of AgCl is

1.26 ×10−5mol/L. Since the initial concentration of Ag+is 0.010 mol/L and

Cl−is 0.01125 mol/L, which are both greater than the molar solubility of AgCl,

a precipitate of AgCl will form.

Question 28

Question

A chemist has 500 mL of a solution containing 0.2 M lead nitrate (Pb(NO3)2).

The chemist wants to precipitate all the lead as lead sulfate (PbSO4) using

sulfuric acid (H2SO4). The balanced equation for the reaction is:

Pb(NO3)2(aq)+H2SO4(aq)→PbSO4(s) + 2HNO3(aq)

Assuming that lead dissolves completely as Pb2+ ions, calculate the volume

of 2 M sulfuric acid required to precipitate all the lead as lead sulfate. (Hint:

PbSO4remains insoluble in an excess of sulfuric acid.)

Solution

Step 1: Write the balanced chemical equation for the reaction between lead

nitrate and sulfuric acid to form lead sulfate and nitric acid:

Pb(NO3)2(aq)+H2SO4(aq)→PbSO4(s) + 2HNO3(aq)

Step 2: Calculate the number of moles of lead nitrate in the solution. Given:

- Concentration of lead nitrate solution = 0.2 M - Volume of lead nitrate solution

= 500 mL = 0.5 L

Number of moles of Pb(NO3)2= Concentration×Volume = 0.2 mol/L×0.5 L = 0.1 mol

Step 3: Use the stoichiometry of the balanced chemical equation to determine

the amount of sulfuric acid needed to precipitate all the lead as lead sulfate.

From the balanced equation, 1 mole of lead nitrate reacts with 1 mole of sulfuric

acid to produce 1 mole of lead sulfate.

Since the molar ratio is 1:1, the number of moles of sulfuric acid required

is equal to the number of moles of lead nitrate. Thus, we need 0.1 moles of

sulfuric acid.

Step 4: Calculate the volume of 2 M sulfuric acid required to supply 0.1

moles of sulfuric acid. Given: - Concentration of sulfuric acid = 2 M

Volume of sulfuric acid = Number of moles of H2SO4

Concentration =0.1 mol

2 mol/L = 0.05 L = 50 mL

Therefore, 50 mL of 2 M sulfuric acid is required to precipitate all the lead

as lead sulfate.

Question 29

Question

Calculate the concentration of sulfate ions in a solution that forms when 150.0

mL of a 0.20 M solution of barium chloride is mixed with 300.0 mL of a 0.10 M

solution of sodium sulfate.

Solution

Step 1: Write the balanced chemical equation for the reaction between barium

chloride and sodium sulfate:

BaCl2+ Na2SO4→BaSO4+ 2NaCl

Step 2: Determine the limiting reagent to find the maximum amount of

precipitate (barium sulfate) that can form. Since barium sulfate (BaSO4) is

insoluble in water, it will precipitate out of the solution. From the equation,

the stoichiometry shows that 1 mole of barium chloride will react with 1 mole

of sodium sulfate to form 1 mole of barium sulfate.

Calculate the moles of barium chloride:

moles of BaCl2= Molarity ×Volume (L) = 0.20 mol/L ×0.150 L = 0.030 mol

and the moles of sodium sulfate:

moles of Na2SO4= Molarity ×Volume (L) = 0.10 mol/L ×0.300 L = 0.030 mol

Since the moles of barium chloride and sodium sulfate are equal, there is

a 1:1 ratio of reactants. Therefore, both reactants will be fully consumed and

there will be no excess reagent.

Step 3: Calculate the concentration of sulfate ions in the final solution after

the precipitation reaction. The moles of sulfate ions in the final solution come

from the sodium sulfate:

moles of sulfate ions = moles of Na2SO4= 0.030 mol

The total volume of the final solution is the sum of the volumes of the two

initial solutions:

Total volume = 150 mL + 300 mL = 450 mL = 0.450 L

Therefore, the concentration of sulfate ions in the final solution is:

Concentration of sulfate ions = moles of sulfate ions

total volume (L) =0.030 mol

0.450 L = 0.067 M

Question 30

Question

A chemistry student is performing a precipitation reaction by mixing 100.0 mL

of 0.150 M silver nitrate (AgNO3) with 150.0 mL of 0.200 M sodium chloride

(NaCl). If the Ksp of silver chloride (AgCl) is 1.8×10−10, will a precipitate

of silver chloride form? If so, calculate the mass of silver chloride that will

precipitate.

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction

between silver nitrate and sodium chloride:

AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)

Step 2: Determine the ions present in solution after mixing the two solu-

tions. Calculate the concentrations of Ag+and Cl−ions in the mixture: - The

initial concentration of Ag+ions = 0.100 L ×0.150 M = 0.0150 mol - The initial

concentration of Cl−ions = 0.150 L ×0.200 M = 0.0300 mol

Step 3: Determine the reaction quotient (Qsp) by multiplying the concen-

trations of the products raised to their stoichiometric coefficients:

Qsp = [Ag+]×[Cl−] = (0.0150)(0.0300) = 0.00045

Step 4: Compare Qsp to Ksp to determine if a precipitate will form: - Since

Qsp < Ksp, a precipitate will not form.

Step 5: The mass of silver chloride that could potentially precipitate can

be calculated using the stoichiometry of the balanced chemical equation: - The

molar mass of AgCl is 107.87 g/mol. - The moles of AgCl that could potentially

precipitate is equal to the minimum of the moles of Ag+and Cl−ions, which

is 0.0150 mol. - Therefore, the mass of silver chloride that could potentially

precipitate is 0.0150 mol ×107.87 g/mol = 1.62 g.

Question 31

Question

A solution contains 20 grams of a compound AB2in 100 mL of water. If the

solubility product constant (Ksp) for AB2is 4.0×10−8, calculate the concen-

tration of AB2ions in the solution, its molar solubility, and the maximum mass

of AB2that can dissolve in 1 L of water.

Solution

Step 1: Calculate the concentration of AB2ions in the solution.

Molar mass of AB2= 2(molar mass of A) + 1(molar mass of B)

= 2(40.0 g/mol) + 1(16.0 g/mol)

= 112.0 g/mol

Number of moles of AB2in 20 g:

Number of moles = Mass in grams

Molar mass

=20 g

112.0 g/mol

≈0.179 mol

The volume of the solution is 100 mL = 0.1 L.

Concentration of AB2ions:

Concentration = Number of moles

Volume in L

=0.179 mol

0.1 L

= 1.79 M

Step 2: Calculate the molar solubility of AB2. Let the molar solubility of

AB2be x.

The solubility-product constant equation is given by:

Ksp = [A][B]2=x(2x)2= 4.0×10−8

Solve for x:

4x3= 4.0×10−8

x3= 1.0×10−8

x= 1.0×10−2M=0.01 M

So, the molar solubility of AB2is 0.01 M.

Step 3: Calculate the maximum mass of AB2that can dissolve in 1 L of

water. Molar mass of AB2is 112.0 g/mol.

Maximum mass that can dissolve in 1 L = molar solubility ×molar mass of

AB2:

Max mass = 0.01 mol/L ×112.0 g/mol = 1.12 g/L

Therefore, the maximum mass of AB2that can dissolve in 1 L of water is

1.12 grams.

Question 32

Question

A certain region receives an average annual precipitation of 850 mm. If the area

of the region is 1200 km2, determine the total volume of water that falls on the

region annually in cubic meters.

Solution

Step 1: Convert the area from km2to m2. Step 2: Convert the precipitation

from mm to m. Step 3: Calculate the total volume of water that falls on the

region annually.

Step 1: To convert the area from km2to m2, we use the conversion factor

1 km2= 1,000,000 m2. So, the area in m2is:

1200 km2×1,000,000 m2/km2= 1,200,000,000 m2

Step 2: To convert the precipitation from mm to m, we use the conversion

factor 1 mm = 0.001 m. So, the precipitation in m is:

850 mm ×0.001 m/mm = 0.85 m

Step 3: To calculate the total volume of water that falls on the region

annually, we multiply the area by the precipitation depth. Therefore, the total

volume of water in cubic meters is:

1,200,000,000 m2×0.85 m = 1,020,000,000 m3

Hence, the total volume of water that falls on the region annually is 1,020,000,000

cubic meters.

Question 33

Question

A solution contains 0.15 M barium chloride. If solid silver nitrate is added to the

solution, what concentration of silver nitrate is needed to completely precipitate

all the chloride ions as silver chloride? The solubility product constant Ksp for

silver chloride is 1.77 ×10−10.

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction:

BaCl2+ 2AgNO3→Ba(NO3)2+ 2AgCl

Step 2: Determine the moles of chloride ions present in the barium chloride

solution:

Moles of Cl−= 0.15 M ×volume of solution

Step 3: Calculate the moles of silver ions needed to precipitate all the chloride

ions: Moles of Ag+=Moles of Cl−

Step 4: Use the mole ratio from the balanced equation to find the moles of

silver nitrate required: Moles of AgNO3= 2 ×Moles of Ag+

Step 5: Calculate the concentration of silver nitrate needed based on the

volume of solution: Concentration of AgNO3=Moles of AgNO3

volume of solution

Step 6: Substitute the given values into the equation to find the concentra-

tion of silver nitrate required.

Question 34

Question

A solution contains 0.15 M barium chloride and 0.20 M silver nitrate. Calculate

the mass of barium sulfate that will precipitate in a volume of 500 mL. (Given:

Ksp of barium sulfate is 1.1×10−10)

Solution

Step 1: Write the balanced chemical equation for the precipitation reaction

between barium chloride and silver nitrate to form barium sulfate and silver

chloride.

BaCl2+ AgNO3→BaSO4+ 2AgCl

Step 2: Determine the limiting reactant by calculating the moles of barium

sulfate that can be formed from each reactant. From the balanced chemical

equation, 1 mole of BaCl2produces 1 mole of BaSO4.

Moles of BaSO4= 0.15 M ×0.5 L = 0.075 mol

From the balanced chemical equation, 1 mole of AgNO3produces 1 mole of

BaSO4.

Moles of BaSO4= 0.20 M ×0.5 L = 0.1 mol

Since 0.075 mol of BaSO4can be formed from BaCl2and 0.1 mol of BaSO4

can be formed from AgNO3, BaCl2is the limiting reactant.

Step 3: Calculate the mass of barium sulfate that will precipitate by finding

the mass of 0.075 mol of BaSO4.

Molar mass of BaSO4= 137.3 g/mol+32.1 g/mol+(4×16.0 g/mol) = 233.3 g/mol

Mass of BaSO4= 0.075 mol ×233.3 g/mol = 17.5 g

Therefore, 17.5 grams of barium sulfate will precipitate in a volume of 500

mL.

Question 35

Question

A sample of water has a concentration of sulfate ions of 0.025 M. If a solution of

barium chloride is added to the water sample, what will be the concentration of

sulfate ions in the solution once the precipitation reaction reaches completion?

The balanced chemical equation for the precipitation reaction is:

BaCl2(aq) + SO2−

4(aq)→BaSO4(s) + 2Cl−(aq)

The solubility product constant Ksp of barium sulfate is 1.1×10−10.

Solution

Step 1: Write the expression for the solubility product constant Ksp. The

solubility product constant Ksp for the reaction is given by:

Ksp = [Ba2+][SO2−

Step 2: Let xbe the concentration of sulfate ions at equilibrium. Since

1 mole of barium sulfate produces 1 mole of sulfate ions in the reaction, the

concentration of sulfate ions ([SO2−

4]) at equilibrium will be equal to x.

Step 3: Set up the equilibrium constant expression and solve for x. Sub-

stitute the known values into the equilibrium constant expression and solve for

Ksp = [Ba2+][SO2−

1.1×10−10 = 2x×0.025

x=1.1×10−10

0.05

x= 2.2×10−9M

Step 4: State the final answer. Therefore, the concentration of sulfate ions

in the solution once the precipitation reaction reaches completion is 2.2×10−9

CHEM 122 - GENERAL CHEMISTRY II Precipitation calculations Question Bank Set 8

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