CHEM 122 - GENERAL CHEMISTRY II Precipitation calculations Question Bank Set 9

CHEM 122 - GENERAL CHEMISTRY

II - Precipitation calculations

Question Bank - Set 9

Liberty University

Question 1

Question

Calculate the concentration of lead (II) iodide, PbI2, that can dissolve in 1.0 L

of water at 25

C. The Ksp of PbI2is 7.1×10−9.

Solution

Step 1: Write the dissociation equation for lead (II) iodide, PbI2. The dissoci-

ation equation for PbI2is:

PbI2⇌Pb2+ + 2I−

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

expression for the solubility product constant, Ksp, is:

Ksp = [Pb2+][I−]2

Step 3: Let the solubility of PbI2be x. Since 1 mol of PbI2dissociates into

1 mol of Pb2+ and 2 mol of I−, the concentrations of Pb2+ and I−will both be

2x. Thus, we have:

Ksp = (2x)(2x)2= 4x3

Step 4: Solve for x. Given that Ksp = 7.1×10−9, we can now solve for x:

4x3= 7.1×10−9

x3=7.1×10−9

x≈3

p1.775 ×10−9

x≈6.43 ×10−4M

Therefore, the concentration of lead (II) iodide that can dissolve in 1.0 L of

water at 25

C is 6.43 ×10−4M.

Question 2

Question

A solution contains 50.0 g of barium nitrate (Ba(NO3)2) dissolved in 100.0 mL

of water. Calculate the mass of the precipitate formed when 50.0 mL of 0.200

M sodium sulfate (Na2SO4) is added to the solution. Assume the reaction goes

to completion.

Solution

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

nitrate and sodium sulfate to determine the precipitate formed. Ba(NO3)2(aq)+

Na2SO4(aq)→BaSO4(s) + 2NaNO3(aq)

Step 2: Determine the limiting reactant based on the stoichiometry of the

reaction. - Calculate the number of moles of barium nitrate present:

mol Ba(NO3)2=mass

molar mass =50.0 g

261.34 g/mol = 0.1914 mol

- Calculate the number of moles of sodium sulfate added:

mol Na2SO4= M ×V=0.200 mol/L ×0.0500 L = 0.0100 mol

Since the molar ratio between barium nitrate and sodium sulfate is 1:1, sodium

sulfate is the limiting reactant.

Step 3: Calculate the theoretical yield of barium sulfate based on the limiting

reactant. - Use the mole ratio from the balanced equation to find the number

of moles of barium sulfate that can be formed: 0.0100 mol - Calculate the mass

of barium sulfate formed:

mass BaSO4= mol ×molar mass = 0.0100 mol ×233.39 g/mol = 2.334 g

Answer: The mass of the precipitate formed when 50.0 mL of 0.200 M

sodium sulfate is added to the solution is 2.334 g.

Question 3

Question

Calculate the concentration of sulfate ions (SO2−

4) in a solution when 100.0 ml

of a 0.20 M solution of sodium sulfate (Na2SO4) is mixed with 50.0 ml of a 0.40

M solution of calcium chloride (CaCl2). Assume the volumes are additive and

that the reactions proceed to completion.

Solution

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

sodium sulfate and calcium chloride to determine the products.

Step 2: Calculate the moles of sulfate ions produced by the reaction.

Step 3: Calculate the total volume of the solution.

Step 4: Determine the concentration of sulfate ions in the final solution.

Step 5: Check the final concentration to ensure it makes sense in the context

of the problem.

Step 1: The balanced chemical equation for the reaction between sodium

sulfate and calcium chloride is as follows:

Na2SO4(aq) + CaCl2(aq)→CaSO4(s) + 2NaCl(aq)

Step 2: From the balanced equation, it is clear that 1 mole of Na2SO4

produces 1 mole of SO2−

4ions. Therefore, the moles of SO2−

4produced from

sodium sulfate can be calculated as:

0.20 M ×0.100 L = 0.020 moles

Step 3: The total volume of the solution is:

100.0 ml + 50.0 ml = 150.0 ml = 0.150 L

Step 4: Now, calculate the concentration of sulfate ions in the final solution:

Concentration = moles of SO2−

total volume =0.020 moles

0.150 L = 0.133 M

Step 5: The final concentration of sulfate ions in the solution is 0.133 M,

which is expected given the dilution effect of mixing the two solutions.

Question 4

Question

A student is conducting a precipitation reaction between silver nitrate (AgNO3)

and sodium chloride (NaCl) to determine the concentration of chloride ions in

a sample. If 50.0 mL of 0.100 M silver nitrate is added to 50.0 mL of 0.150

M sodium chloride solution, how many grams of silver chloride (AgCl) will

precipitate out?

(Hint: The balanced chemical equation for the reaction is AgNO3(aq) +

NaCl(aq)→AgCl(s) + NaNO3(aq))

Solution

Step 1: Calculate the moles of silver nitrate and sodium chloride involved in the

reaction.

Given: Volume of silver nitrate solution = 50.0 mL = 0.0500 L Concentration

of silver nitrate solution = 0.100 M

Number of moles of silver nitrate = Concentration ×Volume = 0.100 mol/L

×0.0500 L = 0.00500 mol

Volume of sodium chloride solution = 50.0 mL = 0.0500 L Concentration of

sodium chloride solution = 0.150 M

Number of moles of sodium chloride = Concentration ×Volume = 0.150

mol/L ×0.0500 L = 0.00750 mol

Step 2: Determine the limiting reactant. From the balanced chemical equa-

tion, we see that the mole ratio between silver nitrate and sodium chloride is

1:1. Since silver nitrate has 0.00500 mol and sodium chloride has 0.00750 mol,

silver nitrate is the limiting reactant.

Step 3: Calculate the moles of silver chloride formed. Since silver nitrate is

the limiting reactant, all of it will react to form silver chloride. The moles of

silver chloride formed will be equal to the moles of silver nitrate used, which is

0.00500 mol.

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

silver chloride (AgCl) is 143.32 g/mol.

Mass of silver chloride formed = Number of moles ×Molar mass = 0.00500

mol ×143.32 g/mol = 0.7166 g

Therefore, 0.7166 grams of silver chloride will precipitate out in this reaction.

Question 5

Question

Calculate the concentration of barium ions (Ba2+) when 150.0 mL of a 0.200

M barium nitrate (Ba(NO3)2) solution is mixed with 300.0 mL of a 0.100 M

sodium sulfate (Na2SO4) solution. The reaction between barium nitrate and

sodium sulfate produces a white precipitate of barium sulfate (BaSO4).

Solution

Step 1: Write the balanced chemical equation for the reaction between Ba(NO3)2

and Na2SO4to determine the stoichiometry of the reaction.

Ba(NO3)2+ Na2SO4→BaSO4↓+2NaNO3

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

of Ba2+ and SO2−

4ions present in each solution. For Ba(NO3)2: Number of

moles of Ba2+ ions = 0.1500 L ×0.200 mol/L = 0.030 mol For Na2SO4: Number

of moles of SO2−

4ions = 0.3000 L ×0.100 mol/L = 0.030 mol

Since both solutions contain the same number of moles of Ba2+ ions and

SO2−

4ions, they are present in stoichiometric quantities. Thus, the limiting

reactant is Ba(NO3)2.

Step 3: Calculate the concentration of Ba2+ ions in the final solution. Total

volume of the mixed solution = 150.0 mL + 300.0 mL = 450.0 mL = 0.4500 L

Number of moles of Ba2+ ions in the final solution = 0.030 mol Concentration

of Ba2+ ions = 0.030 mol

0.4500 L = 0.067 M

Therefore, the concentration of barium ions (Ba2+) in the final solution is

0.067 M.

Question 6

Question

A solution is prepared by dissolving 20.0 g of calcium chloride (CaCl2) in enough

water to make 500.0 mL of solution. What is the mass percent concentration of

calcium chloride in the solution?

Solution

Step 1: Calculate the molarity of the calcium chloride solution. Given: Mass of

CaCl2: 20.0 g Volume of solution: 500.0 mL

First, we convert the volume from milliliters to liters:

500.0 mL ×1 L

1000 mL = 0.500 L

Next, we calculate the molarity using the formula:

Molarity = moles of solute

liters of solution

The molar mass of CaCl2is:

40.078 g/mol (Ca) + 2 ×35.453 g/mol (Cl) = 110.983 g/mol

Now, we calculate the moles of CaCl2:

Moles of CaCl2=20.0 g

110.983 g/mol = 0.180 mol

Therefore, the molarity of the solution is:

Molarity = 0.180 mol

0.500 L = 0.360 M

Step 2: Calculate the mass percent concentration of calcium chloride. The

mass percent is calculated using the formula:

Mass percent = mass of solute

mass of solution ×100%

The mass of the solution is:

20.0 g (CaCl2) + 500.0 g (water) = 520.0 g

Now we can calculate the mass percent concentration of calcium chloride:

Mass percent = 20.0 g

520.0 g ×100% = 3.85%

Therefore, the mass percent concentration of calcium chloride in the solution

is 3.85

Question 7

Question

A solution contains 0.1 M CaCl2and 0.1 M Na2SO4. If solid Na2SO4is added

to the solution until no more precipitate forms, what is the concentration of

Ca2+ ions in the final solution? (Ksp for CaSO4= 1.2×10−4)

Solution

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

CaCl2(aq) + Na2SO4(aq)→CaSO4(s) + 2NaCl(aq)

Step 2: Write the solubility product expression for CaSO4:

Ksp = [Ca2+][SO2−

4]=1.2×10−4

Step 3: Let the concentration of free Ca2+ ions be denoted by x. Since both

CaCl2and Na2SO4are 0.1 M and each Ca2+ ion forms one CaSO4precipitate,

the initial concentration of Ca2+ ions from CaCl2is 0.1 M.

Step 4: Define the change in concentration. Since the reaction is 1:1, xmoles

of CaSO4will form for every 1 mole of Ca2+ ions consumed, hence the change

will be x.

Step 5: Construct the ICE table:

Ca2+ SO2−

Initial (M) 0.1 0.1

Change (M) −x−x

Equilibrium (M) 0.1−x0.1−x

Step 6: Substitute the equilibrium concentrations into the solubility product

expression:

Ksp = (0.1−x)(0.1−x)=1.2×10−4

Step 7: Solve for x:

x2−0.2x+ 0.01 = 1.2×10−4

x2−0.2x+ 0.0098 = 0

Using the quadratic formula, we find x≈0.099 M

Therefore, the concentration of Ca2+ ions in the final solution is approxi-

mately 0.099 M.

Question 8

Question

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

sample. To do this, the chemist adds excess silver nitrate to a 50.0 mL sample

of the water and collects the white precipitate of silver chloride. The precipitate

is filtered, dried, and found to have a mass of 0.287 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 ppm (parts per million).

Solution

Step 1: Calculate the moles of silver chloride formed: The molar mass of silver

chloride is 143.32 g/mol. The mass of silver chloride precipitate collected is

0.287 g.

moles of AgCl =0.287 g

143.32 g/mol = 0.002g

Step 2: Calculate the moles of chloride ions in the sample: In the reaction

between silver nitrate and chloride ions, 1 mole of silver chloride is formed from

1 mole of chloride ions. Thus, the moles of chloride ions in the sample is also

0.002 mol.

Step 3: Calculate the volume of the water sample in liters: The volume of

the water sample used is 50.0 mL = 0.050 L.

Step 4: Calculate the concentration of chloride ions in the water sample in

mol/L:

Concentration of Cl−=0.002 mol

0.050 L = 0.040 mol/L

Step 5: Convert the concentration to ppm: Since 1 ppm = 1 mg/L, we need

to convert the concentration from mol/L to mg/L. The molar mass of chloride

ions (Cl-) is 35.45 g/mol. Converting concentration to mg/L:

0.040 mol/L ×35.45 g/mol ×1000 mg/g = 1418 mg/L

Step 6: Finally, convert the concentration to ppm:

Concentration of Cl−= 1418 ppm

Therefore, the concentration of chloride ions in the water sample is 1418

ppm.

Question 9

Question

Suppose a solution is prepared by mixing 200 mL of 0.4 M calcium chloride

with 300 mL of 0.2 M sodium sulfate. Will a precipitate form when these two

solutions are mixed? If so, what mass of calcium sulfate will be formed?

Given: Molar mass of CaCl2= 111 g/mol, Molar mass of Na2SO4= 142

g/mol, Molar mass of CaSO4= 136 g/mol, Densities of the solutions are the

same and equal to 1 g/mL.

Solution

Step 1: Determine the net ionic equation for the reaction between calcium

chloride and sodium sulfate.

The balanced chemical equation for the reaction is: CaCl2+ Na2SO4→

CaSO4+ 2 NaCl

The net ionic equation for the reaction is: Ca2+ + SO2−

4→CaSO4

Since calcium sulfate is insoluble, a precipitate will form when calcium chlo-

ride and sodium sulfate are mixed.

Step 2: Calculate the molality (moles of solute per kg of solvent) of calcium

sulfate in the solution.

1 L of solution = 1000 mL, Total volume of solution = 200 mL + 300 mL

= 500 mL = 0.5 L

Initial moles of CaCl2= 0.4 mol/L ×0.2 L = 0.08 mol, Initial moles of

Na2SO4= 0.2 mol/L ×0.3 L = 0.06 mol

Since the reaction is 1:1, all of the calcium chloride will react. Thus, after

the reaction, 0.08 moles of CaSO4will be formed.

Molality of CaSO4=moles of CaSO4

kg of solvent The density of the solution is 1 g/mL

so the mass of the solution is 500 g = 0.5 kg, Molality of CaSO4=0.08 mol

0.5 kg =

0.16 mol/kg

Step 3: Calculate the mass of calcium sulfate formed.

Mass of CaSO4= Molality ×Molar mass of CaSO4Mass of CaSO4= 0.16

mol/kg ×136 g/mol = 21.76 g

Therefore, a precipitate will form when calcium chloride and sodium sulfate

are mixed, and 21.76 g of calcium sulfate will be formed.

Question 10

Question

A solution contains 0.25 M silver nitrate (AgNO3) and 0.20 M sodium chloride

(NaCl). Calculate the concentration of Ag+ions after a white precipitate of

silver chloride (AgCl) forms.

Solution

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

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

Step 2: Determine the limiting reactant by calculating the possible products

from each reactant: For AgNO3: 1 mol AgNO3produces 1 mol Ag+and 1 mol

NO−

3For NaCl: 1 mol NaCl produces 1 mol Na+and 1 mol Cl−Since both

reactants produce 1 mol of Ag+per mol, the limiting reactant is the one that

produces the least amount of Ag+. Therefore, NaCl is the limiting reactant.

Step 3: Calculate the concentration of Ag+ions produced: Given:

[NaCl] = 0.20 M

Since 1 mol NaCl produces 1 mol Ag+, the concentration of Ag+ions is also

0.20 M.

Therefore, the concentration of Ag+ions after the precipitation reaction is

0.20 M.

Question 11

Question

Calculate the concentration of a precipitate formed when 200.0 mL of 0.100

M barium chloride solution is mixed with 300.0 mL of 0.150 M sodium sulfate

solution. Assume that barium sulfate is the only precipitate formed and that

the volume of the solutions are additive upon mixing.

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 and calculate the theoretical yield

of barium sulfate. First, calculate the number of moles of barium chloride and

sodium sulfate: For barium chloride:

Moles of BaCl2= Volume (L) ×Molarity = 0.200 L ×0.100 mol/L = 0.020 mol

For sodium sulfate:

Moles of Na2SO4= Volume (L)×Molarity = 0.300 L×0.150 mol/L = 0.045 mol

The limiting reactant is barium chloride since it forms the fewer moles of

the product, which is

0.020 mol BaSO4

Step 3: Calculate the concentration of the precipitate formed. The volume

of the resulting solution is the sum of the initial volumes:

Vtotal = 200.0 mL + 300.0 mL = 500.0 mL = 0.500 L

Since only barium sulfate precipitates out of the solution, the concentration of

the precipitate is:

Concentration of BaSO4=Moles of precipitate

Volume of solution =0.020 mol

0.500 L = 0.040 M

Therefore, the concentration of the precipitate formed when 200.0 mL of

0.100 M barium chloride solution is mixed with 300.0 mL of 0.150 M sodium

sulfate solution is 0.040 M.

Question 12

Question

Calculate the concentration of magnesium ions (Mg2+) in a solution prepared

by mixing 100.0 mL of 0.200 M MgCl2with 300.0 mL of 0.100 M Na2SO4.

Assume complete dissociation of both salts.

Solution

Step 1: Write the balanced chemical equation for the reaction between MgCl2

and Na2SO4to determine the products formed.

MgCl2+ Na2SO4→MgSO4+ 2NaCl

Step 2: Calculate the number of moles of Mg2+ ions in the solution. - Moles

of Mg2+ ions from MgCl2:

Number of moles = Volume ×Molarity = 0.100 L ×0.200 mol/L = 0.0200 mol -

Moles of Mg2+ ions from MgSO4: this is equal to the moles of MgCl2since the

stoichiometry is 1:1.

Total moles of Mg2+ ions = 0.0200 mol

Step 3: Calculate the total volume of the solution. Total volume = 100.0 mL+

300.0 mL = 400.0 mL = 0.400 L

Step 4: Calculate the final concentration of Mg2+ ions in the solution.

Concentration = Total moles of ions

Total volume =0.0200 mol

0.400 L = 0.0500 M

Therefore, the concentration of magnesium ions (Mg2+) in the solution is

0.0500 M.

Question 13

Question

Calculate the mass of ammonium chloride (NH4Cl) that needs to be dissolved

in 500 mL of water at 25

C to reach a saturated solution. The solubility of

ammonium chloride at 25

C is 37 g/100 mL.

Solution

Step 1: Calculate the maximum amount of ammonium chloride that can dissolve

in 500 mL of water at 25

Solubility of NH4Cl = 37 g/100 mL

Maximum solubility of NH4Cl = 37 ×500

100 = 185 g

Step 2: Determine the amount of ammonium chloride remaining to reach

saturation.

Mass of NH4Cl = 185 g

Volume of water = 500 mL = 500 ×1×10−3L=0.5 L

Concentration = 185 g

0.5 L = 370 g/L

Step 3: Calculate the amount of ammonium chloride needed to saturate the

solution.

Amount of NH4Cl needed = Maximum solubility −Amount of NH4Cl present

Amount of NH4Cl present = 370 g/L ×0.5 L = 185 g

Amount of NH4Cl needed = 185 g −185 g = 0 g

Therefore, no additional ammonium chloride needs to be added to reach

saturation in the solution.

Question 14

Question

A solution contains 0.1 M of barium nitrate (Ba(NO3)2) and 0.05 M of sodium

sulfate (Na2SO4). Calculate the concentrations of barium ion (Ba2+) and sulfate

ion (SO2−

4) in the final solution once the precipitation reaction is complete.

Solution

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

between barium nitrate and sodium sulfate:

Ba(NO3)2+ Na2SO4→BaSO4+ 2NaNO3

Step 2: Determine the limiting reactant by comparing the number of moles

of each reactant. Since the initial concentration of barium nitrate is 0.1 M, the

initial concentration of barium ions is also 0.1 M. The initial concentration of

sulfate ions is 0.05 M.

Step 3: Use the balanced chemical equation to find the theoretical yield of

the precipitate (barium sulfate). From the balanced equation, 1 mole of barium

nitrate reacts with 1 mole of sodium sulfate to produce 1 mole of barium sulfate.

This means that the concentration of barium sulfate formed is equal to the initial

concentration of the limiting reactant.

Step 4: Calculate the concentration of barium ions in the final solution. Since

the concentration of barium sulfate formed is equal to the initial concentration

of barium ions, the concentration of barium ions in the final solution is 0.1 M.

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

1 mole of sodium sulfate reacts to form 1 mole of sulfate ions in the product,

the concentration of sulfate ions in the final solution is 0.05 M.

Therefore, the concentration of barium ions (Ba2+) in the final solution is

0.1 M and the concentration of sulfate ions (SO2−

4) in the final solution is 0.05

M once the precipitation reaction is complete.

Question 15

Question

A certain industrial process generates a wastewater stream with a concentration

of 300 mg/L of a pollutant. The stream is to be mixed with a clean water

stream at a rate of 5 L/min to dilute the pollutant concentration to 10 mg/L.

Assuming complete mixing, determine the flow rate of the mixed stream leaving

the process.

Solution

Step 1: Let Vbe the flow rate of the mixed stream leaving the process in L/min.

Step 2: Write the mass balance equation based on the pollutant.

The mass of the pollutant entering the process per minute equals the mass

of the pollutant leaving the process per minute after dilution. This can be

represented as:

300 mg/L ×VL/min = 10 mg/L ×(V+ 5) L/min

Step 3: Solve the equation for V.

300V= 10(V+ 5)

300V= 10V+ 50

290V= 50

V=50

290 ≈0.17 L/min

Therefore, the flow rate of the mixed stream leaving the process is approxi-

mately 0.17 L/min.

Question 16

Question

Calculate the concentration of a PbCl2solution if 0.15 moles of PbCl2is dis-

solved in enough water to make 750 mL of solution. (Ksp of PbCl2is 1.6×10−5)

Solution

Step 1: Write the dissociation reaction for PbCl2.

PbCl2(s)⇌Pb2+(aq) + 2Cl−(aq)

Step 2: Write the expression for Ksp.

Ksp = [Pb2+][Cl−]2

Step 3: Let xbe the molar solubility of PbCl2.

Initially: Pb2+ = 0,Cl−= 0,PbCl2= 0.15 moles

Equilibrium: Pb2+ =x, Cl−= 2x, PbCl2= 0.15 −x

Step 4: Substitute the equilibrium concentrations into the Ksp expression

and solve for x.

1.6×10−5= (x)(2x)2

1.6×10−5= 4x3

x=3

p4.0×10−5

x≈0.032 M

Step 5: Calculate the concentration of PbCl2.

Concentration of PbCl2= 0.15 moles/0.75 L = 0.20 M

Question 17

Question

Calculate the concentration of lead(II) iodide, P bI2, that will precipitate when

100.0 ml of 0.100 M lead(II) nitrate, P b(NO3)2, is mixed with excess potassium

iodide, KI. The Ksp of lead(II) iodide is 7.1×10−9.

Solution

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

P b(NO3)2+ 2KI →P bI2+ 2KNO3

Step 2: Determine the moles of lead(II) nitrate in the solution.

Moles of P b(NO3)2= Volume ×Molarity = 0.100 mol/L ×0.100 L = 0.010 mol

Step 3: Use stoichiometry to find the limiting reactant and calculate the

moles of lead(II) iodide that will precipitate. From the balanced equation, 1

mole of P b(NO3)2produces 1 mole of P bI2.

Moles of P bI2= 0.010 mol

Step 4: Calculate the concentration of lead(II) iodide using the total volume

of the solution.

Volume of solution = 100.0 ml = 0.100 L

Concentration of P bI2=Moles of P bI2

Volume of solution =0.010 mol

0.100 L = 0.100 M

Step 5: Check if a precipitate will form by comparing the Ksp value with the

ion product. The ion product (Qsp) is calculated as [Pb2+][I−]2. Since lead(II)

nitrate is the limiting reagent, the concentration of lead(II) ions is equal to the

concentration of lead(II) iodide.

Qsp = [P b2+][I−]2= (0.100 M) ×(0.100 M)2= 0.0010

Since Qsp > Ksp, a precipitate of lead(II) iodide will form.

Question 18

Question

Calculate the solubility of silver chloride (AgCl) in water at 25

C. The Ksp

(AgCl) is 1.8×10−10.

Solution

Step 1: Write the balanced equilibrium equation for the dissociation of silver

chloride into its ions:

AgCl(s)⇌Ag+(aq) + Cl−(aq)

Step 2: Define the equilibrium constant expression (Ksp) for the dissociation

of silver chloride:

Ksp = [Ag+][Cl−]

Step 3: Since the initial concentration of AgCl is assumed to be 0 (since it

is a solid), let x be the molar solubility of AgCl. Therefore, at equilibrium, the

concentration of Ag+and Cl−ions will be equal to x.

Step 4: Substitute the equilibrium concentrations into the Ksp expression:

Ksp =x×x=x2

Step 5: Given that Ksp = 1.8×10−10, set up and solve the equation for x:

1.8×10−10 =x2

x=p1.8×10−10

x≈1.34 ×10−5

Therefore, the solubility of silver chloride (AgCl) in water at 25

C is approx-

imately 1.34 ×10−5mol/L.

Question 19

Question

A solution contains 0.1 M Zn(NO3)2 and 0.2 M Na2S. Calculate the minimum

volume of Zn(NO3)2 solution needed to completely precipitate all of the metal

ions as ZnS (Ksp = 1 ×10−23).

Solution

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

Zn(N O3)2+Na2S→ZnS + 2N aNO3

Step 2: Determine the limiting reactant by comparing the moles of Zn(NO3)2

and Na2S. Given concentrations:

Zn(N O3)2: 0.1M

Na2S: 0.2M

Step 3: Calculate the moles of Zn(NO3)2 and Na2S. Moles of Zn(NO3)2:

Moles =concentration ×volume = 0.1M×V

Moles of Na2S:

Moles =concentration ×volume = 0.2M×V

Step 4: The mole ratio between Zn(NO3)2 and Na2S is 1:1. So, the limiting

reactant will be the one with fewer moles. Set up an equation to find the volume

of Zn(NO3)2 needed to completely react with Na2S:

0.1M×V= 0.2M×V

0.1V= 0.2V

V= 2 L

Step 5: Therefore, the minimum volume of 0.1 M Zn(NO3)2 solution needed

to completely precipitate all of the metal ions as ZnS is 2 L.

Question 20

Question

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

(NaCl). What is the concentration of silver ions in the solution at equilibrium

after the precipitation of silver chloride (AgCl)? The Ksp of AgCl is 1.8×10−10.

Solution

Step 1: Write the balanced chemical equation for the precipitation of silver

chloride:

Ag+(aq) + Cl−(aq)→AgCl(s)

Step 2: Write the expression for the solubility product constant (Ksp):

Ksp = [Ag+][Cl−]

Step 3: Let x be the concentration of silver ions at equilibrium. Since the

molar ratio of Ag+to AgCl is 1:1, the concentration of AgCl formed will also

be x.

Step 4: Use the given concentrations of silver nitrate and sodium chloride

to determine the initial concentrations of silver and chloride ions: - Initial con-

centration of Ag+= 0.2 M - Initial concentration of Cl−= 0.1 M

Step 5: Set up the ICE (Initial, Change, Equilibrium) table:

Ag+Cl−AgCl

Initial (M) 0.2 0.1 0

Change (M) −x−x+x

Equilibrium (M) 0.2−x0.1−x x

Step 6: Substitute the equilibrium concentrations into the solubility product

expression and solve for x:

1.8×10−10 = (0.2−x)(0.1−x)

Step 7: Expand the right side of the equation and solve for x:

1.8×10−10 = 0.02 −0.3x+x2

x= 1.3×10−5M

Step 8: The concentration of silver ions at equilibrium after the precipitation

of silver chloride is 1.3×10−5M.

Question 21

Question

A solution contains 0.25 M of lead(II) nitrate and 0.15 M of sodium sulfate.

Determine if precipitation will occur when the two solutions are mixed.

Solution

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

nitrate (Pb(NO3)2) and sodium sulfate (Na2SO4):

Pb(NO3)2+ Na2SO4→PbSO4+ 2NaNO3

Step 2: Determine the products of the reaction:

Pb(NO3)2+ Na2SO4→PbSO4+ 2NaNO3

Step 3: Determine the solubility of lead(II) sulfate (PbSO4) and sodium

nitrate (NaNO3): - Lead(II) sulfate is insoluble in water. - Sodium nitrate is

soluble in water.

Step 4: Determine if a precipitate will form: Since lead(II) sulfate is insolu-

ble, a precipitate will form when Pb(NO3)2and Na2SO4are mixed.

Therefore, precipitation will occur when the two solutions are mixed.

Question 22

Question

Calculate the concentration of chloride ions in a solution if 100.0 mL of 0.200

M silver nitrate solution is required to completely precipitate the chloride ions

in 0.500 L of an unknown solution with chloride ions. Assume the reaction goes

to completion and that the only chloride ions present in the unknown solution

come from sodium chloride. (Molar mass: Na = 22.99 g/mol, Cl = 35.45 g/mol)

Solution

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

nitrate (AgNO3) and sodium chloride (NaCl):

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

Step 2: Determine the number of moles of silver nitrate used:

Given: Volume of silver nitrate solution = 100.0 mL = 0.100 L Concentration

of silver nitrate solution = 0.200 M

Using the formula moles = concentration ×volume, we have:

moles of AgNO3moles of AgNO3moles of AgNO3moles of AgNO3= 0.200 mol/L ×0.100 L= 0.020 mol AgN O3

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

the number of moles of chloride ions in the solution:

From the balanced chemical equation, we see that 1 mole of AgNO3reacts

with 1 mole of NaCl to form 1 mole of AgCl. Therefore, the number of moles

of chloride ions is equal to the number of moles of silver nitrate used.

moles of Cl−= 0.020 mol

Step 4: Calculate the concentration of chloride ions in the unknown solution:

Given: Volume of unknown solution = 0.500 L

Concentration of Cl−=moles of Cl−

volume of solution =0.020 mol

0.500 L = 0.040 M

Therefore, the concentration of chloride ions in the unknown solution is 0.040

Question 23

Question

A university lab is conducting an experiment where a silver chloride precipitate

is formed by mixing solutions of silver nitrate and sodium chloride. If 50.0 mL of

0.100 M silver nitrate solution is mixed with 100.0 mL of 0.150 M sodium chlo-

ride solution, what is the mass of silver chloride precipitate formed? (Assume

complete precipitation and that the density of the solutions is 1.00 g/mL)

Solution

Step 1: Determine the limiting reactant.

Given: Volume of silver nitrate solution (VAgN O3) = 50.0 mL Molarity of

silver nitrate solution (MAgN O3) = 0.100 M Volume of sodium chloride solution

(VNaCl) = 100.0 mL Molarity of sodium chloride solution (MNaCl) = 0.150 M

First, convert the volumes of solutions to liters: VAgN O3= 50.0 mL = 0.050

LVNaCl = 100.0 mL = 0.100 L

Next, calculate the moles of each reactant: molesAgN O3=MAgN O3×

VAgNO3= 0.100 mol/L ×0.050 L = 0.005 mol molesN aCl =MNaCl ×VN aCl =

0.150 mol/L ×0.100 L = 0.015 mol

Since silver nitrate and sodium chloride react in a 1:1 ratio, the limiting

reactant will be the one that produces the least amount of product. In this

case, molesAgN O3= 0.005 mol is the limiting reactant.

Step 2: Calculate the mass of silver chloride formed.

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

Using the mole ratio from the balanced chemical equation, calculate the

moles of silver chloride precipitate formed: molesAgCl =molesAgN O3= 0.005 mol

Finally, calculate the mass of silver chloride formed: massAgCl =molesAgCl×

Molar massAgCl = 0.005 mol ×143.32 g/mol = 0.7166 g Therefore, the mass of

silver chloride precipitate formed is 0.7166 g.

Question 24

Question

Calculate the mass of silver chloride (AgCl) that can be obtained by reacting

50.0 mL of 0.200 M silver nitrate (AgNO3) with excess hydrochloric acid (HCl).

The balanced chemical equation for the reaction is:

AgNO3(aq) + HCl(aq)→AgCl(s) + HN O3(aq)

Solution

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

AgNO3(aq) + HCl(aq)→AgCl(s) + HNO3(aq)

Step 2: Determine the moles of silver nitrate (AgNO3) used. Given: Volume

of AgNO3:VAgNO3= 50.0 mL = 0.0500 L Molarity of AgNO3:MAgN O3=

0.200 M

Moles of AgNO3=MAgNO3×VAgN O3= 0.200 mol/L ×0.0500 L = 0.0100 mol

Step 3: Determine the moles of silver chloride (AgCl) formed. From the

balanced chemical equation, the mole ratio between AgN O3and AgCl is 1:1.

Therefore, 0.0100 mol of AgNO3will produce 0.0100 mol of AgCl.

Step 4: Calculate the mass of silver chloride (AgCl). Given: Molar mass of

AgCl:MAgCl = 107.87 g/mol

Mass of AgCl = Moles of AgCl ×MAgCl = 0.0100 mol ×107.87 g/mol = 1.08 g

Therefore, the mass of silver chloride that can be obtained is 1.08 grams.

Question 25

Question

Calculate the concentration of sulfate ions (SO2−

4) in a solution prepared by

mixing 50.0 mL of 0.200 M Na2SO4with 150.0 mL of 0.100 M BaCl2. Assume

complete precipitation of barium sulfate (BaSO4).

Solution

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

between Na2SO4and BaCl2to form BaSO4.

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

Step 2: Determine the limiting reagent by comparing the moles of SO2−

4ions

from each of the reactants. For Na2SO4:

moles of SO2−

4= 0.200 M ×0.0500 L = 0.0100 mol

For BaCl2:

moles of SO2−

4= 0.100 M ×0.150 L ×1 mol Na2SO4

1 mol BaSO4

= 0.0150 mol

Since we need 0.0200 mol of SO2−

4ions to precipitate all barium ions as BaSO4,

Na2SO4is the limiting reagent.

Step 3: Calculate the moles of BaSO4formed using the limiting reagent.

moles of BaSO4= 0.0100 mol

Step 4: Calculate the concentration of SO2−

4ions in the final solution.

Volume of final solution = 0.0500 L + 0.150 L = 0.200 L

Concentration of SO2−

4=0.0100 mol

0.200 L = 0.0500 M

Therefore, the concentration of sulfate ions (SO2−

4) in the final solution is

0.0500 M.

Question 26

Question

A solution contains 0.2 mol/L of barium chloride (BaCl2) and 0.3 mol/L of

sodium sulfate (Na2SO4). What mass of barium sulfate (BaSO4) will precipitate

when these two solutions are mixed together?

(Hint: The balanced chemical equation for the reaction is BaCl2+Na2SO4→

BaSO4+ 2NaCl)

Solution

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

barium chloride and sodium sulfate:

BaCl2+ Na2SO4→BaSO4+ 2NaCl

Step 2: Determine which reactant is limiting by calculating the maximum

amount of product formed by each reactant.

For barium chloride (BaCl2): 1 mol of BaCl2produces 1 mol of BaSO4,

Therefore, 0.2 mol/L of BaCl2will produce 0.2 mol/L of BaSO4.

For sodium sulfate (Na2SO4): 1 mol of Na2SO4produces 1 mol of BaSO4,

Therefore, 0.3 mol/L of Na2SO4will produce 0.3 mol/L of BaSO4.

Step 3: Since the formation of barium sulfate is limited by the amount of

barium chloride, the amount of barium sulfate formed will be 0.2 mol/L.

Step 4: Calculate the mass of barium sulfate precipitated: The molar mass

of BaSO4can be calculated as:

molar mass of BaSO4= 1×molar mass of Ba+1×molar mass of S+4×molar mass of O

The molar mass of BaSO4is 137.3 g/mol + 32.1 g/mol + (4 ×16.0 g/mol)

= 233.3 g/mol

Now, we can calculate the mass of BaSO4precipitated:

Mass = moles ×molar mass = 0.2 mol/L ×233.3 g/mol = 46.66 g

Therefore, 46.66 grams of barium sulfate will precipitate when 0.2 mol/L of

barium chloride and 0.3 mol/L of sodium sulfate are mixed together.

Question 27

Question

A chemist is analyzing the precipitation reaction between lead nitrate (Pb(NO3)2)

and potassium iodide (KI) to form lead iodide (PbI2) and potassium nitrate

(KNO3). If 200 mL of a 0.5 M lead nitrate solution is mixed with 300 mL of

a 0.3 M potassium iodide solution, what mass of lead iodide will be produced?

Assume the reaction goes to completion.

Solution

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

Pb(NO3)2+ 2KI →PbI2+ 2KNO3

Step 2: Determine the limiting reactant by comparing the number of moles

of each reactant.

For lead nitrate: Number of moles = concentration ×volume Number of

moles = 0.5 M ×0.2 L Number of moles = 0.1 mol

For potassium iodide: Number of moles = concentration ×volume Number

of moles = 0.3 M ×0.3 L Number of moles = 0.09 mol

Since 2 moles of potassium iodide are needed for every mole of lead nitrate,

potassium iodide is the limiting reactant.

Step 3: Calculate the theoretical yield of lead iodide using the limiting re-

actant.

Number of moles of lead iodide = 0.09 mol ×1 mol PbI2

2 mol KI Number of moles of

lead iodide = 0.045 mol PbI2

Step 4: Calculate the mass of lead iodide produced using its molar mass.

Molar mass of lead iodide (PbI2) = 207.2 g/mol + 2 ×126.9 g/mol Molar

mass of lead iodide = 459 g/mol

Mass of lead iodide = 0.045 mol ×459 g/mol Mass of lead iodide = 20.67 g

Therefore, the mass of lead iodide produced is 20.67 g.

Question 28

Question

Calculate the solubility of silver chloride in a 0.10 M solution of sodium chloride

at 25

C. The Ksp of silver chloride is 1.77 ×10−10.

Solution

Step 1: Write the solubility equilibrium equation for silver chloride: The solu-

bility equilibrium equation for silver chloride is:

AgCl(s)⇌Ag+(aq) + Cl−(aq)

Step 2: Write the equilibrium expression: The equilibrium expression for

the dissociation of silver chloride is:

Ksp = [Ag+][Cl−]

Step 3: Set up an ICE table for the dissociation of silver chloride: Let the

solubility of silver chloride be represented as x.

Ag+Cl−

Initial (M) 0 0.10

Change (M) +x+x

Equilibrium (M) x0.10 + x

Step 4: Substitute the equilibrium concentrations into the equilibrium ex-

pression and solve for x: Substitute the equilibrium concentrations into the

equilibrium expression:

Ksp =x(0.10 + x)

Given that Ksp = 1.77 ×10−10, solve for x:

1.77 ×10−10 =x(0.10 + x)

1.77 ×10−10 = 0.10x+x2

x2+ 0.10x−1.77 ×10−10 = 0

Solve the quadratic equation to find the value of x.

Step 5: Calculate the solubility of silver chloride: After solving the quadratic

equation, determine the value of xrepresenting the solubility of silver chloride.

Remember to check if the assumption 0.10 + x≈0.10 holds true.

Therefore, the solubility of silver chloride in a 0.10 M solution of sodium

chloride at 25

C is the calculated value of x.

Question 29

Question

A chemistry student is conducting an experiment where two solutions are mixed

together to form a precipitate. If 50.0 mL of a 0.200 M solution of calcium

chloride is mixed with 75.0 mL of a 0.150 M solution of sodium carbonate,

what mass of calcium carbonate will be formed? (Assume the reaction goes to

completion and that the volume of the solutions is additive.)

Solution

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

chloride (CaCl2) and sodium carbonate (Na2CO3):

CaCl2(aq)+Na2CO3(aq)→CaCO3(s)+ 2N aCl(aq)

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

of each reactant using the given concentrations and volumes: For calcium chlo-

ride (CaCl2):

nCaCl2=CCaCl2×VCaCl2

nCaCl2= 0.200 M ×0.0500 L

nCaCl2= 0.0100 mol

For sodium carbonate (Na2CO3):

nNa2CO3=CN a2CO3×VN a2CO3

nNa2CO3= 0.150 M ×0.0750 L

nNa2CO3= 0.0113 mol

Since calcium chloride is the limiting reactant (0.0100 mol), the reaction will

consume all 0.0100 mol of CaCl2.

Step 3: Use stoichiometry to calculate the moles of calcium carbonate formed:

From the balanced equation, 1 mol of CaCl2produces 1 mol of CaCO3. So,

0.0100 mol of CaCl2will produce 0.0100 mol of CaCO3.

Step 4: Calculate the mass of calcium carbonate formed: Using the molar

mass of CaCO3(100.09 g/mol):

mCaCO3=nCaCO3×Molar mass of CaCO3

mCaCO3= 0.0100 mol ×100.09 g/mol

mCaCO3= 1.00 g

Therefore, the mass of calcium carbonate formed is 1.00 g.

Question 30

Question

Calculate the concentration of sulfate ion (SO2−

4) in a solution if 45.0 mL of

1.25 M barium chloride (BaCl2) is required to completely precipitate the sulfate

ion from the solution. Assume the reaction goes to completion. The molecular

weight of BaCl2is 208.23 g/mol.

Solution

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

between barium chloride and sulfate ion. Step 2: Determine the mole ratio

between barium chloride and sulfate ion. Step 3: Calculate the moles of sulfate

ion in the solution. Step 4: Determine the volume of the solution to find the

concentration of sulfate ion.

Step 1: The balanced chemical equation for the precipitation reaction is:

BaCl2+ SO2−

4→BaSO4↓+2Cl−

Step 2: From the balanced chemical equation, the mole ratio between bar-

ium chloride and sulfate ion is 1:1. This means 1 mole of barium chloride reacts

with 1 mole of sulfate ion.

Step 3: - Given that 45.0 mL of 1.25 M barium chloride is used, calculate

the moles of barium chloride:

Moles of BaCl2= Volume (L)×Molarity = 0.0450 L×1.25 mol/L = 0.0563 mol

- Since the mole ratio between barium chloride and sulfate ion is 1:1, the moles

of sulfate ion are also 0.0563 mol.

Step 4: - Calculate the volume of the solution by dividing the moles of

sulfate ion by its concentration:

Volume (L) = Moles

Molarity =0.0563 mol

0.0450 L = 1.25 M

Therefore, the concentration of sulfate ion in the solution is 1.25 M.

Question 31

Question

A solution is prepared by mixing 50.0 mL of 0.200 M silver nitrate with 75.0 mL

of 0.150 M sodium chloride. Assuming complete reaction, 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 limiting reactant. First, calculate the moles of silver

nitrate and sodium chloride:

Moles of AgNO3= Volume ×Molarity

= 0.0500 L ×0.200 mol/L

= 0.0100 mol

Moles of NaCl = 0.0750 L ×0.150 mol/L

= 0.0113 mol

Since the AgNO3:NaCl mole ratio is 1:1, sodium chloride is the limiting

reactant because it forms less moles of product.

Step 3: Calculate the mass of silver chloride precipitated. The molar mass

of AgCl is 143.32 g/mol.

Mass of AgCl = Moles of AgCl ×Molar mass of AgCl

= 0.0113 mol ×143.32 g/mol

= 1.62 g

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

Question 32

Question

A student is conducting an experiment to determine the concentration of chlo-

ride ions in a water sample. The student adds an excess of silver nitrate solution

to 100.0 mL of the water sample to precipitate the chloride ions as silver chlo-

ride. After collecting and drying the precipitate, the student finds that it weighs

0.569 g.

Calculate the concentration of chloride ions in the original water sample in

units of mol/L.

(Note: The molar mass of AgCl is 143.32 g/mol.)

Solution

Step 1: Calculate the moles of silver chloride precipitate.

Moles of AgCl = Mass of AgCl

Molar mass of AgCl

Moles of AgCl = 0.569 g

143.32 g/mol = 0.00397 mol

Step 2: Use the stoichiometry of the reaction to determine the moles of

chloride ions. The balanced chemical equation for the reaction is:

Ag++ Cl−→AgCl(s)

Since 1 mole of silver chloride is formed per mole of chloride ions, the moles

of chloride ions is also 0.00397 mol.

Step 3: Calculate the concentration of chloride ions in the original water

sample.

Volume of water sample = 100.0 mL = 0.1000 L

Concentration of chloride ions = Moles of Cl−

Volume of solution in L

Concentration of chloride ions = 0.00397 mol

0.1000 L = 0.0397 mol/L

Therefore, the concentration of chloride ions in the original water sample is

0.0397 mol/L.

Question 33

Question

A solution contains 0.025 M of calcium chloride and 0.015 M of sodium sulfate.

Determine whether a precipitate will form when the two solutions are mixed,

and if so, calculate the mass of the precipitate formed.

Solution

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

calcium chloride and sodium sulfate:

CaCl2+ Na2SO4→CaSO4+ 2NaCl

Step 2: Determine the possible products of the reaction: The possible prod-

ucts are calcium sulfate (CaSO) and sodium chloride (NaCl).

Step 3: Determine the solubility of the products: - Calcium sulfate is slightly

soluble in water. - Sodium chloride is highly soluble in water.

Step 4: Determine if a precipitate will form: Since calcium sulfate is slightly

soluble, it will form a precipitate when calcium chloride and sodium sulfate are

mixed.

Step 5: Write the net ionic equation for the reaction:

Ca2+ + SO2−

4→CaSO4

Step 6: Determine the limiting reactant: - From the balanced chemical

equation, we see that 1 mole of calcium chloride reacts with 1 mole of sodium

sulfate to produce 1 mole of calcium sulfate. - Calculate the moles of cal-

cium chloride: 0.025 M ×volume in L - Calculate the moles of sodium sulfate:

0.015 M ×volume in L - Identify the limiting reactant.

Step 7: Calculate the mass of calcium sulfate precipitate formed: - Once

the limiting reactant is determined, use the mole ratio in the balanced chemical

equation to calculate the moles of calcium sulfate formed. - Convert the moles

of calcium sulfate to grams using the molar mass of calcium sulfate.

This calculation will provide the mass of the precipitate formed when calcium

chloride and sodium sulfate solutions are mixed.

Question 34

Question

A solution is prepared by dissolving 5.00 g of silver nitrate in 100.0 mL of

water at 25

C. This solution was added dropwise to a solution containing 4.00

g of potassium iodide in 150.0 mL of water at 25

C. Calculate the mass of the

precipitate formed.

(Hint: The reaction that occurs is the double displacement reaction between

silver nitrate and potassium iodide to form silver iodide.)

Solution

Step 1: Find the moles of each reactant. The molar mass of silver nitrate

(AgNO3) is 169.87 g/mol. The molar mass of potassium iodide (KI) is 166.00

g/mol.

Moles of silver nitrate:

moles = mass

molar mass =5.00 g

169.87 g/mol = 0.0294 mol

Moles of potassium iodide:

moles = mass

molar mass =4.00 g

166.00 g/mol = 0.0241 mol

Step 2: Determine the limiting reactant. The balanced chemical equation

for the reaction is:

AgNO3+KI →AgI +KN O3

From the balanced equation, 1 mol of AgNO3reacts with 1 mol of KI to

produce 1 mol of AgI. Therefore, the moles ratio of AgN O3to KI is 1:1.

Since the moles ratio is 1:1, the limiting reactant is the one with the smaller

number of moles, which is potassium iodide.

Step 3: Calculate the mass of silver iodide precipitate. The molar mass of

silver iodide (AgI) is 234.77 g/mol. The theoretical yield of silver iodide can be

calculated using the moles of KI:

mass = moles ×molar mass = 0.0241 mol ×234.77 g/mol = 5.66 g

Therefore, the mass of the precipitate formed is 5.66 g.

Question 35

Question

Calculate the mass of barium sulfate (BaSO4) that can be formed by reacting

175 mL of 0.300 M barium chloride (BaCl2) with excess sulfuric acid (H2SO4).

The balanced chemical equation for the reaction is:

BaCl2+ H2SO4→BaSO4+ 2HCl

Solution

Step 1: Determine the moles of barium chloride. Given: Volume of BaCl2=

175 mL = 0.175 L Molarity of BaCl2= 0.300 M

Using the formula:

Molarity = moles

volume in liters

we can rearrange it to solve for moles:

moles = Molarity ×volume in liters

Plugging in the values:

moles of BaCl2= 0.300 M ×0.175 L = 0.0525 moles

Step 2: Find the limiting reactant. From the balanced chemical equation,

we see that the mole ratio between BaCl2and BaSO4is 1:1. Therefore, since

there are 0.0525 moles of BaCl2, this will also be the moles of BaSO4formed.

Step 3: Calculate the mass of barium sulfate. The molar mass of BaSO4can

be calculated as:

molar mass of BaSO4= molar mass of Ba+molar mass of S+4×molar mass of O

= 137.33 g/mol + 32.07 g/mol + 4 ×16.00 g/mol = 233.33 g/mol

Question 2

Question

A solution contains 50.0 g of barium nitrate (Ba(NO3)2) dissolved in 100.0 mL

of water. Calculate the mass of the precipitate formed when 50.0 mL of 0.200

M sodium sulfate (Na2SO4) is added to the solution. Assume the reaction goes

to completion.

Solution

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

nitrate and sodium sulfate to determine the precipitate formed. Ba(NO3)2(aq)+

Na2SO4(aq)→BaSO4(s) + 2NaNO3(aq)

Step 2: Determine the limiting reactant based on the stoichiometry of the

reaction. - Calculate the number of moles of barium nitrate present:

mol Ba(NO3)2=mass

molar mass =50.0 g

261.34 g/mol = 0.1914 mol

- Calculate the number of moles of sodium sulfate added:

mol Na2SO4= M ×V=0.200 mol/L ×0.0500 L = 0.0100 mol

Since the molar ratio between barium nitrate and sodium sulfate is 1:1, sodium

sulfate is the limiting reactant.

Step 3: Calculate the theoretical yield of barium sulfate based on the limiting

reactant. - Use the mole ratio from the balanced equation to find the number

of moles of barium sulfate that can be formed: 0.0100 mol - Calculate the mass

of barium sulfate formed:

mass BaSO4= mol ×molar mass = 0.0100 mol ×233.39 g/mol = 2.334 g

Answer: The mass of the precipitate formed when 50.0 mL of 0.200 M

sodium sulfate is added to the solution is 2.334 g.

Question 3

Question

Calculate the concentration of sulfate ions (SO2−

4) in a solution when 100.0 ml

of a 0.20 M solution of sodium sulfate (Na2SO4) is mixed with 50.0 ml of a 0.40

M solution of calcium chloride (CaCl2). Assume the volumes are additive and

that the reactions proceed to completion.

Solution

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

sodium sulfate and calcium chloride to determine the products.

Step 2: Calculate the moles of sulfate ions produced by the reaction.

Step 3: Calculate the total volume of the solution.

Step 4: Determine the concentration of sulfate ions in the final solution.

Step 5: Check the final concentration to ensure it makes sense in the context

of the problem.

Step 1: The balanced chemical equation for the reaction between sodium

sulfate and calcium chloride is as follows:

Na2SO4(aq) + CaCl2(aq)→CaSO4(s) + 2NaCl(aq)

Step 2: From the balanced equation, it is clear that 1 mole of Na2SO4

produces 1 mole of SO2−

4ions. Therefore, the moles of SO2−

4produced from

sodium sulfate can be calculated as:

0.20 M ×0.100 L = 0.020 moles

Step 3: The total volume of the solution is:

100.0 ml + 50.0 ml = 150.0 ml = 0.150 L

Step 4: Now, calculate the concentration of sulfate ions in the final solution:

Concentration = moles of SO2−

total volume =0.020 moles

0.150 L = 0.133 M

Step 5: The final concentration of sulfate ions in the solution is 0.133 M,

which is expected given the dilution effect of mixing the two solutions.

Question 4

Question

A student is conducting a precipitation reaction between silver nitrate (AgNO3)

and sodium chloride (NaCl) to determine the concentration of chloride ions in

a sample. If 50.0 mL of 0.100 M silver nitrate is added to 50.0 mL of 0.150

M sodium chloride solution, how many grams of silver chloride (AgCl) will

precipitate out?

(Hint: The balanced chemical equation for the reaction is AgNO3(aq) +

NaCl(aq)→AgCl(s) + NaNO3(aq))

Solution

Step 1: Calculate the moles of silver nitrate and sodium chloride involved in the

reaction.

Given: Volume of silver nitrate solution = 50.0 mL = 0.0500 L Concentration

of silver nitrate solution = 0.100 M

Number of moles of silver nitrate = Concentration ×Volume = 0.100 mol/L

×0.0500 L = 0.00500 mol

Volume of sodium chloride solution = 50.0 mL = 0.0500 L Concentration of

sodium chloride solution = 0.150 M

Number of moles of sodium chloride = Concentration ×Volume = 0.150

mol/L ×0.0500 L = 0.00750 mol

Step 2: Determine the limiting reactant. From the balanced chemical equa-

tion, we see that the mole ratio between silver nitrate and sodium chloride is

1:1. Since silver nitrate has 0.00500 mol and sodium chloride has 0.00750 mol,

silver nitrate is the limiting reactant.

Step 3: Calculate the moles of silver chloride formed. Since silver nitrate is

the limiting reactant, all of it will react to form silver chloride. The moles of

silver chloride formed will be equal to the moles of silver nitrate used, which is

0.00500 mol.

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

silver chloride (AgCl) is 143.32 g/mol.

Mass of silver chloride formed = Number of moles ×Molar mass = 0.00500

mol ×143.32 g/mol = 0.7166 g

Therefore, 0.7166 grams of silver chloride will precipitate out in this reaction.

Question 5

Question

Calculate the concentration of barium ions (Ba2+) when 150.0 mL of a 0.200

M barium nitrate (Ba(NO3)2) solution is mixed with 300.0 mL of a 0.100 M

sodium sulfate (Na2SO4) solution. The reaction between barium nitrate and

sodium sulfate produces a white precipitate of barium sulfate (BaSO4).

Solution

Step 1: Write the balanced chemical equation for the reaction between Ba(NO3)2

and Na2SO4to determine the stoichiometry of the reaction.

Ba(NO3)2+ Na2SO4→BaSO4↓+2NaNO3

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

of Ba2+ and SO2−

4ions present in each solution. For Ba(NO3)2: Number of

moles of Ba2+ ions = 0.1500 L ×0.200 mol/L = 0.030 mol For Na2SO4: Number

of moles of SO2−

4ions = 0.3000 L ×0.100 mol/L = 0.030 mol

Since both solutions contain the same number of moles of Ba2+ ions and

SO2−

4ions, they are present in stoichiometric quantities. Thus, the limiting

reactant is Ba(NO3)2.

Step 3: Calculate the concentration of Ba2+ ions in the final solution. Total

volume of the mixed solution = 150.0 mL + 300.0 mL = 450.0 mL = 0.4500 L

Number of moles of Ba2+ ions in the final solution = 0.030 mol Concentration

of Ba2+ ions = 0.030 mol

0.4500 L = 0.067 M

Therefore, the concentration of barium ions (Ba2+) in the final solution is

0.067 M.

Question 6

Question

A solution is prepared by dissolving 20.0 g of calcium chloride (CaCl2) in enough

water to make 500.0 mL of solution. What is the mass percent concentration of

calcium chloride in the solution?

Solution

Step 1: Calculate the molarity of the calcium chloride solution. Given: Mass of

CaCl2: 20.0 g Volume of solution: 500.0 mL

First, we convert the volume from milliliters to liters:

500.0 mL ×1 L

1000 mL = 0.500 L

Next, we calculate the molarity using the formula:

Molarity = moles of solute

liters of solution

The molar mass of CaCl2is:

40.078 g/mol (Ca) + 2 ×35.453 g/mol (Cl) = 110.983 g/mol

Now, we calculate the moles of CaCl2:

Moles of CaCl2=20.0 g

110.983 g/mol = 0.180 mol

Therefore, the molarity of the solution is:

Molarity = 0.180 mol

0.500 L = 0.360 M

Step 2: Calculate the mass percent concentration of calcium chloride. The

mass percent is calculated using the formula:

Mass percent = mass of solute

mass of solution ×100%

The mass of the solution is:

20.0 g (CaCl2) + 500.0 g (water) = 520.0 g

Now we can calculate the mass percent concentration of calcium chloride:

Mass percent = 20.0 g

520.0 g ×100% = 3.85%

Therefore, the mass percent concentration of calcium chloride in the solution

is 3.85

Question 7

Question

A solution contains 0.1 M CaCl2and 0.1 M Na2SO4. If solid Na2SO4is added

to the solution until no more precipitate forms, what is the concentration of

Ca2+ ions in the final solution? (Ksp for CaSO4= 1.2×10−4)

Solution

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

CaCl2(aq) + Na2SO4(aq)→CaSO4(s) + 2NaCl(aq)

Step 2: Write the solubility product expression for CaSO4:

Ksp = [Ca2+][SO2−

4]=1.2×10−4

Step 3: Let the concentration of free Ca2+ ions be denoted by x. Since both

CaCl2and Na2SO4are 0.1 M and each Ca2+ ion forms one CaSO4precipitate,

the initial concentration of Ca2+ ions from CaCl2is 0.1 M.

Step 4: Define the change in concentration. Since the reaction is 1:1, xmoles

of CaSO4will form for every 1 mole of Ca2+ ions consumed, hence the change

will be x.

Step 5: Construct the ICE table:

Ca2+ SO2−

Initial (M) 0.1 0.1

Change (M) −x−x

Equilibrium (M) 0.1−x0.1−x

Step 6: Substitute the equilibrium concentrations into the solubility product

expression:

Ksp = (0.1−x)(0.1−x)=1.2×10−4

Step 7: Solve for x:

x2−0.2x+ 0.01 = 1.2×10−4

x2−0.2x+ 0.0098 = 0

Using the quadratic formula, we find x≈0.099 M

Therefore, the concentration of Ca2+ ions in the final solution is approxi-

mately 0.099 M.

Question 8

Question

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

sample. To do this, the chemist adds excess silver nitrate to a 50.0 mL sample

of the water and collects the white precipitate of silver chloride. The precipitate

is filtered, dried, and found to have a mass of 0.287 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 ppm (parts per million).

Solution

Step 1: Calculate the moles of silver chloride formed: The molar mass of silver

chloride is 143.32 g/mol. The mass of silver chloride precipitate collected is

0.287 g.

moles of AgCl =0.287 g

143.32 g/mol = 0.002g

Step 2: Calculate the moles of chloride ions in the sample: In the reaction

between silver nitrate and chloride ions, 1 mole of silver chloride is formed from

1 mole of chloride ions. Thus, the moles of chloride ions in the sample is also

0.002 mol.

Step 3: Calculate the volume of the water sample in liters: The volume of

the water sample used is 50.0 mL = 0.050 L.

Step 4: Calculate the concentration of chloride ions in the water sample in

mol/L:

Concentration of Cl−=0.002 mol

0.050 L = 0.040 mol/L

Step 5: Convert the concentration to ppm: Since 1 ppm = 1 mg/L, we need

to convert the concentration from mol/L to mg/L. The molar mass of chloride

ions (Cl-) is 35.45 g/mol. Converting concentration to mg/L:

0.040 mol/L ×35.45 g/mol ×1000 mg/g = 1418 mg/L

Step 6: Finally, convert the concentration to ppm:

Concentration of Cl−= 1418 ppm

Therefore, the concentration of chloride ions in the water sample is 1418

ppm.

Question 9

Question

Suppose a solution is prepared by mixing 200 mL of 0.4 M calcium chloride

with 300 mL of 0.2 M sodium sulfate. Will a precipitate form when these two

solutions are mixed? If so, what mass of calcium sulfate will be formed?

Given: Molar mass of CaCl2= 111 g/mol, Molar mass of Na2SO4= 142

g/mol, Molar mass of CaSO4= 136 g/mol, Densities of the solutions are the

same and equal to 1 g/mL.

Solution

Step 1: Determine the net ionic equation for the reaction between calcium

chloride and sodium sulfate.

The balanced chemical equation for the reaction is: CaCl2+ Na2SO4→

CaSO4+ 2 NaCl

The net ionic equation for the reaction is: Ca2+ + SO2−

4→CaSO4

Since calcium sulfate is insoluble, a precipitate will form when calcium chlo-

ride and sodium sulfate are mixed.

Step 2: Calculate the molality (moles of solute per kg of solvent) of calcium

sulfate in the solution.

1 L of solution = 1000 mL, Total volume of solution = 200 mL + 300 mL

= 500 mL = 0.5 L

Initial moles of CaCl2= 0.4 mol/L ×0.2 L = 0.08 mol, Initial moles of

Na2SO4= 0.2 mol/L ×0.3 L = 0.06 mol

Since the reaction is 1:1, all of the calcium chloride will react. Thus, after

the reaction, 0.08 moles of CaSO4will be formed.

Molality of CaSO4=moles of CaSO4

kg of solvent The density of the solution is 1 g/mL

so the mass of the solution is 500 g = 0.5 kg, Molality of CaSO4=0.08 mol

0.5 kg =

0.16 mol/kg

Step 3: Calculate the mass of calcium sulfate formed.

Mass of CaSO4= Molality ×Molar mass of CaSO4Mass of CaSO4= 0.16

mol/kg ×136 g/mol = 21.76 g

Therefore, a precipitate will form when calcium chloride and sodium sulfate

are mixed, and 21.76 g of calcium sulfate will be formed.

Question 10

Question

A solution contains 0.25 M silver nitrate (AgNO3) and 0.20 M sodium chloride

(NaCl). Calculate the concentration of Ag+ions after a white precipitate of

silver chloride (AgCl) forms.

Solution

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

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

Step 2: Determine the limiting reactant by calculating the possible products

from each reactant: For AgNO3: 1 mol AgNO3produces 1 mol Ag+and 1 mol

NO−

3For NaCl: 1 mol NaCl produces 1 mol Na+and 1 mol Cl−Since both

reactants produce 1 mol of Ag+per mol, the limiting reactant is the one that

produces the least amount of Ag+. Therefore, NaCl is the limiting reactant.

Step 3: Calculate the concentration of Ag+ions produced: Given:

[NaCl] = 0.20 M

Since 1 mol NaCl produces 1 mol Ag+, the concentration of Ag+ions is also

0.20 M.

Therefore, the concentration of Ag+ions after the precipitation reaction is

0.20 M.

Question 11

Question

Calculate the concentration of a precipitate formed when 200.0 mL of 0.100

M barium chloride solution is mixed with 300.0 mL of 0.150 M sodium sulfate

solution. Assume that barium sulfate is the only precipitate formed and that

the volume of the solutions are additive upon mixing.

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 and calculate the theoretical yield

of barium sulfate. First, calculate the number of moles of barium chloride and

sodium sulfate: For barium chloride:

Moles of BaCl2= Volume (L) ×Molarity = 0.200 L ×0.100 mol/L = 0.020 mol

For sodium sulfate:

Moles of Na2SO4= Volume (L)×Molarity = 0.300 L×0.150 mol/L = 0.045 mol

The limiting reactant is barium chloride since it forms the fewer moles of

the product, which is

0.020 mol BaSO4

Step 3: Calculate the concentration of the precipitate formed. The volume

of the resulting solution is the sum of the initial volumes:

Vtotal = 200.0 mL + 300.0 mL = 500.0 mL = 0.500 L

Since only barium sulfate precipitates out of the solution, the concentration of

the precipitate is:

Concentration of BaSO4=Moles of precipitate

Volume of solution =0.020 mol

0.500 L = 0.040 M

Therefore, the concentration of the precipitate formed when 200.0 mL of

0.100 M barium chloride solution is mixed with 300.0 mL of 0.150 M sodium

sulfate solution is 0.040 M.

Question 12

Question

Calculate the concentration of magnesium ions (Mg2+) in a solution prepared

by mixing 100.0 mL of 0.200 M MgCl2with 300.0 mL of 0.100 M Na2SO4.

Assume complete dissociation of both salts.

Solution

Step 1: Write the balanced chemical equation for the reaction between MgCl2

and Na2SO4to determine the products formed.

MgCl2+ Na2SO4→MgSO4+ 2NaCl

Step 2: Calculate the number of moles of Mg2+ ions in the solution. - Moles

of Mg2+ ions from MgCl2:

Number of moles = Volume ×Molarity = 0.100 L ×0.200 mol/L = 0.0200 mol -

Moles of Mg2+ ions from MgSO4: this is equal to the moles of MgCl2since the

stoichiometry is 1:1.

Total moles of Mg2+ ions = 0.0200 mol

Step 3: Calculate the total volume of the solution. Total volume = 100.0 mL+

300.0 mL = 400.0 mL = 0.400 L

Step 4: Calculate the final concentration of Mg2+ ions in the solution.

Concentration = Total moles of ions

Total volume =0.0200 mol

0.400 L = 0.0500 M

Therefore, the concentration of magnesium ions (Mg2+) in the solution is

0.0500 M.

Question 13

Question

Calculate the mass of ammonium chloride (NH4Cl) that needs to be dissolved

in 500 mL of water at 25

C to reach a saturated solution. The solubility of

ammonium chloride at 25

C is 37 g/100 mL.

Solution

Step 1: Calculate the maximum amount of ammonium chloride that can dissolve

in 500 mL of water at 25

Solubility of NH4Cl = 37 g/100 mL

Maximum solubility of NH4Cl = 37 ×500

100 = 185 g

Step 2: Determine the amount of ammonium chloride remaining to reach

saturation.

Mass of NH4Cl = 185 g

Volume of water = 500 mL = 500 ×1×10−3L=0.5 L

Concentration = 185 g

0.5 L = 370 g/L

Step 3: Calculate the amount of ammonium chloride needed to saturate the

solution.

Amount of NH4Cl needed = Maximum solubility −Amount of NH4Cl present

Amount of NH4Cl present = 370 g/L ×0.5 L = 185 g

Amount of NH4Cl needed = 185 g −185 g = 0 g

Therefore, no additional ammonium chloride needs to be added to reach

saturation in the solution.

Question 14

Question

A solution contains 0.1 M of barium nitrate (Ba(NO3)2) and 0.05 M of sodium

sulfate (Na2SO4). Calculate the concentrations of barium ion (Ba2+) and sulfate

ion (SO2−

4) in the final solution once the precipitation reaction is complete.

Solution

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

between barium nitrate and sodium sulfate:

Ba(NO3)2+ Na2SO4→BaSO4+ 2NaNO3

Step 2: Determine the limiting reactant by comparing the number of moles

of each reactant. Since the initial concentration of barium nitrate is 0.1 M, the

initial concentration of barium ions is also 0.1 M. The initial concentration of

sulfate ions is 0.05 M.

Step 3: Use the balanced chemical equation to find the theoretical yield of

the precipitate (barium sulfate). From the balanced equation, 1 mole of barium

nitrate reacts with 1 mole of sodium sulfate to produce 1 mole of barium sulfate.

This means that the concentration of barium sulfate formed is equal to the initial

concentration of the limiting reactant.

Step 4: Calculate the concentration of barium ions in the final solution. Since

the concentration of barium sulfate formed is equal to the initial concentration

of barium ions, the concentration of barium ions in the final solution is 0.1 M.

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

1 mole of sodium sulfate reacts to form 1 mole of sulfate ions in the product,

the concentration of sulfate ions in the final solution is 0.05 M.

Therefore, the concentration of barium ions (Ba2+) in the final solution is

0.1 M and the concentration of sulfate ions (SO2−

4) in the final solution is 0.05

M once the precipitation reaction is complete.

Question 15

Question

A certain industrial process generates a wastewater stream with a concentration

of 300 mg/L of a pollutant. The stream is to be mixed with a clean water

stream at a rate of 5 L/min to dilute the pollutant concentration to 10 mg/L.

Assuming complete mixing, determine the flow rate of the mixed stream leaving

the process.

Solution

Step 1: Let Vbe the flow rate of the mixed stream leaving the process in L/min.

Step 2: Write the mass balance equation based on the pollutant.

The mass of the pollutant entering the process per minute equals the mass

of the pollutant leaving the process per minute after dilution. This can be

represented as:

300 mg/L ×VL/min = 10 mg/L ×(V+ 5) L/min

Step 3: Solve the equation for V.

300V= 10(V+ 5)

300V= 10V+ 50

290V= 50

V=50

290 ≈0.17 L/min

Therefore, the flow rate of the mixed stream leaving the process is approxi-

mately 0.17 L/min.

Question 16

Question

Calculate the concentration of a PbCl2solution if 0.15 moles of PbCl2is dis-

solved in enough water to make 750 mL of solution. (Ksp of PbCl2is 1.6×10−5)

Solution

Step 1: Write the dissociation reaction for PbCl2.

PbCl2(s)⇌Pb2+(aq) + 2Cl−(aq)

Step 2: Write the expression for Ksp.

Ksp = [Pb2+][Cl−]2

Step 3: Let xbe the molar solubility of PbCl2.

Initially: Pb2+ = 0,Cl−= 0,PbCl2= 0.15 moles

Equilibrium: Pb2+ =x, Cl−= 2x, PbCl2= 0.15 −x

Step 4: Substitute the equilibrium concentrations into the Ksp expression

and solve for x.

1.6×10−5= (x)(2x)2

1.6×10−5= 4x3

x=3

p4.0×10−5

x≈0.032 M

Step 5: Calculate the concentration of PbCl2.

Concentration of PbCl2= 0.15 moles/0.75 L = 0.20 M

Question 17

Question

Calculate the concentration of lead(II) iodide, P bI2, that will precipitate when

100.0 ml of 0.100 M lead(II) nitrate, P b(NO3)2, is mixed with excess potassium

iodide, KI. The Ksp of lead(II) iodide is 7.1×10−9.

Solution

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

P b(NO3)2+ 2KI →P bI2+ 2KNO3

Step 2: Determine the moles of lead(II) nitrate in the solution.

Moles of P b(NO3)2= Volume ×Molarity = 0.100 mol/L ×0.100 L = 0.010 mol

Step 3: Use stoichiometry to find the limiting reactant and calculate the

moles of lead(II) iodide that will precipitate. From the balanced equation, 1

mole of P b(NO3)2produces 1 mole of P bI2.

Moles of P bI2= 0.010 mol

Step 4: Calculate the concentration of lead(II) iodide using the total volume

of the solution.

Volume of solution = 100.0 ml = 0.100 L

Concentration of P bI2=Moles of P bI2

Volume of solution =0.010 mol

0.100 L = 0.100 M

Step 5: Check if a precipitate will form by comparing the Ksp value with the

ion product. The ion product (Qsp) is calculated as [Pb2+][I−]2. Since lead(II)

nitrate is the limiting reagent, the concentration of lead(II) ions is equal to the

concentration of lead(II) iodide.

Qsp = [P b2+][I−]2= (0.100 M) ×(0.100 M)2= 0.0010

Since Qsp > Ksp, a precipitate of lead(II) iodide will form.

Question 18

Question

Calculate the solubility of silver chloride (AgCl) in water at 25

C. The Ksp

(AgCl) is 1.8×10−10.

Solution

Step 1: Write the balanced equilibrium equation for the dissociation of silver

chloride into its ions:

AgCl(s)⇌Ag+(aq) + Cl−(aq)

Step 2: Define the equilibrium constant expression (Ksp) for the dissociation

of silver chloride:

Ksp = [Ag+][Cl−]

Step 3: Since the initial concentration of AgCl is assumed to be 0 (since it

is a solid), let x be the molar solubility of AgCl. Therefore, at equilibrium, the

concentration of Ag+and Cl−ions will be equal to x.

Step 4: Substitute the equilibrium concentrations into the Ksp expression:

Ksp =x×x=x2

Step 5: Given that Ksp = 1.8×10−10, set up and solve the equation for x:

1.8×10−10 =x2

x=p1.8×10−10

x≈1.34 ×10−5

Therefore, the solubility of silver chloride (AgCl) in water at 25

C is approx-

imately 1.34 ×10−5mol/L.

Question 19

Question

A solution contains 0.1 M Zn(NO3)2 and 0.2 M Na2S. Calculate the minimum

volume of Zn(NO3)2 solution needed to completely precipitate all of the metal

ions as ZnS (Ksp = 1 ×10−23).

Solution

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

Zn(N O3)2+Na2S→ZnS + 2N aNO3

Step 2: Determine the limiting reactant by comparing the moles of Zn(NO3)2

and Na2S. Given concentrations:

Zn(N O3)2: 0.1M

Na2S: 0.2M

Step 3: Calculate the moles of Zn(NO3)2 and Na2S. Moles of Zn(NO3)2:

Moles =concentration ×volume = 0.1M×V

Moles of Na2S:

Moles =concentration ×volume = 0.2M×V

Step 4: The mole ratio between Zn(NO3)2 and Na2S is 1:1. So, the limiting

reactant will be the one with fewer moles. Set up an equation to find the volume

of Zn(NO3)2 needed to completely react with Na2S:

0.1M×V= 0.2M×V

0.1V= 0.2V

V= 2 L

Step 5: Therefore, the minimum volume of 0.1 M Zn(NO3)2 solution needed

to completely precipitate all of the metal ions as ZnS is 2 L.

Question 20

Question

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

(NaCl). What is the concentration of silver ions in the solution at equilibrium

after the precipitation of silver chloride (AgCl)? The Ksp of AgCl is 1.8×10−10.

Solution

Step 1: Write the balanced chemical equation for the precipitation of silver

chloride:

Ag+(aq) + Cl−(aq)→AgCl(s)

Step 2: Write the expression for the solubility product constant (Ksp):

Ksp = [Ag+][Cl−]

Step 3: Let x be the concentration of silver ions at equilibrium. Since the

molar ratio of Ag+to AgCl is 1:1, the concentration of AgCl formed will also

be x.

Step 4: Use the given concentrations of silver nitrate and sodium chloride

to determine the initial concentrations of silver and chloride ions: - Initial con-

centration of Ag+= 0.2 M - Initial concentration of Cl−= 0.1 M

Step 5: Set up the ICE (Initial, Change, Equilibrium) table:

Ag+Cl−AgCl

Initial (M) 0.2 0.1 0

Change (M) −x−x+x

Equilibrium (M) 0.2−x0.1−x x

Step 6: Substitute the equilibrium concentrations into the solubility product

expression and solve for x:

1.8×10−10 = (0.2−x)(0.1−x)

Step 7: Expand the right side of the equation and solve for x:

1.8×10−10 = 0.02 −0.3x+x2

x= 1.3×10−5M

Step 8: The concentration of silver ions at equilibrium after the precipitation

of silver chloride is 1.3×10−5M.

Question 21

Question

A solution contains 0.25 M of lead(II) nitrate and 0.15 M of sodium sulfate.

Determine if precipitation will occur when the two solutions are mixed.

Solution

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

nitrate (Pb(NO3)2) and sodium sulfate (Na2SO4):

Pb(NO3)2+ Na2SO4→PbSO4+ 2NaNO3

Step 2: Determine the products of the reaction:

Pb(NO3)2+ Na2SO4→PbSO4+ 2NaNO3

Step 3: Determine the solubility of lead(II) sulfate (PbSO4) and sodium

nitrate (NaNO3): - Lead(II) sulfate is insoluble in water. - Sodium nitrate is

soluble in water.

Step 4: Determine if a precipitate will form: Since lead(II) sulfate is insolu-

ble, a precipitate will form when Pb(NO3)2and Na2SO4are mixed.

Therefore, precipitation will occur when the two solutions are mixed.

Question 22

Question

Calculate the concentration of chloride ions in a solution if 100.0 mL of 0.200

M silver nitrate solution is required to completely precipitate the chloride ions

in 0.500 L of an unknown solution with chloride ions. Assume the reaction goes

to completion and that the only chloride ions present in the unknown solution

come from sodium chloride. (Molar mass: Na = 22.99 g/mol, Cl = 35.45 g/mol)

Solution

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

nitrate (AgNO3) and sodium chloride (NaCl):

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

Step 2: Determine the number of moles of silver nitrate used:

Given: Volume of silver nitrate solution = 100.0 mL = 0.100 L Concentration

of silver nitrate solution = 0.200 M

Using the formula moles = concentration ×volume, we have:

moles of AgNO3moles of AgNO3moles of AgNO3moles of AgNO3= 0.200 mol/L ×0.100 L= 0.020 mol AgN O3

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

the number of moles of chloride ions in the solution:

From the balanced chemical equation, we see that 1 mole of AgNO3reacts

with 1 mole of NaCl to form 1 mole of AgCl. Therefore, the number of moles

of chloride ions is equal to the number of moles of silver nitrate used.

moles of Cl−= 0.020 mol

Step 4: Calculate the concentration of chloride ions in the unknown solution:

Given: Volume of unknown solution = 0.500 L

Concentration of Cl−=moles of Cl−

volume of solution =0.020 mol

0.500 L = 0.040 M

Therefore, the concentration of chloride ions in the unknown solution is 0.040

Question 23

Question

A university lab is conducting an experiment where a silver chloride precipitate

is formed by mixing solutions of silver nitrate and sodium chloride. If 50.0 mL of

0.100 M silver nitrate solution is mixed with 100.0 mL of 0.150 M sodium chlo-

ride solution, what is the mass of silver chloride precipitate formed? (Assume

complete precipitation and that the density of the solutions is 1.00 g/mL)

Solution

Step 1: Determine the limiting reactant.

Given: Volume of silver nitrate solution (VAgN O3) = 50.0 mL Molarity of

silver nitrate solution (MAgN O3) = 0.100 M Volume of sodium chloride solution

(VNaCl) = 100.0 mL Molarity of sodium chloride solution (MNaCl) = 0.150 M

First, convert the volumes of solutions to liters: VAgN O3= 50.0 mL = 0.050

LVNaCl = 100.0 mL = 0.100 L

Next, calculate the moles of each reactant: molesAgN O3=MAgN O3×

VAgNO3= 0.100 mol/L ×0.050 L = 0.005 mol molesN aCl =MNaCl ×VN aCl =

0.150 mol/L ×0.100 L = 0.015 mol

Since silver nitrate and sodium chloride react in a 1:1 ratio, the limiting

reactant will be the one that produces the least amount of product. In this

case, molesAgN O3= 0.005 mol is the limiting reactant.

Step 2: Calculate the mass of silver chloride formed.

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

Using the mole ratio from the balanced chemical equation, calculate the

moles of silver chloride precipitate formed: molesAgCl =molesAgN O3= 0.005 mol

Finally, calculate the mass of silver chloride formed: massAgCl =molesAgCl×

Molar massAgCl = 0.005 mol ×143.32 g/mol = 0.7166 g Therefore, the mass of

silver chloride precipitate formed is 0.7166 g.

Question 24

Question

Calculate the mass of silver chloride (AgCl) that can be obtained by reacting

50.0 mL of 0.200 M silver nitrate (AgNO3) with excess hydrochloric acid (HCl).

The balanced chemical equation for the reaction is:

AgNO3(aq) + HCl(aq)→AgCl(s) + HN O3(aq)

Solution

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

AgNO3(aq) + HCl(aq)→AgCl(s) + HNO3(aq)

Step 2: Determine the moles of silver nitrate (AgNO3) used. Given: Volume

of AgNO3:VAgNO3= 50.0 mL = 0.0500 L Molarity of AgNO3:MAgN O3=

0.200 M

Moles of AgNO3=MAgNO3×VAgN O3= 0.200 mol/L ×0.0500 L = 0.0100 mol

Step 3: Determine the moles of silver chloride (AgCl) formed. From the

balanced chemical equation, the mole ratio between AgN O3and AgCl is 1:1.

Therefore, 0.0100 mol of AgNO3will produce 0.0100 mol of AgCl.

Step 4: Calculate the mass of silver chloride (AgCl). Given: Molar mass of

AgCl:MAgCl = 107.87 g/mol

Mass of AgCl = Moles of AgCl ×MAgCl = 0.0100 mol ×107.87 g/mol = 1.08 g

Therefore, the mass of silver chloride that can be obtained is 1.08 grams.

Question 25

Question

Calculate the concentration of sulfate ions (SO2−

4) in a solution prepared by

mixing 50.0 mL of 0.200 M Na2SO4with 150.0 mL of 0.100 M BaCl2. Assume

complete precipitation of barium sulfate (BaSO4).

Solution

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

between Na2SO4and BaCl2to form BaSO4.

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

Step 2: Determine the limiting reagent by comparing the moles of SO2−

4ions

from each of the reactants. For Na2SO4:

moles of SO2−

4= 0.200 M ×0.0500 L = 0.0100 mol

For BaCl2:

moles of SO2−

4= 0.100 M ×0.150 L ×1 mol Na2SO4

1 mol BaSO4

= 0.0150 mol

Since we need 0.0200 mol of SO2−

4ions to precipitate all barium ions as BaSO4,

Na2SO4is the limiting reagent.

Step 3: Calculate the moles of BaSO4formed using the limiting reagent.

moles of BaSO4= 0.0100 mol

Step 4: Calculate the concentration of SO2−

4ions in the final solution.

Volume of final solution = 0.0500 L + 0.150 L = 0.200 L

Concentration of SO2−

4=0.0100 mol

0.200 L = 0.0500 M

Therefore, the concentration of sulfate ions (SO2−

4) in the final solution is

0.0500 M.

Question 26

Question

A solution contains 0.2 mol/L of barium chloride (BaCl2) and 0.3 mol/L of

sodium sulfate (Na2SO4). What mass of barium sulfate (BaSO4) will precipitate

when these two solutions are mixed together?

(Hint: The balanced chemical equation for the reaction is BaCl2+Na2SO4→

BaSO4+ 2NaCl)

Solution

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

barium chloride and sodium sulfate:

BaCl2+ Na2SO4→BaSO4+ 2NaCl

Step 2: Determine which reactant is limiting by calculating the maximum

amount of product formed by each reactant.

For barium chloride (BaCl2): 1 mol of BaCl2produces 1 mol of BaSO4,

Therefore, 0.2 mol/L of BaCl2will produce 0.2 mol/L of BaSO4.

For sodium sulfate (Na2SO4): 1 mol of Na2SO4produces 1 mol of BaSO4,

Therefore, 0.3 mol/L of Na2SO4will produce 0.3 mol/L of BaSO4.

Step 3: Since the formation of barium sulfate is limited by the amount of

barium chloride, the amount of barium sulfate formed will be 0.2 mol/L.

Step 4: Calculate the mass of barium sulfate precipitated: The molar mass

of BaSO4can be calculated as:

molar mass of BaSO4= 1×molar mass of Ba+1×molar mass of S+4×molar mass of O

The molar mass of BaSO4is 137.3 g/mol + 32.1 g/mol + (4 ×16.0 g/mol)

= 233.3 g/mol

Now, we can calculate the mass of BaSO4precipitated:

Mass = moles ×molar mass = 0.2 mol/L ×233.3 g/mol = 46.66 g

Therefore, 46.66 grams of barium sulfate will precipitate when 0.2 mol/L of

barium chloride and 0.3 mol/L of sodium sulfate are mixed together.

Question 27

Question

A chemist is analyzing the precipitation reaction between lead nitrate (Pb(NO3)2)

and potassium iodide (KI) to form lead iodide (PbI2) and potassium nitrate

(KNO3). If 200 mL of a 0.5 M lead nitrate solution is mixed with 300 mL of

a 0.3 M potassium iodide solution, what mass of lead iodide will be produced?

Assume the reaction goes to completion.

Solution

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

Pb(NO3)2+ 2KI →PbI2+ 2KNO3

Step 2: Determine the limiting reactant by comparing the number of moles

of each reactant.

For lead nitrate: Number of moles = concentration ×volume Number of

moles = 0.5 M ×0.2 L Number of moles = 0.1 mol

For potassium iodide: Number of moles = concentration ×volume Number

of moles = 0.3 M ×0.3 L Number of moles = 0.09 mol

Since 2 moles of potassium iodide are needed for every mole of lead nitrate,

potassium iodide is the limiting reactant.

Step 3: Calculate the theoretical yield of lead iodide using the limiting re-

actant.

Number of moles of lead iodide = 0.09 mol ×1 mol PbI2

2 mol KI Number of moles of

lead iodide = 0.045 mol PbI2

Step 4: Calculate the mass of lead iodide produced using its molar mass.

Molar mass of lead iodide (PbI2) = 207.2 g/mol + 2 ×126.9 g/mol Molar

mass of lead iodide = 459 g/mol

Mass of lead iodide = 0.045 mol ×459 g/mol Mass of lead iodide = 20.67 g

Therefore, the mass of lead iodide produced is 20.67 g.

Question 28

Question

Calculate the solubility of silver chloride in a 0.10 M solution of sodium chloride

at 25

C. The Ksp of silver chloride is 1.77 ×10−10.

Solution

Step 1: Write the solubility equilibrium equation for silver chloride: The solu-

bility equilibrium equation for silver chloride is:

AgCl(s)⇌Ag+(aq) + Cl−(aq)

Step 2: Write the equilibrium expression: The equilibrium expression for

the dissociation of silver chloride is:

Ksp = [Ag+][Cl−]

Step 3: Set up an ICE table for the dissociation of silver chloride: Let the

solubility of silver chloride be represented as x.

Ag+Cl−

Initial (M) 0 0.10

Change (M) +x+x

Equilibrium (M) x0.10 + x

Step 4: Substitute the equilibrium concentrations into the equilibrium ex-

pression and solve for x: Substitute the equilibrium concentrations into the

equilibrium expression:

Ksp =x(0.10 + x)

Given that Ksp = 1.77 ×10−10, solve for x:

1.77 ×10−10 =x(0.10 + x)

1.77 ×10−10 = 0.10x+x2

x2+ 0.10x−1.77 ×10−10 = 0

Solve the quadratic equation to find the value of x.

Step 5: Calculate the solubility of silver chloride: After solving the quadratic

equation, determine the value of xrepresenting the solubility of silver chloride.

Remember to check if the assumption 0.10 + x≈0.10 holds true.

Therefore, the solubility of silver chloride in a 0.10 M solution of sodium

chloride at 25

C is the calculated value of x.

Question 29

Question

A chemistry student is conducting an experiment where two solutions are mixed

together to form a precipitate. If 50.0 mL of a 0.200 M solution of calcium

chloride is mixed with 75.0 mL of a 0.150 M solution of sodium carbonate,

what mass of calcium carbonate will be formed? (Assume the reaction goes to

completion and that the volume of the solutions is additive.)

Solution

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

chloride (CaCl2) and sodium carbonate (Na2CO3):

CaCl2(aq)+Na2CO3(aq)→CaCO3(s)+ 2N aCl(aq)

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

of each reactant using the given concentrations and volumes: For calcium chlo-

ride (CaCl2):

nCaCl2=CCaCl2×VCaCl2

nCaCl2= 0.200 M ×0.0500 L

nCaCl2= 0.0100 mol

For sodium carbonate (Na2CO3):

nNa2CO3=CN a2CO3×VN a2CO3

nNa2CO3= 0.150 M ×0.0750 L

nNa2CO3= 0.0113 mol

Since calcium chloride is the limiting reactant (0.0100 mol), the reaction will

consume all 0.0100 mol of CaCl2.

Step 3: Use stoichiometry to calculate the moles of calcium carbonate formed:

From the balanced equation, 1 mol of CaCl2produces 1 mol of CaCO3. So,

0.0100 mol of CaCl2will produce 0.0100 mol of CaCO3.

Step 4: Calculate the mass of calcium carbonate formed: Using the molar

mass of CaCO3(100.09 g/mol):

mCaCO3=nCaCO3×Molar mass of CaCO3

mCaCO3= 0.0100 mol ×100.09 g/mol

mCaCO3= 1.00 g

Therefore, the mass of calcium carbonate formed is 1.00 g.

Question 30

Question

Calculate the concentration of sulfate ion (SO2−

4) in a solution if 45.0 mL of

1.25 M barium chloride (BaCl2) is required to completely precipitate the sulfate

ion from the solution. Assume the reaction goes to completion. The molecular

weight of BaCl2is 208.23 g/mol.

Solution

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

between barium chloride and sulfate ion. Step 2: Determine the mole ratio

between barium chloride and sulfate ion. Step 3: Calculate the moles of sulfate

ion in the solution. Step 4: Determine the volume of the solution to find the

concentration of sulfate ion.

Step 1: The balanced chemical equation for the precipitation reaction is:

BaCl2+ SO2−

4→BaSO4↓+2Cl−

Step 2: From the balanced chemical equation, the mole ratio between bar-

ium chloride and sulfate ion is 1:1. This means 1 mole of barium chloride reacts

with 1 mole of sulfate ion.

Step 3: - Given that 45.0 mL of 1.25 M barium chloride is used, calculate

the moles of barium chloride:

Moles of BaCl2= Volume (L)×Molarity = 0.0450 L×1.25 mol/L = 0.0563 mol

- Since the mole ratio between barium chloride and sulfate ion is 1:1, the moles

of sulfate ion are also 0.0563 mol.

Step 4: - Calculate the volume of the solution by dividing the moles of

sulfate ion by its concentration:

Volume (L) = Moles

Molarity =0.0563 mol

0.0450 L = 1.25 M

Therefore, the concentration of sulfate ion in the solution is 1.25 M.

Question 31

Question

A solution is prepared by mixing 50.0 mL of 0.200 M silver nitrate with 75.0 mL

of 0.150 M sodium chloride. Assuming complete reaction, 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 limiting reactant. First, calculate the moles of silver

nitrate and sodium chloride:

Moles of AgNO3= Volume ×Molarity

= 0.0500 L ×0.200 mol/L

= 0.0100 mol

Moles of NaCl = 0.0750 L ×0.150 mol/L

= 0.0113 mol

Since the AgNO3:NaCl mole ratio is 1:1, sodium chloride is the limiting

reactant because it forms less moles of product.

Step 3: Calculate the mass of silver chloride precipitated. The molar mass

of AgCl is 143.32 g/mol.

Mass of AgCl = Moles of AgCl ×Molar mass of AgCl

= 0.0113 mol ×143.32 g/mol

= 1.62 g

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

Question 32

Question

A student is conducting an experiment to determine the concentration of chlo-

ride ions in a water sample. The student adds an excess of silver nitrate solution

to 100.0 mL of the water sample to precipitate the chloride ions as silver chlo-

ride. After collecting and drying the precipitate, the student finds that it weighs

0.569 g.

Calculate the concentration of chloride ions in the original water sample in

units of mol/L.

(Note: The molar mass of AgCl is 143.32 g/mol.)

Solution

Step 1: Calculate the moles of silver chloride precipitate.

Moles of AgCl = Mass of AgCl

Molar mass of AgCl

Moles of AgCl = 0.569 g

143.32 g/mol = 0.00397 mol

Step 2: Use the stoichiometry of the reaction to determine the moles of

chloride ions. The balanced chemical equation for the reaction is:

Ag++ Cl−→AgCl(s)

Since 1 mole of silver chloride is formed per mole of chloride ions, the moles

of chloride ions is also 0.00397 mol.

Step 3: Calculate the concentration of chloride ions in the original water

sample.

Volume of water sample = 100.0 mL = 0.1000 L

Concentration of chloride ions = Moles of Cl−

Volume of solution in L

Concentration of chloride ions = 0.00397 mol

0.1000 L = 0.0397 mol/L

Therefore, the concentration of chloride ions in the original water sample is

0.0397 mol/L.

Question 33

Question

A solution contains 0.025 M of calcium chloride and 0.015 M of sodium sulfate.

Determine whether a precipitate will form when the two solutions are mixed,

and if so, calculate the mass of the precipitate formed.

Solution

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

calcium chloride and sodium sulfate:

CaCl2+ Na2SO4→CaSO4+ 2NaCl

Step 2: Determine the possible products of the reaction: The possible prod-

ucts are calcium sulfate (CaSO) and sodium chloride (NaCl).

Step 3: Determine the solubility of the products: - Calcium sulfate is slightly

soluble in water. - Sodium chloride is highly soluble in water.

Step 4: Determine if a precipitate will form: Since calcium sulfate is slightly

soluble, it will form a precipitate when calcium chloride and sodium sulfate are

mixed.

Step 5: Write the net ionic equation for the reaction:

Ca2+ + SO2−

4→CaSO4

Step 6: Determine the limiting reactant: - From the balanced chemical

equation, we see that 1 mole of calcium chloride reacts with 1 mole of sodium

sulfate to produce 1 mole of calcium sulfate. - Calculate the moles of cal-

cium chloride: 0.025 M ×volume in L - Calculate the moles of sodium sulfate:

0.015 M ×volume in L - Identify the limiting reactant.

Step 7: Calculate the mass of calcium sulfate precipitate formed: - Once

the limiting reactant is determined, use the mole ratio in the balanced chemical

equation to calculate the moles of calcium sulfate formed. - Convert the moles

of calcium sulfate to grams using the molar mass of calcium sulfate.

This calculation will provide the mass of the precipitate formed when calcium

chloride and sodium sulfate solutions are mixed.

Question 34

Question

A solution is prepared by dissolving 5.00 g of silver nitrate in 100.0 mL of

water at 25

C. This solution was added dropwise to a solution containing 4.00

g of potassium iodide in 150.0 mL of water at 25

C. Calculate the mass of the

precipitate formed.

(Hint: The reaction that occurs is the double displacement reaction between

silver nitrate and potassium iodide to form silver iodide.)

Solution

Step 1: Find the moles of each reactant. The molar mass of silver nitrate

(AgNO3) is 169.87 g/mol. The molar mass of potassium iodide (KI) is 166.00

g/mol.

Moles of silver nitrate:

moles = mass

molar mass =5.00 g

169.87 g/mol = 0.0294 mol

Moles of potassium iodide:

moles = mass

molar mass =4.00 g

166.00 g/mol = 0.0241 mol

Step 2: Determine the limiting reactant. The balanced chemical equation

for the reaction is:

AgNO3+KI →AgI +KN O3

From the balanced equation, 1 mol of AgNO3reacts with 1 mol of KI to

produce 1 mol of AgI. Therefore, the moles ratio of AgN O3to KI is 1:1.

Since the moles ratio is 1:1, the limiting reactant is the one with the smaller

number of moles, which is potassium iodide.

Step 3: Calculate the mass of silver iodide precipitate. The molar mass of

silver iodide (AgI) is 234.77 g/mol. The theoretical yield of silver iodide can be

calculated using the moles of KI:

mass = moles ×molar mass = 0.0241 mol ×234.77 g/mol = 5.66 g

Therefore, the mass of the precipitate formed is 5.66 g.

Question 35

Question

Calculate the mass of barium sulfate (BaSO4) that can be formed by reacting

175 mL of 0.300 M barium chloride (BaCl2) with excess sulfuric acid (H2SO4).

The balanced chemical equation for the reaction is:

BaCl2+ H2SO4→BaSO4+ 2HCl

Solution

Step 1: Determine the moles of barium chloride. Given: Volume of BaCl2=

175 mL = 0.175 L Molarity of BaCl2= 0.300 M

Using the formula:

Molarity = moles

volume in liters

we can rearrange it to solve for moles:

moles = Molarity ×volume in liters

Plugging in the values:

moles of BaCl2= 0.300 M ×0.175 L = 0.0525 moles

Step 2: Find the limiting reactant. From the balanced chemical equation,

we see that the mole ratio between BaCl2and BaSO4is 1:1. Therefore, since

there are 0.0525 moles of BaCl2, this will also be the moles of BaSO4formed.

Step 3: Calculate the mass of barium sulfate. The molar mass of BaSO4can

be calculated as:

molar mass of BaSO4= molar mass of Ba+molar mass of S+4×molar mass of O

= 137.33 g/mol + 32.07 g/mol + 4 ×16.00 g/mol = 233.33 g/mol

Finally, the mass of barium sulfate can be calculated using the formula:

mass = moles ×molar mass

mass = 0.0525 moles ×233.33 g/mol = 12.26 g

Therefore, the mass of barium sulfate that can be formed is 12.26 grams.

CHEM 122 - GENERAL CHEMISTRY II Precipitation calculations Question Bank Set 9

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