CHEM 122 - GENERAL CHEMISTRY
II - Precipitation calculations
Question Bank - Set 6
Liberty University
Question 1
Question
Calculate the mass of lead(II) iodide (PbI2) that can be formed by mixing 75.0
mL of 0.200 M lead(II) nitrate (Pb(NO3)2) with excess potassium iodide at
standard temperature and pressure. (Assume the reaction goes to completion
and the molar mass of lead(II) iodide is 461.01 g/mol)
Solution
Step 1: Write the balanced chemical equation for the reaction between lead(II)
nitrate and potassium iodide to form lead(II) iodide.
2Pb(NO3)2+ 4KI →2PbI2+ 4KNO3
Step 2: Calculate the moles of lead(II) nitrate used in the reaction. Given:
Volume of lead(II) nitrate solution, V= 75.0 mL = 0.0750 L Concentration of
lead(II) nitrate solution, C= 0.200 M
Using the formula n=C×V, we have nPb(NO3)2= 0.200 mol/L×0.0750 L =
0.0150 mol
Step 3: Determine the limiting reactant to find the moles of lead(II) iodide
formed. We need to find the moles of lead(II) iodide formed from the moles of
lead(II) nitrate.
From the balanced chemical equation, we see that 2 moles of lead(II) nitrate
produce 2 moles of lead(II) iodide. Thus, 0.0150 mol of lead(II) nitrate will
produce 0.0150 mol of lead(II) iodide.
Step 4: Calculate the mass of lead(II) iodide formed. Given molar mass of
lead(II) iodide, M= 461.01 g/mol
Using the formula m=n×M, we have mPbI2= 0.0150 mol×461.01 g/mol =
6.915 g
Therefore, the mass of lead(II) iodide that can be formed is 6.915 g.
Question 2
Question
Calculate the concentration of sulfate ions in a solution when 50 mL of 0.2
M barium sulfate solution is mixed with excess sodium sulfate solution. The
resulting barium sulfate precipitate is then filtered off and found to have a mass
of 0.345 g. (Molar mass of BaSO4= 233.4 g/mol)
Solution
Step 1: Determine the moles of barium sulfate.
Given: Volume of barium sulfate solution, VBaSO4= 50 mL = 0.05 L Molar-
ity of barium sulfate solution, MBaSO4= 0.2 M
The moles of barium sulfate can be calculated using the formula:
moles of BaSO4=MBaSO4×VBaSO4
moles of BaSO4= 0.2 mol/L ×0.05 L = 0.01 mol
Therefore, there are 0.01 moles of barium sulfate present.
Step 2: Calculate the moles of sulfate ions.
Since barium sulfate reacts in a 1:1 molar ratio with sulfate ions, the moles
of sulfate ions present is also 0.01 moles.
Step 3: Calculate the total volume of solution.
Since the sodium sulfate solution is in excess, the volume of the mixture is
the sum of the volumes of both solutions. Assuming the volume of the sodium
sulfate solution is large compared to 50 mL, we can consider the total volume
to be 50 mL + the volume of sodium sulfate solution.
Step 4: Calculate the concentration of sulfate ions.
The concentration of sulfate ions can be calculated using the formula:
Concentration of sulfate ions = moles of sulfate ions
total volume of solution
Since the moles of sulfate ions is 0.01 and the total volume is 0.05 L:
Concentration of sulfate ions = 0.01 mol
0.05 L = 0.2 M
Therefore, the concentration of sulfate ions in the solution is 0.2 M.
Question 3
Question
A solution contains 0.2 mol/L of lead(II) nitrate (Pb(NO3)2) and 0.1 mol/L of
sodium sulfate (Na2SO4). Determine the maximum amount of lead(II) sulfate
(PbSO4) that can precipitate out in grams from a 500 mL solution.
2
(Hint: The solubility product constant for lead(II) sulfate is 1.2×10−8at
25
°
C)
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between lead(II) nitrate and sodium sulfate:
Pb(NO3)2+ Na2SO4→PbSO4+ 2NaNO3
Step 2: Calculate the number of moles of lead(II) nitrate and sodium sulfate
that will react. Since the volume of the solution is 500 mL, the moles of each
substance can be calculated:
Moles of Pb(NO3)2= 0.2 mol/L ×0.5 L = 0.1 mol
Moles of Na2SO4= 0.1 mol/L ×0.5 L = 0.05 mol
Step 3: Determine the limiting reactant in the reaction based on the sto-
ichiometry of the balanced chemical equation. Since the stoichiometry is 1:1,
the limiting reactant is the one with the lower number of moles. Here, sodium
sulfate is the limiting reactant as it has fewer moles.
Step 4: Calculate the number of moles of lead(II) sulfate that can be formed
using the limiting reactant (sodium sulfate):
Moles of PbSO4= 0.05 mol
Step 5: Calculate the mass of lead(II) sulfate formed using its molar mass:
Molar mass of PbSO4= 207.2 g/mol + 32.1 g/mol + 4(16 g/mol) = 303.3 g/mol
Mass of PbSO4= 0.05 mol ×303.3 g/mol = 15.165 g
Therefore, the maximum amount of lead(II) sulfate that can precipitate out
from the solution is 15.165 grams.
Question 4
Question
A chemical reaction takes place in a beaker where 200 mL of a 0.1 M solution of
lead(II) nitrate (Pb(NO3)2) is mixed with 100 mL of a 0.2 M solution of sodium
iodide (NaI).
Calculate the maximum amount of lead(II) iodide (PbI2) that can precipitate
in grams. (Assume the reaction goes to completion and that the density of the
solutions is 1 g/mL.)
3
Solution
Step 1: Write the balanced chemical equation for the reaction between lead(II)
nitrate (Pb(NO3)2) and sodium iodide (NaI):
Pb(NO3)2+ 2NaI →PbI2+ 2NaNO3
Step 2: Calculate the moles of lead(II) nitrate (Pb(NO3)2) and sodium iodide
(NaI) used: For lead(II) nitrate:
Moles Pb(NO3)2= Volume ×Molarity = 0.2 L ×0.1 mol/L = 0.02 mol
For sodium iodide:
Moles NaI = Volume ×Molarity = 0.1 L ×0.2 mol/L = 0.02 mol
Step 3: Determine the limiting reactant. The limiting reactant in this case
is lead(II) nitrate (Pb(NO3)2) since it is completely consumed when 0.02 mol
of sodium iodide is used.
Step 4: Calculate the theoretical yield of lead(II) iodide (PbI2) that can be
produced:
Moles PbI2= Moles Pb(NO3)2= 0.02 mol
Step 5: Calculate the molar mass of lead(II) iodide (PbI2):
Molar mass of PbI2= Atomic mass of Pb+2×Atomic mass of I = 207.2 g/mol+2×126.9 g/mol = 460 g/mol
Step 6: Convert moles of lead(II) iodide to grams:
Mass of PbI2= Moles PbI2×Molar mass of PbI2= 0.02 mol×460 g/mol = 9.2 g
Therefore, the maximum amount of lead(II) iodide that can precipitate is
9.2 grams.
Question 5
Question
Calculate the solubility of lead(II) iodide (PbI2) in a 0.25 M potassium iodide
(KI) solution at 25◦C. The solubility product constant of lead(II) iodide is 7.1×
10−9.
Solution
Step 1: Write the equilibrium expression for the dissolution of lead(II) iodide.
The equilibrium expression for the dissolution of PbI2is:
PbI2(s)⇌Pb2+(aq) + 2I−(aq)
4
Step 2: Define the variables. Let xbe the molar solubility of lead(II) iodide
in mol/L.
Step 3: Write the equilibrium constant expression. The solubility product
constant (Ksp) expression for PbI2is:
Ksp = [Pb2+][I−]2
Substitute the expressions for the ions in terms of x:
Ksp =x×(2x)2= 4x3
Step 4: Substitute known values. Given that Ksp = 7.1×10−9, substitute
this into the expression:
7.1×10−9= 4x3
Step 5: Solve for x.
x=3
r7.1×10−9
4= 0.00061 mol/L
Therefore, the solubility of lead(II) iodide in a 0.25 M potassium iodide
solution at 25◦C is 0.00061 mol/L.
Question 6
Question
A sample of water contains 2.5×10−3M of calcium ions (Ca2+). If CaCO3is
added to the solution until the calcium ion concentration is 1.5×10−3M, how
many grams of CaCO3were added? (Assume the only source of CaCO3in the
solution is from the added solid.)
Given: Ksp of CaCO3= 4.8×10−9.
Solution
Step 1: Write the balanced equation for the dissociation of CaCO3:
CaCO3→Ca2+ + CO2−
3
Step 2: Write the equilibrium constant expression based on the balanced
equation:
Ksp = [Ca2+][CO2−
3]
Step 3: Calculate the initial concentration of CO2−
3from the initial con-
centration of Ca2+: Given: [Ca2+] = 2.5×10−3M. Since 1 mole of CaCO3
produces 1 mole of Ca2+ and 1 mole of CO2−
3, initial [CO2−
3]=2.5×10−3M.
Step 4: Calculate the final concentration of CO2−
3using the concentration
of Ca2+ after CaCO3is added: Given: new [Ca2+]=1.5×10−3M. Since the
5
initial [Ca2+] = [CO2−
3] before the CaCO3is added, the decrease in [Ca2+] is
equal to the increase in [CO2−
3: [CO2−
3]=2.5×10−3+(2.5×10−3−1.5×10−3) =
3.5×10−3M.
Step 5: Use the equilibrium constant expression to determine the concen-
tration of Ca2+ once all the CaCO3has dissolved: 4.8×10−9= (1.5×10−3)×
(3.5×10−3)
Step 6: Calculate the number of moles of CaCO3added: Since 1 mole of
CaCO3produces 1 mole of Ca2+, moles of CaCO3added = 1.5×10−3moles.
Step 7: Calculate the molar mass of CaCO3(mass of Ca + 1 mol of C + 3
mol of O): 40.08 g/mol + 12.01 g/mol + (3 ×16.00 g/mol) = 100.08 g/mol
Step 8: Calculate the mass of CaCO3added: mass = moles ×molar mass =
1.5×10−3×100.08 = 0.15012 g
Therefore, 0.15012 grams of CaCO3were added to the solution.
Question 7
Question
In a precipitation reaction, 50.0 mL of 0.100 M of silver nitrate (AgNO3) is
mixed with 25.0 mL of 0.200 M sodium chloride (NaCl) solution. What is
the molarity of silver ions (Ag+) after the reaction has occurred? 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 precipitation reaction
between silver nitrate and sodium chloride:
AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)
Step 2: Determine the limiting reactant by calculating the moles of each
reactant: For silver nitrate:
Moles of AgNO3= Volume ×Molarity
= (50.0×10−3L) ×0.100 M
= 5.00 ×10−3mol
For sodium chloride:
Moles of NaCl = Volume ×Molarity
= (25.0×10−3L) ×0.200 M
= 5.00 ×10−3mol
Since both reactants result in the same number of moles of product, silver
nitrate is the limiting reactant.
6
Step 3: Calculate the moles of silver ions formed: From the balanced chemi-
cal equation, every mole of silver nitrate produces one mole of silver ions. Thus,
the moles of silver ions formed is also 5.00 ×10−3mol.
Step 4: Calculate the new volume of the solution after the reaction: The
total volume of the solution after mixing the two solutions is 50.0×10−3L +
25.0×10−3L = 75.0×10−3L.
Step 5: Calculate the molarity of silver ions:
Molarity of Ag+=Moles of Ag+
Volume
=5.00 ×10−3mol
75.0×10−3L
= 0.067 M
Therefore, the molarity of silver ions (Ag+) after the reaction is 0.067 M.
Question 8
Question
A 250 mL solution contains 0.15 M barium chloride (BaCl2) and 0.20 M sodium
sulfate (Na2SO4). Calculate the concentrations of Ba2+ and SO2−
4ions after
precipitation reaction occurs between barium chloride and sodium sulfate.
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between BaCl2and Na2SO4.
BaCl2(aq) + Na2SO4(aq)→BaSO4(s) + 2NaCl(aq)
Step 2: Determine the limiting reactant and calculate the amount of BaSO4
formed. Since BaCl2and Na2SO4react in a 1:1 molar ratio, the limiting reactant
will be the one that runs out first. To find the limiting reactant, we compare
the moles of each reactant:
Moles of BaCl2: 0.15 M ×0.25 L = 0.0375 mol
Moles of Na2SO4: 0.20 M ×0.25 L = 0.05 mol
Since BaCl2is the limiting reactant (having fewer moles), the amount of BaSO4
formed will be 0.0375 mol.
Step 3: Calculate the concentration of Ba2+ ions after precipitation. Ini-
tially, the concentration of Ba2+ ions was 0.15 M, but since 0.0375 mol of BaCl2
precipitated, the new concentration is:
0.15 mol
0.25 L = 0.6 M Ba2+
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Step 4: Calculate the concentration of SO2−
4ions after precipitation. Since
the reaction went to completion, all Na2SO4reacts to form NaCl and BaSO4.
The SO2−
4ions come from the Na2SO4, so the concentration will be:
0.05 mol
0.25 L = 0.2 M SO2−
4
Question 9
Question
A chemist is performing a precipitation reaction by mixing 50.0 mL of a 0.200
M lead(II) nitrate solution with 75.0 mL of a 0.150 M sodium iodide solution.
If lead(II) iodide is formed as a precipitate, calculate the mass of lead(II) iodide
that forms.
(Note: The density of water is 1.00 g/mL.)
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between lead(II) nitrate and sodium iodide:
P b(NO3)2+ 2NaI →P bI2+ 2NaNO3
Step 2: Determine the limiting reactant by using the stoichiometry of the
balanced equation.
From the equation: 1 mole of lead(II) nitrate reacts with 2 moles of sodium
iodide to produce 1 mole of lead(II) iodide.
Calculate the moles of lead(II) nitrate and sodium iodide:
moles of Pb(NO3)2= 0.050 L ×0.200 mol/L = 0.0100 mol
moles of NaI = 0.075 L ×0.150 mol/L = 0.0113 mol
Since the reaction equation is 1:2 for Pb(NO3)2and NaI, NaI is the limiting
reactant.
Step 3: Calculate the theoretical yield of lead(II) iodide based on the limiting
reactant.
Using the mole ratio from the balanced equation, the moles of lead(II) iodide
formed will be half of the moles of sodium iodide:
moles of PbI2= 0.0113 mol ×1
2= 0.00565 mol
Step 4: Calculate the mass of lead(II) iodide formed using its molar mass.
The molar mass of lead(II) iodide (PbI2) is:
207.2 g/mol for Pb + 2 ×126.9 g/mol for I = 460.1 g/mol
Therefore, the mass of lead(II) iodide formed is:
0.00565 mol ×460.1 g/mol = 2.60 g
8
Question 10
Question
Calculate the mass of lead(II) iodide (PbI2) that can be precipitated when 50.0
mL of 0.200 M lead(II) nitrate (Pb(NO3)2) is mixed with 50.0 mL of 0.150
M potassium iodide (KI). (Assume the reaction goes to completion and that
lead(II) iodide is insoluble.)
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction be-
tween lead(II) nitrate and potassium iodide to form lead(II) iodide and potas-
sium nitrate.
Pb(NO3)2(aq) + 2KI(aq)→PbI2(s) + 2KNO3(aq)
Step 2: Determine the limiting reactant. Calculate the moles of each reactant
using the formula n= M ×V. For lead(II) nitrate:
nPb(NO3)2= 0.200 M ×0.0500 L = 0.0100 mol
For potassium iodide:
nKI = 0.150 M ×0.0500 L = 0.00750 mol
Since lead(II) nitrate produces 0.0100 mol of lead(II) iodide and potassium
iodide produces 0.00750 mol of lead(II) iodide, potassium iodide is the limiting
reactant.
Step 3: Calculate the mass of lead(II) iodide precipitated. The molar mass of
lead(II) iodide (PbI2) is approximately 461 g/mol. The amount of lead(II) iodide
formed will be 0.00750 mol. Therefore, the mass of lead(II) iodide precipitated
is:
Mass = moles ×molar mass = 0.00750 mol ×461 g/mol = 3.46 g
Thus, approximately 3.46 grams of lead(II) iodide can be precipitated.
Question 11
Question
Calculate the solubility of silver chromate (Ag2CrO4) in a 0.10 M solution of
silver nitrate (AgN O3). The Ksp of silver chromate is 1.1×10−12.
9
Solution
Step 1: Write the balanced dissociation equation for silver chromate. The dis-
sociation equation for silver chromate is:
Ag2CrO4(s)⇌2Ag+(aq) + CrO2−
4(aq)
Step 2: Write the expression for the solubility product constant, Ksp. The
solubility product constant, Ksp, is the product of the concentrations of the ions
raised to their stoichiometric coefficients. Therefore,
Ksp = [Ag+]2[CrO2−
4]
Substitute the given Ksp value into the equation:
1.1×10−12 = (2x)2(x)
Step 3: Solve for the value of x.
4x3= 1.1×10−12
x3=1.1×10−12
4
x≈5.23 ×10−5
Step 4: Calculate the solubility of silver chromate in the solution. The
solubility of silver chromate is equal to the concentration of CrO2−
4ions, which
is approximately 5.23 ×10−5M.
Question 12
Question
Calculate the solubility of silver chloride (AgCl) in water at 25◦C. The Ksp of
AgCl is 1.8×10−10.
Solution
Step 1: Write the equation for the dissociation of silver chloride in water:
AgCl(s)⇌Ag+(aq) + Cl−(aq)
Step 2: Write the expression for the equilibrium constant Ksp:
Ksp = [Ag+][Cl−]
Step 3: Let xbe the solubility of AgCl in mol/L. Since 1 mol of AgCl
dissociates into 1 mol of Ag+and 1 mol of Cl−, we have:
[Ag+] = xmol/L
10
[Cl−] = xmol/L
Step 4: Substitute the concentrations into the Ksp expression:
Ksp = (x)(x)
1.8×10−10 =x2
Step 5: Solve for x:
x=p1.8×10−10
x= 1.34 ×10−5mol/L
Therefore, the solubility of silver chloride in water at 25◦C is 1.34 ×10−5
mol/L.
Question 13
Question
A chemistry student is performing an experiment that involves the precipitation
of lead chloride (PbCl2) from a solution of lead nitrate (Pb(NO3)2) and sodium
chloride (NaCl). If 25.0 mL of 0.25 M lead nitrate solution is mixed with excess
sodium chloride solution, how many grams of lead chloride will precipitate?
(Assume the reaction goes to completion)
Solution
Step 1: Write out the balanced chemical equation for the precipitation reaction:
Pb(NO3)2+ 2NaCl →PbCl2+ 2N aN O3
Step 2: Calculate the moles of lead nitrate:
Moles of Pb(NO3)2= Volume (L) ×Molarity
Moles of Pb(NO3)2= 0.025 L ×0.25 mol/L = 0.00625 mol
Step 3: Determine the limiting reactant. Since sodium chloride is in excess,
lead nitrate is the limiting reactant. This means all of the lead nitrate will react.
Step 4: Use the mole ratio from the balanced equation to find the moles of
lead chloride formed:
Moles of PbCl2= Moles of Pb(NO3)2×1 mol PbCl2
1 mol Pb(N O3)2
Moles of PbCl2= 0.00625 mol ×1 mol PbCl2
1 mol Pb(N O3)2
= 0.00625 mol
Step 5: Calculate the mass of lead chloride precipitated:
Mass of PbCl2= Moles of PbCl2×Molar mass of PbCl2
Molar mass of PbCl2= 207.2 g/mol + 2(35.5 g/mol) = 278.2 g/mol
Mass of PbCl2= 0.00625 mol ×278.2 g/mol = 1.738 g
Therefore, 1.738 grams of lead chloride will precipitate in the reaction.
11
Question 14
Question
Determine the concentration of sulfate ions in a solution that forms a white pre-
cipitate with barium chloride in a 25.0 mL sample, where 0.500 g of precipitate
was formed. The precipitation reaction is:
BaCl2(aq) + Na2SO4(aq)→BaSO4(s) + 2NaCl(aq)
(The molar mass of BaSO4 is 233.39 g/mol)
Solution
Step 1: Calculate the moles of BaSO4 formed. Step 2: Use the stoichiometry of
the reaction to find the moles of sulfate ions. Step 3: Calculate the concentration
of sulfate ions in the solution.
Step 1: The molar mass of BaSO4 is 233.39 g/mol. Given that 0.500 g of
BaSO4 precipitate was formed:
Moles of BaSO4=0.500 g
233.39 g/mol = 0.002141 mol
Step 2: From the balanced equation, 1 mole of BaSO4 is formed for every
mole of Na2SO4. Therefore, the moles of sulfate ions present in the solution is
also 0.002141 mol.
Step 3: The volume of the sample is 25.0 mL. Converting this volume to
liters:
Volume in Liters = 25.0 mL ×1 L
1000 mL = 0.0250 L
The concentration of sulfate ions is given by:
Concentration (mol/L) = Moles
Volume (L) =0.002141 mol
0.0250 L = 0.0856 mol/L
Therefore, the concentration of sulfate ions in the solution is 0.0856 mol/L.
Question 15
Question
Calculate the concentration of iodide ions (I−) in a solution when 25.0 mL of
0.020 M lead (II) iodide (P bI2) is mixed with 45.0 mL of 0.015 M potassium
iodide (KI). Assume the reaction goes to completion.
12
Solution
Step 1: Write the balanced chemical equation for the reaction between lead (II)
iodide and potassium iodide:
P bI2+ 2KI →P bI2(s)+2K++ 2I−
Step 2: Determine the limiting reactant to calculate the amount of iodide
ions produced. To find the limiting reactant, we compare the moles of lead (II)
iodide and potassium iodide present: - Moles of P bI2: 0.0250 L ×0.020 mol/L =
0.0005 mol - Moles of KI: 0.0450 L ×0.015 mol/L = 0.000675 mol
Since lead (II) iodide (0.0005 mol) is less than potassium iodide (0.000675
mol), lead (II) iodide is the limiting reactant.
Step 3: Use stoichiometry to determine the number of moles of iodide ions
produced from lead (II) iodide: From the balanced equation, 1 mole of P bI2pro-
duces 2 moles of I−. So, moles of I−produced = 0.0005 mol P bI2×2 mol I−
1 mol P bI2=
0.0010 mol I−
Step 4: Calculate the concentration of iodide ions in the solution: Total
volume of solution = 25.0 mL + 45.0 mL = 70.0 mL = 0.0700 L Concentration
of iodide ions (I−) = 0.0010 mol
0.0700 L = 0.0143 M
Therefore, the concentration of iodide ions in the solution is 0.0143 M.
Question 16
Question
A scientist is studying the precipitation levels in a certain region. The average
annual precipitation in the region is 85 inches. If the precipitation data for the
past 5 years are as follows: 90 inches, 80 inches, 95 inches, 88 inches, and 82
inches, calculate the coefficient of variation for the precipitation levels in this
region.
Solution
Step 1: Find the mean precipitation level. The mean precipitation level is
calculated by finding the average of the precipitation data.
Mean = 90 + 80 + 95 + 88 + 82
5=435
5= 87 inches
Step 2: Find the standard deviation of the precipitation levels. First, find
the squared differences between each precipitation level and the mean:
(90−87)2= 9,(80−87)2= 49,(95−87)2= 64,(88−87)2= 1,(82−87)2= 25
Next, find the variance by calculating the average of these squared differ-
ences:
Variance = 9 + 49 + 64 + 1 + 25
5=148
5= 29.6
13
Finally, find the standard deviation by taking the square root of the variance:
Standard Deviation = √29.6≈5.44
Step 3: Calculate the coefficient of variation (CV). The coefficient of vari-
ation is calculated by dividing the standard deviation by the mean and then
multiplying by 100%:
CV = 5.44
87 ×100% ≈6.25%
Therefore, the coefficient of variation for the precipitation levels in this region
is approximately 6.25%.
Question 17
Question
A solution contains 0.1 mol/L of silver nitrate (AgNO3) and 0.2 mol/L of potas-
sium chloride (KCl). What mass of silver chloride (AgCl) will precipitate when
the two solutions are mixed together?
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction:
AgNO3(aq) + KCl(aq)→AgCl(s) + KNO3(aq)
Step 2: Determine the limiting reactant by finding the number of moles of
each reactant available. We will assume we have 1 L of each solution to make
calculations easier. For silver nitrate (AgNO3): Number of moles = concentra-
tion ×volume = 0.1 mol/L ×1 L = 0.1 mol For potassium chloride (KCl):
Number of moles = concentration ×volume = 0.2 mol/L ×1 L = 0.2 mol
Step 3: Determine the limiting reactant by looking at the stoichiometry of
the reaction. Since AgNO3and KCl react in a 1:1 molar ratio, AgNO3is the
limiting reactant.
Step 4: Calculate the theoretical yield of AgCl precipitate in grams. To find
the mass of AgCl, we need to use the molar mass of AgCl which is 107.87 +
35.45 = 143.32 g/mol. Number of moles of AgCl = number of moles of limiting
reactant = 0.1 mol Mass of AgCl = number of moles ×molar mass = 0.1 mol
×143.32 g/mol = 14.332 g
Therefore, when the two solutions are mixed together, 14.332 g of AgCl will
precipitate.
14
Question 18
Question
A chemistry student is conducting an experiment that involves the precipitation
of silver chloride (AgCl). They mix 50.0 mL of 0.100 M silver nitrate (AgNO3)
with 50.0 mL of 0.200 M sodium chloride (NaCl). Given the equilibrium con-
stant Ksp for the dissolution of silver chloride (AgCl) is 1.8×10−10, determine
if a precipitate will form in this experiment.
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction:
AgN O3+N aCl →AgCl +NaNO3
Step 2: Calculate the initial concentration of silver ions (Ag+) and chloride ions
(Cl−) in the solution. Initial concentration of Ag+:
[Ag+]=0.100 M
Initial concentration of Cl−:
[Cl−]=0.200 M
Step 3: Determine the change in concentration of each ion at equilibrium. Let x
be the concentration of AgCl that dissolves/dissociates. The change in concen-
tration of Ag+and Cl−will be +x, and the change in concentration of AgCl
will be -x. At equilibrium:
[Ag+]=0.100 + x
[Cl−]=0.200 + x
[AgCl] = x
Step 4: Write the expression for the solubility product constant:
Ksp = [Ag+][Cl−]
Substitute the equilibrium concentrations:
Ksp = (0.100 + x)(0.200 + x)
Step 5: Solve for x using the given Ksp value:
1.8×10−10 = (0.100 + x)(0.200 + x)
1.8×10−10 = 0.020 + 0.1x+ 0.2x+x2
15
Step 6: Simplify and solve for x using the quadratic formula:
x2+ 0.3x−1.8×10−2= 0
Solving the quadratic equation, we get:
x≈4×10−3M
Step 7: Determine if a precipitate will form by comparing x to the initial con-
centrations of Ag+and Cl−: Since 4 ×10−3M<0.100 M and 4 ×10−3M<
0.200 M, the concentration of AgCl that dissolves is small and will not exceed
the initial concentrations of Ag+and Cl−. Therefore, no precipitate will form
in this experiment.
Question 19
Question
A solution was prepared by dissolving 0.250 moles of silver nitrate in enough
water to make 0.500 L of solution. The solution was then mixed with 0.300 L
of 0.120 M sodium chloride solution. Calculate the mass of the precipitate that
forms.
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between silver nitrate (AgNO3) and sodium chloride (NaCl):
AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)
Step 2: Determine the limiting reagent by calculating the moles of each
reactant: For silver nitrate:
moles of AgNO3= Molarity ×Volume = 0.250 mol
For sodium chloride:
moles of NaCl = Molarity ×Volume = 0.120 ×0.300 = 0.036 mol
Since silver nitrate is the limiting reagent (0.250 mol ¡ 0.036 mol), we will base
our calculations on silver nitrate.
Step 3: Use stoichiometry to find the moles of silver chloride (AgCl) formed:
From the balanced equation: 1 mole of AgNO3produces 1 mole of AgCl. There-
fore, moles of AgCl formed = moles of AgNO3= 0.250 mol
Step 4: Calculate the mass of silver chloride formed: The molar mass of
AgCl is 143.32 g/mol.
Mass of AgCl = moles of AgCl×molar mass of AgCl = 0.250 mol×143.32 g/mol = 35.83 g
Therefore, the mass of the precipitate (silver chloride) that forms is 35.83
grams.
16
Question 20
Question
A solution is prepared by dissolving 25.0 g of calcium nitrate, Ca(NO3)2, in
100.0 mL of water. What is the concentration of calcium ions, in mol/L, in the
solution? (Molar mass of Ca(NO3)2= 164.10 g/mol)
Solution
Step 1: Calculate the number of moles of calcium nitrate.
Moles of Ca(NO3)2=Mass
Molar mass =25.0 g
164.10 g/mol
Moles of Ca(NO3)2= 0.152 mol
Step 2: Determine the number of moles of calcium ions present in the so-
lution. Calcium nitrate, Ca(NO3)2, dissociates into one calcium ion, Ca2+, for
every one formula unit. Therefore, the number of moles of calcium ions is the
same as the number of moles of calcium nitrate.
Moles of Ca2+ = 0.152 mol
Step 3: Calculate the concentration of calcium ions in the solution. The
volume of the solution is given as 100.0 mL or 0.100 L.
Concentration of Ca2+ =Moles of Ca2+
Volume of solution =0.152 mol
0.100 L
Concentration of Ca2+ = 1.52 mol/L
Answer: The concentration of calcium ions in the solution is 1.52 mol/L.
Question 21
Question
Calculate the concentration of sulfate ions in a solution that forms when 50.0
mL of 0.200 M sulfuric acid (H2SO4) is mixed with 150.0 mL of 0.100 M barium
hydroxide (Ba(OH)2). The balanced chemical equation for this reaction is:
H2SO4(aq) + Ba(OH)2(aq)→BaSO4(s) + 2H2O(l)
(Hint: After determining the limiting reagent, use stoichiometry to find the
amount of sulfate ions produced.)
17
Solution
Step 1: Write the balanced chemical equation and determine the limiting reagent.
The balanced equation shows that one mole of sulfuric acid reacts with one
mole of barium hydroxide to produce one mole of barium sulfate.
H2SO4(aq) + Ba(OH)2(aq)→BaSO4(s) + 2H2O(l)
To determine the limiting reagent, calculate the moles of each reactant:
Moles of H2SO4= Volume ×Molarity
= 0.0500 L ×0.200 mol/L
= 0.0100 mol
Moles of Ba(OH)2= Volume ×Molarity
= 0.1500 L ×0.100 mol/L
= 0.0150 mol
Since the reaction requires 1 mole of H2SO4for every 1 mole of Ba(OH)2,
sulfuric acid is the limiting reagent.
Step 2: Calculate the amount of sulfate ions produced.
From the balanced equation, it is clear that 1 mole of barium sulfate produces
1 mole of sulfate ions. Therefore, the moles of sulfate ions produced are equal
to the moles of barium sulfate formed, which is equal to the initial moles of
sulfuric acid:
Moles of sulfate ions = 0.0100 mol
Step 3: Calculate the concentration of sulfate ions.
The final volume of the solution is the sum of the initial volumes of the two
solutions:
Volume of final solution = 0.0500 L + 0.1500 L = 0.2000 L
Finally, the concentration of sulfate ions is given by:
Concentration of sulfate ions = Moles of sulfate ions
Volume of final solution =0.0100 mol
0.2000 L = 0.0500 mol/L
Therefore, the concentration of sulfate ions in the final solution is 0.0500 mol/L,
or 0.0500 M.
Question 22
Question
A solution is prepared by dissolving 45.0 g of potassium bromide (KBr) in 155.0
g of water. Determine if precipitation will occur when 20.0 mL of 0.130 M
lead(II) nitrate (Pb(NO3)2) solution is added to the solution. The solubility
constants for KBr and PbBr2are 6.3×10−2and 6.6×10−6, respectively.
18
Solution
Step 1: Write the chemical equation for potential precipitation to determine if
the given reactants will produce an insoluble product. The possible chemical
reaction is:
Pb(NO3)2+ 2KBr →PbBr2+ 2KNO3
Step 2: Calculate the concentrations of the ions in the solution before mixing
them. First, find the moles of KBr:
moles of KBr = 45.0 g
119.0 g/mol = 0.378 mol
Next, calculate the moles of water:
moles of water = 155.0 g
18.0 g/mol = 8.61 mol
Then, determine the initial concentration of KBr:
initial concentration of KBr = 0.378 mol
8.61 mol = 0.0439 M
Step 3: Determine the solubility of PbBr2in water using the solubility prod-
uct constant. The solubility product constant for PbBr2is 6.6×10−6.
Step 4: Calculate the initial concentration of bromide ions, [Br−]. Since
2 moles of Br−ions are formed for every mole of KBr dissolved, the initial
concentration of Br−ions can be calculated as:
[Br−] = 2 ×0.0439 = 0.0877 M
Step 5: Determine the initial concentration of lead(II) ions, [Pb2+]. Since
the initial concentration of lead(II) nitrate is 0.130 M, the initial concentration
of lead(II) ions is also 0.130 M.
Given concentration of Pb2+ = 0.130 M
Step 6: Determine the ion product, Q, and compare it to the solubility
product constant, Ksp. The ion product, Q, for PbBr2is given by:
Q= [Pb2+][Br−]2
Substitute the values:
Q= (0.130)(0.0877)2= 0.00107
Step 7: Analyze whether precipitation will occur. Since Q(0.00107) >
Ksp(6.6×10−6), precipitation of lead(II) bromide (PbBr2) will occur when 20.0
mL of the lead(II) nitrate solution is added to the potassium bromide solution.
19
Question 23
Question
A solution contains 0.1 M barium chloride (BaCl2) and 0.15 M sodium sulfate
(Na2SO4). Calculate the concentration of the sulfate ion after a precipitation
reaction, where all sulfate reacts with barium to form insoluble barium sulfate
(BaSO4). The solubility product (Ksp) of barium sulfate is 1.1×10−10.
Solution
Step 1: Write the balanced equation for the precipitation reaction between
barium chloride and sodium sulfate:
BaCl2+ Na2SO4→BaSO4+ 2NaCl
Step 2: Determine the limiting reactant. Since barium sulfate is insoluble, it
will precipitate out until one of the ions runs out. Calculate the moles of each ion
in the solution: - Moles of sulfate ion from sodium sulfate: 0.15 M ×2 mol/L =
0.30 mol/L - Moles of barium ion from barium chloride: 0.10 M ×1 mol/L =
0.10 mol/L
Since there are more moles of sulfate ion (limiting reactant), barium will be
completely consumed.
Step 3: Calculate the concentration of sulfate ion after the reaction. Since all
sulfate reacts with barium to form insoluble barium sulfate, the concentration
of sulfate ion after the reaction will be zero.
Therefore, the concentration of the sulfate ion after the precipitation reaction
is 0 M .
Question 24
Question
A solution contains 0.15 M of silver ion (Ag+) and 0.13 M of sulfate ion (SO2−
4).
What is the minimum concentration of sodium chloride (NaCl) that needs to be
added to the solution in order to precipitate all of the silver ion as silver chloride
(AgCl) in a 1.0 L solution? (Hint: The Ksp of silver chloride is 1.77 ×10−10)
Solution
Step 1: Write the balanced chemical equation for the precipitation of silver
chloride:
Ag++ Cl−→AgCl
Step 2: Write the equation for the equilibrium constant (Ksp):
Ksp = [Ag+][Cl−]
20
Step 3: Substitute the initial concentrations into the Ksp expression. Since
Cl−is coming from NaCl, we assume that the concentration of NaCl is equal
to that of Cl−:
Ksp = (0.15)(x)
1.77 ×10−10 = 0.15x
Step 4: Solve for xto find the minimum concentration of NaCl needed:
x=1.77 ×10−10
0.15 = 1.18 ×10−9M
Therefore, the minimum concentration of NaCl that needs to be added to
the solution is 1.18 ×10−9M.
Question 25
Question
At a weather station, the annual precipitation can be modeled by the function
P(t)=5t2−30t+ 40, where trepresents the month of the year with January
as t= 1 and December as t= 12. Calculate the total precipitation for the year.
Solution
Step 1: To calculate the total precipitation for the year, we need to find the
integral of the precipitation function P(t) over the interval [1,12].
Step 2: The integral of the precipitation function P(t)=5t2−30t+ 40 is
given by
Z12
1
(5t2−30t+ 40) dt
Step 3: To find the integral, we first calculate the antiderivative of each
term:
Z5t2dt =5
3t3+C, Z−30t dt =−15t2+C, Z40 dt = 40t+C
Step 4: Applying the antiderivative to the integral, we have
Z12
1
(5t2−30t+ 40) dt =5
3t3−15t2+ 40t12
1
Step 5: Evaluate the antiderivative at the upper and lower limits:
=5
3(12)3−15(12)2+ 40(12)−5
3(1)3−15(1)2+ 40(1)
Step 6: Simplifying further, we get
= (960 −2160 + 480) −5
3−15 + 40
21
Step 7: Finally, compute the total precipitation for the year:
= 280 −5
3+ 15 −40 = 254.33 units
Therefore, the total precipitation for the year is 254.33 units.
Question 26
Question
A solution contains 20 grams of calcium chloride (CaCl2) in 200 mL of water.
If sodium carbonate (Na2CO3) solution is added to this solution, how many
grams of CaCO3will precipitate out?
Solution
Step 1: Write the balanced chemical equation for the reaction between calcium
chloride and sodium carbonate.
CaCl2(aq) + Na2CO3(aq)−→ CaCO3(s)+2NaCl(aq)
Step 2: Calculate the number of moles of calcium chloride present in the
solution. Given: Mass of CaCl2: 20 grams Molar mass of CaCl2: 40.08 (Ca) +
2(35.45) (Cl) = 110.98 g/mol
Number of moles of CaCl2=20 g
110.98 g/mol ≈0.18 mol
Step 3: Calculate the number of moles of CaCO3that can be formed. From
the balanced chemical equation, 1 mole of CaCl2produces 1 mole of CaCO3.
Therefore, number of moles of CaCO3= number of moles of CaCl2= 0.18
mol
Step 4: Calculate the mass of CaCO3formed. Molar mass of CaCO3: 40.08
(Ca) + 12.01 (C) + 3(16.00) (O) = 100.09 g/mol
Mass of CaCO3= number of moles of CaCO3×molar mass of CaCO3Mass
of CaCO3= 0.18 mol ×100.09 g/mol = 18.02 grams
Therefore, 18.02 grams of CaCO3will precipitate out.
Question 27
Question
A solution contains 0.15 M calcium chloride (CaCl2) and 0.20 M silver nitrate
(AgNO3). Determine if a precipitate forms when these two solutions are mixed.
If so, calculate the mass of precipitate formed when 500.0 mL of each solution
are combined.
22
Solution
Step 1: Write the chemical equation for the possible precipitation reaction be-
tween calcium chloride and silver nitrate:
CaCl2+ 2AgNO3−→ Ca(NO3)2+ 2AgCl
Step 2: Determine the possible products of the reaction: - The cation from
CaCl2is calcium (Ca2+). - The anion from AgNO3is nitrate (NO−
3). - The
possible precipitate is silver chloride (AgCl).
Step 3: Determine the net ionic equation for the reaction:
Ca2+ + 2NO−
3+ 2Ag++ 2Cl−−→ Ca2+ + 2NO−
3+ 2AgCl
Step 4: Determine if a precipitate will form by examining the solubility rules.
Silver chloride is insoluble in water, so a white precipitate of silver chloride will
form when these solutions are mixed.
Step 5: Calculate the moles of each reactant: - For calcium chloride:
moles of CaCl2= 0.15 M ×0.5 L = 0.075 mol
- For silver nitrate:
moles of AgNO3= 0.20 M ×0.5 L = 0.100 mol
Step 6: Determine the limiting reactant by looking at the stoichiometry of
the reaction. Since the reaction consumes 2 moles of silver nitrate for every 1
mole of calcium chloride, calcium chloride is the limiting reactant.
Step 7: Calculate the mass of precipitate formed (silver chloride):
moles of AgCl = 0.075 mol ×2 mol AgCl
1 mol CaCl2×143.32 g/mol = 21.41 g
Therefore, when 500.0 mL of 0.15 M calcium chloride and 0.20 M silver
nitrate are mixed, a white precipitate of silver chloride weighing 21.41 g will
form.
Question 28
Question
Calculate the concentration of sulfate ions (SO2−
4) in a solution prepared by
mixing 100.0 mL of 0.200 M sodium sulfate (Na2SO4) with 150.0 mL of 0.400
M barium chloride (BaCl2). Assume complete precipitation of barium sulfate
(BaSO4) occurs.
23
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between sodium sulfate and barium chloride:
Na2SO4(aq) + BaCl2(aq)→BaSO4(s) + 2NaCl(aq)
Step 2: Determine the limiting reactant. Since Na2SO4and BaCl2are in a
1:1 molar ratio, we need to find out which reactant will run out first.
Calculate the moles of Na2SO4:
Moles of Na2SO4= Volume of Na2SO4×Molarity of Na2SO4
= 0.100 L ×0.200 mol/L = 0.020 mol
Calculate the moles of BaCl2:
Moles of BaCl2= Volume of BaCl2×Molarity of BaCl2
= 0.150 L ×0.400 mol/L = 0.060 mol
Since 0.020 mol of Na2SO4is less than 0.060 mol of BaCl2, Na2SO4is the
limiting reactant.
Step 3: Calculate the moles of barium sulfate formed using the limiting
reactant:
Moles of BaSO4= Moles of Na2SO4= 0.020 mol
Step 4: Calculate the concentration of sulfate ions present in the solution
after precipitation:
Volume of final solution = 0.100L+ 0.150L= 0.250L
Concentration of sulfate ions = Moles of sulfate ions
Volume of final solution
=0.020 mol
0.250 L = 0.080 mol/L
Therefore, the concentration of sulfate ions in the final solution is 0.080
mol/L.
Question 29
Question
Calculate the concentration of sulfate ions in a solution after adding excess
barium chloride to 100.0 mL of a 0.200 M sodium sulfate solution. Assume
complete precipitation of barium sulfate.
24
Solution
Step 1: Write the balanced chemical equation for the reaction between sodium
sulfate and barium chloride.
Na2SO4+ BaCl2→BaSO4+ 2NaCl
Step 2: Determine the limiting reactant. Since barium chloride is added in
excess, the limiting reactant is sodium sulfate.
Step 3: Use stoichiometry to find the moles of sulfate ions in the solution.
Moles of Na2SO4= Volume ×Molarity = 0.100 L ×0.200 mol/L = 0.020 mol
Step 4: Since each mole of Na2SO4produces one mole of SO2−
4ions, the
concentration of sulfate ions in the solution is the same as the concentration of
sodium sulfate.
Concentration of SO2−
4= 0.200 M
Question 30
Question
A chemist is tasked with determining the amount of precipitate formed when
100.0 mL of 0.200 M silver nitrate solution is added to 150.0 mL of 0.100 M
sodium chloride solution. Assuming that a precipitate forms according to the
reaction:
AgN O3(aq) + N aCl(aq)→AgCl(s) + NaNO3(aq)
Calculate the mass of silver chloride (AgCl) that will precipitate out.
(Hint: First, determine the limiting reagent in the reaction to find the
amount of precipitate formed.)
Solution
Step 1: Write the balanced chemical equation for the reaction and determine
the limiting reagent. The balanced chemical equation is:
AgN O3(aq) + N aCl(aq)→AgCl(s) + NaNO3(aq)
From the balanced equation, the molar ratio between silver nitrate (AgNO3)
and silver chloride (AgCl) is 1:1, and between sodium chloride (NaCl) and silver
chloride (AgCl) is 1:1.
Calculate the moles of silver nitrate and sodium chloride:
Moles of AgN O3= (0.100 mol/L) * (0.100 L) = 0.020 mol
Moles of NaCl = (0.150 mol/L) * (0.200 L) = 0.030 mol
Since the molar ratios are 1:1, silver nitrate is the limiting reagent because
it produces fewer moles of silver chloride.
Step 2: Calculate the mass of silver chloride precipitate formed. The molar
mass of AgCl is 143.32 g/mol.
25
Calculate the mass of AgCl precipitated out using the moles of AgNO3:
Mass of AgCl = moles of AgN O3* molar mass of AgCl
Mass of AgCl = 0.020 mol * 143.32 g/mol = 2.87 g
Therefore, the mass of silver chloride that will precipitate out is 2.87 grams.
Question 31
Question
A chemist is conducting an experiment that involves mixing two solutions to
form a precipitate. The chemist mixes 150.0 mL of a 0.200 M solution of calcium
chloride with 200.0 mL of a 0.150 M solution of sodium sulfate. Calculate the
mass of calcium sulfate (CaSO4) that will precipitate out of solution.
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction:
CaCl2+N a2SO4→CaSO4+ 2N aCl
Step 2: Determine the limiting reactant by calculating the moles of each
reactant.
Moles of calcium chloride (CaCl2):
0.150 mol/L ×0.150 L = 0.0300 mol
Moles of sodium sulfate (Na2SO4):
0.200 mol/L ×0.150 L = 0.0300 mol
Both reactants have the same number of moles, so CaCl2is the limiting
reactant.
Step 3: Calculate the moles of CaSO4formed using the mole ratio from the
balanced chemical equation.
Moles of CaSO4= 0.0300 mol ×1 mol CaSO4
1 mol CaCl2
= 0.0300 mol
Step 4: Convert moles of CaSO4to grams.
Molar mass of CaSO4:
40.08 g/mol + 32.06 g/mol + 4(16.00 g/mol) = 136.06 g/mol
Mass of CaSO4:
0.0300 mol ×136.06 g/mol = 4.08 g
Therefore, the mass of calcium sulfate precipitated out of solution is 4.08
grams.
26
Question 32
Question
Calculate the molarity of chloride ions in a solution that contains 0.250 moles
of calcium chloride, CaCl2, dissolved in 500.0 mL of solution.
Solution
Step 1: Determine the number of moles of chloride ions present in calcium
chloride. Given that calcium chloride, CaCl2, dissociates into three ions when
dissolved in water, the number of moles of chloride ions is 3 times the moles
of calcium chloride. Number of moles of Cl−ions = 3 ×0.250 moles = 0.750
moles
Step 2: Calculate the total volume of the solution in liters. The volume
provided is 500.0 mL, which is equivalent to 0.500 L.
Step 3: Calculate the molarity of chloride ions. Molarity (M) is defined
as the moles of solute divided by the volume of solution in liters. Therefore,
molarity of chloride ions = 0.750 moles
0.500 L = 1.50 M
Therefore, the molarity of chloride ions in the solution is 1.50 M.
Question 33
Question
Calculate the concentration of sulfate ions in a solution when 100.0 mL of 0.200
M silver sulfate solution is mixed with 150.0 mL of 0.500 M potassium chloride
solution. Assume complete precipitation of silver sulfate.
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between silver sulfate and potassium chloride:
Ag2SO4+ 2KCl −→ 2AgCl +K2SO4
Step 2: Determine the limiting reactant. Since the reaction goes to comple-
tion and all the silver sulfate precipitates, we need to find out how much of the
sulfate ions comes from the silver sulfate:
Moles of Ag2SO4= 100.0 mL ×0.200 mol/L = 0.0200 mol
Moles of SO2−
4= 0.0200 mol ×1 mol SO2−
4
1 mol Ag2SO4
= 0.0200 mol
Step 3: Calculate the final volume of the solution:
Vfinal = 100.0 mL + 150.0 mL = 250.0 mL = 0.250 L
27
Step 4: Calculate the concentration of the sulfate ions in the final solution:
Concentration of SO2−
4=0.0200 mol
0.250 L = 0.0800 M
Therefore, the concentration of sulfate ions in the solution is 0.0800 M.
Question 34
Question
A student is conducting an experiment to determine the concentration of chlo-
ride ions in a water sample by precipitation with silver ions. The student mixes
50.0 mL of the water sample with an excess of silver nitrate solution. The re-
sulting precipitate of silver chloride is then filtered, dried, and found to have a
mass of 0.527 g.
Calculate the concentration of chloride ions (in mol/L) in the original water
sample.
(Given: molar mass of AgCl = 143.32 g/mol)
Solution
Step 1: Calculate the moles of silver chloride precipitate formed.
Moles of AgCl = Mass of AgCl
Molar mass of AgCl
Moles of AgCl = 0.527 g
143.32 g/mol = 0.003675 mol
Step 2: Since the reaction between chloride ions and silver ions is a 1:1 ratio,
the moles of chloride ions is equal to the moles of silver chloride formed.
Moles of chloride ions = 0.003675 mol
Step 3: Calculate the volume of the water sample in liters.
Volume of water sample (L) = 50.0 mL
1000 mL/L = 0.0500 L
Step 4: Calculate the concentration of chloride ions in the original water
sample.
Concentration of chloride ions (mol/L) = Moles of chloride ions
Volume of water sample (L)
Concentration of chloride ions (mol/L) = 0.003675 mol
0.0500 L = 0.0735 mol/L
Therefore, the concentration of chloride ions in the original water sample is
0.0735 mol/L.
28
Question 35
Question
Given that the solubility product constant (Ksp) of lead(II) chloride (PbCl2) is
1.6×10−5at 25◦C, calculate the molar solubility of lead(II) chloride in a 0.050
M hydrochloric acid solution. Assume that the dissolution of PbCl2does not
significantly impact the concentration of Cl−ions in the solution.
Solution
Step 1: Write the equilibrium equation for the dissolution of lead(II) chloride.
The equilibrium equation for the dissolution of PbCl2is:
PbCl2⇌Pb2+ + 2Cl−
Step 2: Write the expression for the solubility product constant (Ksp). The
solubility product constant can be expressed as:
Ksp = [Pb2+][Cl−]2
Step 3: Let xbe the molar solubility of lead(II) chloride in the solution.
After dissociation, we have:
[Pb2+] = x
[Cl−]=2x
Step 4: Substitute the values into the Ksp expression. Plugging in the values,
we get:
1.6×10−5= (x)(2x)2
Step 5: Solve for x.
1.6×10−5= 4x3
x=3
r1.6×10−5
4
x= 0.020 M
Therefore, the molar solubility of lead(II) chloride in a 0.050 M hydrochloric
acid solution is 0.020 M.
29
Question 2
Question
Calculate the concentration of sulfate ions in a solution when 50 mL of 0.2
M barium sulfate solution is mixed with excess sodium sulfate solution. The
resulting barium sulfate precipitate is then filtered off and found to have a mass
of 0.345 g. (Molar mass of BaSO4= 233.4 g/mol)
Solution
Step 1: Determine the moles of barium sulfate.
Given: Volume of barium sulfate solution, VBaSO4= 50 mL = 0.05 L Molar-
ity of barium sulfate solution, MBaSO4= 0.2 M
The moles of barium sulfate can be calculated using the formula:
moles of BaSO4=MBaSO4×VBaSO4
moles of BaSO4= 0.2 mol/L ×0.05 L = 0.01 mol
Therefore, there are 0.01 moles of barium sulfate present.
Step 2: Calculate the moles of sulfate ions.
Since barium sulfate reacts in a 1:1 molar ratio with sulfate ions, the moles
of sulfate ions present is also 0.01 moles.
Step 3: Calculate the total volume of solution.
Since the sodium sulfate solution is in excess, the volume of the mixture is
the sum of the volumes of both solutions. Assuming the volume of the sodium
sulfate solution is large compared to 50 mL, we can consider the total volume
to be 50 mL + the volume of sodium sulfate solution.
Step 4: Calculate the concentration of sulfate ions.
The concentration of sulfate ions can be calculated using the formula:
Concentration of sulfate ions = moles of sulfate ions
total volume of solution
Since the moles of sulfate ions is 0.01 and the total volume is 0.05 L:
Concentration of sulfate ions = 0.01 mol
0.05 L = 0.2 M
Therefore, the concentration of sulfate ions in the solution is 0.2 M.
Question 3
Question
A solution contains 0.2 mol/L of lead(II) nitrate (Pb(NO3)2) and 0.1 mol/L of
sodium sulfate (Na2SO4). Determine the maximum amount of lead(II) sulfate
(PbSO4) that can precipitate out in grams from a 500 mL solution.
2
(Hint: The solubility product constant for lead(II) sulfate is 1.2×10−8at
25
°
C)
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between lead(II) nitrate and sodium sulfate:
Pb(NO3)2+ Na2SO4→PbSO4+ 2NaNO3
Step 2: Calculate the number of moles of lead(II) nitrate and sodium sulfate
that will react. Since the volume of the solution is 500 mL, the moles of each
substance can be calculated:
Moles of Pb(NO3)2= 0.2 mol/L ×0.5 L = 0.1 mol
Moles of Na2SO4= 0.1 mol/L ×0.5 L = 0.05 mol
Step 3: Determine the limiting reactant in the reaction based on the sto-
ichiometry of the balanced chemical equation. Since the stoichiometry is 1:1,
the limiting reactant is the one with the lower number of moles. Here, sodium
sulfate is the limiting reactant as it has fewer moles.
Step 4: Calculate the number of moles of lead(II) sulfate that can be formed
using the limiting reactant (sodium sulfate):
Moles of PbSO4= 0.05 mol
Step 5: Calculate the mass of lead(II) sulfate formed using its molar mass:
Molar mass of PbSO4= 207.2 g/mol + 32.1 g/mol + 4(16 g/mol) = 303.3 g/mol
Mass of PbSO4= 0.05 mol ×303.3 g/mol = 15.165 g
Therefore, the maximum amount of lead(II) sulfate that can precipitate out
from the solution is 15.165 grams.
Question 4
Question
A chemical reaction takes place in a beaker where 200 mL of a 0.1 M solution of
lead(II) nitrate (Pb(NO3)2) is mixed with 100 mL of a 0.2 M solution of sodium
iodide (NaI).
Calculate the maximum amount of lead(II) iodide (PbI2) that can precipitate
in grams. (Assume the reaction goes to completion and that the density of the
solutions is 1 g/mL.)
3
Solution
Step 1: Write the balanced chemical equation for the reaction between lead(II)
nitrate (Pb(NO3)2) and sodium iodide (NaI):
Pb(NO3)2+ 2NaI →PbI2+ 2NaNO3
Step 2: Calculate the moles of lead(II) nitrate (Pb(NO3)2) and sodium iodide
(NaI) used: For lead(II) nitrate:
Moles Pb(NO3)2= Volume ×Molarity = 0.2 L ×0.1 mol/L = 0.02 mol
For sodium iodide:
Moles NaI = Volume ×Molarity = 0.1 L ×0.2 mol/L = 0.02 mol
Step 3: Determine the limiting reactant. The limiting reactant in this case
is lead(II) nitrate (Pb(NO3)2) since it is completely consumed when 0.02 mol
of sodium iodide is used.
Step 4: Calculate the theoretical yield of lead(II) iodide (PbI2) that can be
produced:
Moles PbI2= Moles Pb(NO3)2= 0.02 mol
Step 5: Calculate the molar mass of lead(II) iodide (PbI2):
Molar mass of PbI2= Atomic mass of Pb+2×Atomic mass of I = 207.2 g/mol+2×126.9 g/mol = 460 g/mol
Step 6: Convert moles of lead(II) iodide to grams:
Mass of PbI2= Moles PbI2×Molar mass of PbI2= 0.02 mol×460 g/mol = 9.2 g
Therefore, the maximum amount of lead(II) iodide that can precipitate is
9.2 grams.
Question 5
Question
Calculate the solubility of lead(II) iodide (PbI2) in a 0.25 M potassium iodide
(KI) solution at 25◦C. The solubility product constant of lead(II) iodide is 7.1×
10−9.
Solution
Step 1: Write the equilibrium expression for the dissolution of lead(II) iodide.
The equilibrium expression for the dissolution of PbI2is:
PbI2(s)⇌Pb2+(aq) + 2I−(aq)
4
Step 2: Define the variables. Let xbe the molar solubility of lead(II) iodide
in mol/L.
Step 3: Write the equilibrium constant expression. The solubility product
constant (Ksp) expression for PbI2is:
Ksp = [Pb2+][I−]2
Substitute the expressions for the ions in terms of x:
Ksp =x×(2x)2= 4x3
Step 4: Substitute known values. Given that Ksp = 7.1×10−9, substitute
this into the expression:
7.1×10−9= 4x3
Step 5: Solve for x.
x=3
r7.1×10−9
4= 0.00061 mol/L
Therefore, the solubility of lead(II) iodide in a 0.25 M potassium iodide
solution at 25◦C is 0.00061 mol/L.
Question 6
Question
A sample of water contains 2.5×10−3M of calcium ions (Ca2+). If CaCO3is
added to the solution until the calcium ion concentration is 1.5×10−3M, how
many grams of CaCO3were added? (Assume the only source of CaCO3in the
solution is from the added solid.)
Given: Ksp of CaCO3= 4.8×10−9.
Solution
Step 1: Write the balanced equation for the dissociation of CaCO3:
CaCO3→Ca2+ + CO2−
3
Step 2: Write the equilibrium constant expression based on the balanced
equation:
Ksp = [Ca2+][CO2−
3]
Step 3: Calculate the initial concentration of CO2−
3from the initial con-
centration of Ca2+: Given: [Ca2+] = 2.5×10−3M. Since 1 mole of CaCO3
produces 1 mole of Ca2+ and 1 mole of CO2−
3, initial [CO2−
3]=2.5×10−3M.
Step 4: Calculate the final concentration of CO2−
3using the concentration
of Ca2+ after CaCO3is added: Given: new [Ca2+]=1.5×10−3M. Since the
5
initial [Ca2+] = [CO2−
3] before the CaCO3is added, the decrease in [Ca2+] is
equal to the increase in [CO2−
3: [CO2−
3]=2.5×10−3+(2.5×10−3−1.5×10−3) =
3.5×10−3M.
Step 5: Use the equilibrium constant expression to determine the concen-
tration of Ca2+ once all the CaCO3has dissolved: 4.8×10−9= (1.5×10−3)×
(3.5×10−3)
Step 6: Calculate the number of moles of CaCO3added: Since 1 mole of
CaCO3produces 1 mole of Ca2+, moles of CaCO3added = 1.5×10−3moles.
Step 7: Calculate the molar mass of CaCO3(mass of Ca + 1 mol of C + 3
mol of O): 40.08 g/mol + 12.01 g/mol + (3 ×16.00 g/mol) = 100.08 g/mol
Step 8: Calculate the mass of CaCO3added: mass = moles ×molar mass =
1.5×10−3×100.08 = 0.15012 g
Therefore, 0.15012 grams of CaCO3were added to the solution.
Question 7
Question
In a precipitation reaction, 50.0 mL of 0.100 M of silver nitrate (AgNO3) is
mixed with 25.0 mL of 0.200 M sodium chloride (NaCl) solution. What is
the molarity of silver ions (Ag+) after the reaction has occurred? 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 precipitation reaction
between silver nitrate and sodium chloride:
AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)
Step 2: Determine the limiting reactant by calculating the moles of each
reactant: For silver nitrate:
Moles of AgNO3= Volume ×Molarity
= (50.0×10−3L) ×0.100 M
= 5.00 ×10−3mol
For sodium chloride:
Moles of NaCl = Volume ×Molarity
= (25.0×10−3L) ×0.200 M
= 5.00 ×10−3mol
Since both reactants result in the same number of moles of product, silver
nitrate is the limiting reactant.
6
Step 3: Calculate the moles of silver ions formed: From the balanced chemi-
cal equation, every mole of silver nitrate produces one mole of silver ions. Thus,
the moles of silver ions formed is also 5.00 ×10−3mol.
Step 4: Calculate the new volume of the solution after the reaction: The
total volume of the solution after mixing the two solutions is 50.0×10−3L +
25.0×10−3L = 75.0×10−3L.
Step 5: Calculate the molarity of silver ions:
Molarity of Ag+=Moles of Ag+
Volume
=5.00 ×10−3mol
75.0×10−3L
= 0.067 M
Therefore, the molarity of silver ions (Ag+) after the reaction is 0.067 M.
Question 8
Question
A 250 mL solution contains 0.15 M barium chloride (BaCl2) and 0.20 M sodium
sulfate (Na2SO4). Calculate the concentrations of Ba2+ and SO2−
4ions after
precipitation reaction occurs between barium chloride and sodium sulfate.
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between BaCl2and Na2SO4.
BaCl2(aq) + Na2SO4(aq)→BaSO4(s) + 2NaCl(aq)
Step 2: Determine the limiting reactant and calculate the amount of BaSO4
formed. Since BaCl2and Na2SO4react in a 1:1 molar ratio, the limiting reactant
will be the one that runs out first. To find the limiting reactant, we compare
the moles of each reactant:
Moles of BaCl2: 0.15 M ×0.25 L = 0.0375 mol
Moles of Na2SO4: 0.20 M ×0.25 L = 0.05 mol
Since BaCl2is the limiting reactant (having fewer moles), the amount of BaSO4
formed will be 0.0375 mol.
Step 3: Calculate the concentration of Ba2+ ions after precipitation. Ini-
tially, the concentration of Ba2+ ions was 0.15 M, but since 0.0375 mol of BaCl2
precipitated, the new concentration is:
0.15 mol
0.25 L = 0.6 M Ba2+
7
Step 4: Calculate the concentration of SO2−
4ions after precipitation. Since
the reaction went to completion, all Na2SO4reacts to form NaCl and BaSO4.
The SO2−
4ions come from the Na2SO4, so the concentration will be:
0.05 mol
0.25 L = 0.2 M SO2−
4
Question 9
Question
A chemist is performing a precipitation reaction by mixing 50.0 mL of a 0.200
M lead(II) nitrate solution with 75.0 mL of a 0.150 M sodium iodide solution.
If lead(II) iodide is formed as a precipitate, calculate the mass of lead(II) iodide
that forms.
(Note: The density of water is 1.00 g/mL.)
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between lead(II) nitrate and sodium iodide:
P b(NO3)2+ 2NaI →P bI2+ 2NaNO3
Step 2: Determine the limiting reactant by using the stoichiometry of the
balanced equation.
From the equation: 1 mole of lead(II) nitrate reacts with 2 moles of sodium
iodide to produce 1 mole of lead(II) iodide.
Calculate the moles of lead(II) nitrate and sodium iodide:
moles of Pb(NO3)2= 0.050 L ×0.200 mol/L = 0.0100 mol
moles of NaI = 0.075 L ×0.150 mol/L = 0.0113 mol
Since the reaction equation is 1:2 for Pb(NO3)2and NaI, NaI is the limiting
reactant.
Step 3: Calculate the theoretical yield of lead(II) iodide based on the limiting
reactant.
Using the mole ratio from the balanced equation, the moles of lead(II) iodide
formed will be half of the moles of sodium iodide:
moles of PbI2= 0.0113 mol ×1
2= 0.00565 mol
Step 4: Calculate the mass of lead(II) iodide formed using its molar mass.
The molar mass of lead(II) iodide (PbI2) is:
207.2 g/mol for Pb + 2 ×126.9 g/mol for I = 460.1 g/mol
Therefore, the mass of lead(II) iodide formed is:
0.00565 mol ×460.1 g/mol = 2.60 g
8
Question 10
Question
Calculate the mass of lead(II) iodide (PbI2) that can be precipitated when 50.0
mL of 0.200 M lead(II) nitrate (Pb(NO3)2) is mixed with 50.0 mL of 0.150
M potassium iodide (KI). (Assume the reaction goes to completion and that
lead(II) iodide is insoluble.)
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction be-
tween lead(II) nitrate and potassium iodide to form lead(II) iodide and potas-
sium nitrate.
Pb(NO3)2(aq) + 2KI(aq)→PbI2(s) + 2KNO3(aq)
Step 2: Determine the limiting reactant. Calculate the moles of each reactant
using the formula n= M ×V. For lead(II) nitrate:
nPb(NO3)2= 0.200 M ×0.0500 L = 0.0100 mol
For potassium iodide:
nKI = 0.150 M ×0.0500 L = 0.00750 mol
Since lead(II) nitrate produces 0.0100 mol of lead(II) iodide and potassium
iodide produces 0.00750 mol of lead(II) iodide, potassium iodide is the limiting
reactant.
Step 3: Calculate the mass of lead(II) iodide precipitated. The molar mass of
lead(II) iodide (PbI2) is approximately 461 g/mol. The amount of lead(II) iodide
formed will be 0.00750 mol. Therefore, the mass of lead(II) iodide precipitated
is:
Mass = moles ×molar mass = 0.00750 mol ×461 g/mol = 3.46 g
Thus, approximately 3.46 grams of lead(II) iodide can be precipitated.
Question 11
Question
Calculate the solubility of silver chromate (Ag2CrO4) in a 0.10 M solution of
silver nitrate (AgN O3). The Ksp of silver chromate is 1.1×10−12.
9
Solution
Step 1: Write the balanced dissociation equation for silver chromate. The dis-
sociation equation for silver chromate is:
Ag2CrO4(s)⇌2Ag+(aq) + CrO2−
4(aq)
Step 2: Write the expression for the solubility product constant, Ksp. The
solubility product constant, Ksp, is the product of the concentrations of the ions
raised to their stoichiometric coefficients. Therefore,
Ksp = [Ag+]2[CrO2−
4]
Substitute the given Ksp value into the equation:
1.1×10−12 = (2x)2(x)
Step 3: Solve for the value of x.
4x3= 1.1×10−12
x3=1.1×10−12
4
x≈5.23 ×10−5
Step 4: Calculate the solubility of silver chromate in the solution. The
solubility of silver chromate is equal to the concentration of CrO2−
4ions, which
is approximately 5.23 ×10−5M.
Question 12
Question
Calculate the solubility of silver chloride (AgCl) in water at 25◦C. The Ksp of
AgCl is 1.8×10−10.
Solution
Step 1: Write the equation for the dissociation of silver chloride in water:
AgCl(s)⇌Ag+(aq) + Cl−(aq)
Step 2: Write the expression for the equilibrium constant Ksp:
Ksp = [Ag+][Cl−]
Step 3: Let xbe the solubility of AgCl in mol/L. Since 1 mol of AgCl
dissociates into 1 mol of Ag+and 1 mol of Cl−, we have:
[Ag+] = xmol/L
10
[Cl−] = xmol/L
Step 4: Substitute the concentrations into the Ksp expression:
Ksp = (x)(x)
1.8×10−10 =x2
Step 5: Solve for x:
x=p1.8×10−10
x= 1.34 ×10−5mol/L
Therefore, the solubility of silver chloride in water at 25◦C is 1.34 ×10−5
mol/L.
Question 13
Question
A chemistry student is performing an experiment that involves the precipitation
of lead chloride (PbCl2) from a solution of lead nitrate (Pb(NO3)2) and sodium
chloride (NaCl). If 25.0 mL of 0.25 M lead nitrate solution is mixed with excess
sodium chloride solution, how many grams of lead chloride will precipitate?
(Assume the reaction goes to completion)
Solution
Step 1: Write out the balanced chemical equation for the precipitation reaction:
Pb(NO3)2+ 2NaCl →PbCl2+ 2N aN O3
Step 2: Calculate the moles of lead nitrate:
Moles of Pb(NO3)2= Volume (L) ×Molarity
Moles of Pb(NO3)2= 0.025 L ×0.25 mol/L = 0.00625 mol
Step 3: Determine the limiting reactant. Since sodium chloride is in excess,
lead nitrate is the limiting reactant. This means all of the lead nitrate will react.
Step 4: Use the mole ratio from the balanced equation to find the moles of
lead chloride formed:
Moles of PbCl2= Moles of Pb(NO3)2×1 mol PbCl2
1 mol Pb(N O3)2
Moles of PbCl2= 0.00625 mol ×1 mol PbCl2
1 mol Pb(N O3)2
= 0.00625 mol
Step 5: Calculate the mass of lead chloride precipitated:
Mass of PbCl2= Moles of PbCl2×Molar mass of PbCl2
Molar mass of PbCl2= 207.2 g/mol + 2(35.5 g/mol) = 278.2 g/mol
Mass of PbCl2= 0.00625 mol ×278.2 g/mol = 1.738 g
Therefore, 1.738 grams of lead chloride will precipitate in the reaction.
11
Question 14
Question
Determine the concentration of sulfate ions in a solution that forms a white pre-
cipitate with barium chloride in a 25.0 mL sample, where 0.500 g of precipitate
was formed. The precipitation reaction is:
BaCl2(aq) + Na2SO4(aq)→BaSO4(s) + 2NaCl(aq)
(The molar mass of BaSO4 is 233.39 g/mol)
Solution
Step 1: Calculate the moles of BaSO4 formed. Step 2: Use the stoichiometry of
the reaction to find the moles of sulfate ions. Step 3: Calculate the concentration
of sulfate ions in the solution.
Step 1: The molar mass of BaSO4 is 233.39 g/mol. Given that 0.500 g of
BaSO4 precipitate was formed:
Moles of BaSO4=0.500 g
233.39 g/mol = 0.002141 mol
Step 2: From the balanced equation, 1 mole of BaSO4 is formed for every
mole of Na2SO4. Therefore, the moles of sulfate ions present in the solution is
also 0.002141 mol.
Step 3: The volume of the sample is 25.0 mL. Converting this volume to
liters:
Volume in Liters = 25.0 mL ×1 L
1000 mL = 0.0250 L
The concentration of sulfate ions is given by:
Concentration (mol/L) = Moles
Volume (L) =0.002141 mol
0.0250 L = 0.0856 mol/L
Therefore, the concentration of sulfate ions in the solution is 0.0856 mol/L.
Question 15
Question
Calculate the concentration of iodide ions (I−) in a solution when 25.0 mL of
0.020 M lead (II) iodide (P bI2) is mixed with 45.0 mL of 0.015 M potassium
iodide (KI). Assume the reaction goes to completion.
12
Solution
Step 1: Write the balanced chemical equation for the reaction between lead (II)
iodide and potassium iodide:
P bI2+ 2KI →P bI2(s)+2K++ 2I−
Step 2: Determine the limiting reactant to calculate the amount of iodide
ions produced. To find the limiting reactant, we compare the moles of lead (II)
iodide and potassium iodide present: - Moles of P bI2: 0.0250 L ×0.020 mol/L =
0.0005 mol - Moles of KI: 0.0450 L ×0.015 mol/L = 0.000675 mol
Since lead (II) iodide (0.0005 mol) is less than potassium iodide (0.000675
mol), lead (II) iodide is the limiting reactant.
Step 3: Use stoichiometry to determine the number of moles of iodide ions
produced from lead (II) iodide: From the balanced equation, 1 mole of P bI2pro-
duces 2 moles of I−. So, moles of I−produced = 0.0005 mol P bI2×2 mol I−
1 mol P bI2=
0.0010 mol I−
Step 4: Calculate the concentration of iodide ions in the solution: Total
volume of solution = 25.0 mL + 45.0 mL = 70.0 mL = 0.0700 L Concentration
of iodide ions (I−) = 0.0010 mol
0.0700 L = 0.0143 M
Therefore, the concentration of iodide ions in the solution is 0.0143 M.
Question 16
Question
A scientist is studying the precipitation levels in a certain region. The average
annual precipitation in the region is 85 inches. If the precipitation data for the
past 5 years are as follows: 90 inches, 80 inches, 95 inches, 88 inches, and 82
inches, calculate the coefficient of variation for the precipitation levels in this
region.
Solution
Step 1: Find the mean precipitation level. The mean precipitation level is
calculated by finding the average of the precipitation data.
Mean = 90 + 80 + 95 + 88 + 82
5=435
5= 87 inches
Step 2: Find the standard deviation of the precipitation levels. First, find
the squared differences between each precipitation level and the mean:
(90−87)2= 9,(80−87)2= 49,(95−87)2= 64,(88−87)2= 1,(82−87)2= 25
Next, find the variance by calculating the average of these squared differ-
ences:
Variance = 9 + 49 + 64 + 1 + 25
5=148
5= 29.6
13
Finally, find the standard deviation by taking the square root of the variance:
Standard Deviation = √29.6≈5.44
Step 3: Calculate the coefficient of variation (CV). The coefficient of vari-
ation is calculated by dividing the standard deviation by the mean and then
multiplying by 100%:
CV = 5.44
87 ×100% ≈6.25%
Therefore, the coefficient of variation for the precipitation levels in this region
is approximately 6.25%.
Question 17
Question
A solution contains 0.1 mol/L of silver nitrate (AgNO3) and 0.2 mol/L of potas-
sium chloride (KCl). What mass of silver chloride (AgCl) will precipitate when
the two solutions are mixed together?
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction:
AgNO3(aq) + KCl(aq)→AgCl(s) + KNO3(aq)
Step 2: Determine the limiting reactant by finding the number of moles of
each reactant available. We will assume we have 1 L of each solution to make
calculations easier. For silver nitrate (AgNO3): Number of moles = concentra-
tion ×volume = 0.1 mol/L ×1 L = 0.1 mol For potassium chloride (KCl):
Number of moles = concentration ×volume = 0.2 mol/L ×1 L = 0.2 mol
Step 3: Determine the limiting reactant by looking at the stoichiometry of
the reaction. Since AgNO3and KCl react in a 1:1 molar ratio, AgNO3is the
limiting reactant.
Step 4: Calculate the theoretical yield of AgCl precipitate in grams. To find
the mass of AgCl, we need to use the molar mass of AgCl which is 107.87 +
35.45 = 143.32 g/mol. Number of moles of AgCl = number of moles of limiting
reactant = 0.1 mol Mass of AgCl = number of moles ×molar mass = 0.1 mol
×143.32 g/mol = 14.332 g
Therefore, when the two solutions are mixed together, 14.332 g of AgCl will
precipitate.
14
Question 18
Question
A chemistry student is conducting an experiment that involves the precipitation
of silver chloride (AgCl). They mix 50.0 mL of 0.100 M silver nitrate (AgNO3)
with 50.0 mL of 0.200 M sodium chloride (NaCl). Given the equilibrium con-
stant Ksp for the dissolution of silver chloride (AgCl) is 1.8×10−10, determine
if a precipitate will form in this experiment.
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction:
AgN O3+N aCl →AgCl +NaNO3
Step 2: Calculate the initial concentration of silver ions (Ag+) and chloride ions
(Cl−) in the solution. Initial concentration of Ag+:
[Ag+]=0.100 M
Initial concentration of Cl−:
[Cl−]=0.200 M
Step 3: Determine the change in concentration of each ion at equilibrium. Let x
be the concentration of AgCl that dissolves/dissociates. The change in concen-
tration of Ag+and Cl−will be +x, and the change in concentration of AgCl
will be -x. At equilibrium:
[Ag+]=0.100 + x
[Cl−]=0.200 + x
[AgCl] = x
Step 4: Write the expression for the solubility product constant:
Ksp = [Ag+][Cl−]
Substitute the equilibrium concentrations:
Ksp = (0.100 + x)(0.200 + x)
Step 5: Solve for x using the given Ksp value:
1.8×10−10 = (0.100 + x)(0.200 + x)
1.8×10−10 = 0.020 + 0.1x+ 0.2x+x2
15
Step 6: Simplify and solve for x using the quadratic formula:
x2+ 0.3x−1.8×10−2= 0
Solving the quadratic equation, we get:
x≈4×10−3M
Step 7: Determine if a precipitate will form by comparing x to the initial con-
centrations of Ag+and Cl−: Since 4 ×10−3M<0.100 M and 4 ×10−3M<
0.200 M, the concentration of AgCl that dissolves is small and will not exceed
the initial concentrations of Ag+and Cl−. Therefore, no precipitate will form
in this experiment.
Question 19
Question
A solution was prepared by dissolving 0.250 moles of silver nitrate in enough
water to make 0.500 L of solution. The solution was then mixed with 0.300 L
of 0.120 M sodium chloride solution. Calculate the mass of the precipitate that
forms.
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between silver nitrate (AgNO3) and sodium chloride (NaCl):
AgNO3(aq) + NaCl(aq)→AgCl(s) + NaNO3(aq)
Step 2: Determine the limiting reagent by calculating the moles of each
reactant: For silver nitrate:
moles of AgNO3= Molarity ×Volume = 0.250 mol
For sodium chloride:
moles of NaCl = Molarity ×Volume = 0.120 ×0.300 = 0.036 mol
Since silver nitrate is the limiting reagent (0.250 mol ¡ 0.036 mol), we will base
our calculations on silver nitrate.
Step 3: Use stoichiometry to find the moles of silver chloride (AgCl) formed:
From the balanced equation: 1 mole of AgNO3produces 1 mole of AgCl. There-
fore, moles of AgCl formed = moles of AgNO3= 0.250 mol
Step 4: Calculate the mass of silver chloride formed: The molar mass of
AgCl is 143.32 g/mol.
Mass of AgCl = moles of AgCl×molar mass of AgCl = 0.250 mol×143.32 g/mol = 35.83 g
Therefore, the mass of the precipitate (silver chloride) that forms is 35.83
grams.
16
Question 20
Question
A solution is prepared by dissolving 25.0 g of calcium nitrate, Ca(NO3)2, in
100.0 mL of water. What is the concentration of calcium ions, in mol/L, in the
solution? (Molar mass of Ca(NO3)2= 164.10 g/mol)
Solution
Step 1: Calculate the number of moles of calcium nitrate.
Moles of Ca(NO3)2=Mass
Molar mass =25.0 g
164.10 g/mol
Moles of Ca(NO3)2= 0.152 mol
Step 2: Determine the number of moles of calcium ions present in the so-
lution. Calcium nitrate, Ca(NO3)2, dissociates into one calcium ion, Ca2+, for
every one formula unit. Therefore, the number of moles of calcium ions is the
same as the number of moles of calcium nitrate.
Moles of Ca2+ = 0.152 mol
Step 3: Calculate the concentration of calcium ions in the solution. The
volume of the solution is given as 100.0 mL or 0.100 L.
Concentration of Ca2+ =Moles of Ca2+
Volume of solution =0.152 mol
0.100 L
Concentration of Ca2+ = 1.52 mol/L
Answer: The concentration of calcium ions in the solution is 1.52 mol/L.
Question 21
Question
Calculate the concentration of sulfate ions in a solution that forms when 50.0
mL of 0.200 M sulfuric acid (H2SO4) is mixed with 150.0 mL of 0.100 M barium
hydroxide (Ba(OH)2). The balanced chemical equation for this reaction is:
H2SO4(aq) + Ba(OH)2(aq)→BaSO4(s) + 2H2O(l)
(Hint: After determining the limiting reagent, use stoichiometry to find the
amount of sulfate ions produced.)
17
Solution
Step 1: Write the balanced chemical equation and determine the limiting reagent.
The balanced equation shows that one mole of sulfuric acid reacts with one
mole of barium hydroxide to produce one mole of barium sulfate.
H2SO4(aq) + Ba(OH)2(aq)→BaSO4(s) + 2H2O(l)
To determine the limiting reagent, calculate the moles of each reactant:
Moles of H2SO4= Volume ×Molarity
= 0.0500 L ×0.200 mol/L
= 0.0100 mol
Moles of Ba(OH)2= Volume ×Molarity
= 0.1500 L ×0.100 mol/L
= 0.0150 mol
Since the reaction requires 1 mole of H2SO4for every 1 mole of Ba(OH)2,
sulfuric acid is the limiting reagent.
Step 2: Calculate the amount of sulfate ions produced.
From the balanced equation, it is clear that 1 mole of barium sulfate produces
1 mole of sulfate ions. Therefore, the moles of sulfate ions produced are equal
to the moles of barium sulfate formed, which is equal to the initial moles of
sulfuric acid:
Moles of sulfate ions = 0.0100 mol
Step 3: Calculate the concentration of sulfate ions.
The final volume of the solution is the sum of the initial volumes of the two
solutions:
Volume of final solution = 0.0500 L + 0.1500 L = 0.2000 L
Finally, the concentration of sulfate ions is given by:
Concentration of sulfate ions = Moles of sulfate ions
Volume of final solution =0.0100 mol
0.2000 L = 0.0500 mol/L
Therefore, the concentration of sulfate ions in the final solution is 0.0500 mol/L,
or 0.0500 M.
Question 22
Question
A solution is prepared by dissolving 45.0 g of potassium bromide (KBr) in 155.0
g of water. Determine if precipitation will occur when 20.0 mL of 0.130 M
lead(II) nitrate (Pb(NO3)2) solution is added to the solution. The solubility
constants for KBr and PbBr2are 6.3×10−2and 6.6×10−6, respectively.
18
Solution
Step 1: Write the chemical equation for potential precipitation to determine if
the given reactants will produce an insoluble product. The possible chemical
reaction is:
Pb(NO3)2+ 2KBr →PbBr2+ 2KNO3
Step 2: Calculate the concentrations of the ions in the solution before mixing
them. First, find the moles of KBr:
moles of KBr = 45.0 g
119.0 g/mol = 0.378 mol
Next, calculate the moles of water:
moles of water = 155.0 g
18.0 g/mol = 8.61 mol
Then, determine the initial concentration of KBr:
initial concentration of KBr = 0.378 mol
8.61 mol = 0.0439 M
Step 3: Determine the solubility of PbBr2in water using the solubility prod-
uct constant. The solubility product constant for PbBr2is 6.6×10−6.
Step 4: Calculate the initial concentration of bromide ions, [Br−]. Since
2 moles of Br−ions are formed for every mole of KBr dissolved, the initial
concentration of Br−ions can be calculated as:
[Br−] = 2 ×0.0439 = 0.0877 M
Step 5: Determine the initial concentration of lead(II) ions, [Pb2+]. Since
the initial concentration of lead(II) nitrate is 0.130 M, the initial concentration
of lead(II) ions is also 0.130 M.
Given concentration of Pb2+ = 0.130 M
Step 6: Determine the ion product, Q, and compare it to the solubility
product constant, Ksp. The ion product, Q, for PbBr2is given by:
Q= [Pb2+][Br−]2
Substitute the values:
Q= (0.130)(0.0877)2= 0.00107
Step 7: Analyze whether precipitation will occur. Since Q(0.00107) >
Ksp(6.6×10−6), precipitation of lead(II) bromide (PbBr2) will occur when 20.0
mL of the lead(II) nitrate solution is added to the potassium bromide solution.
19
Question 23
Question
A solution contains 0.1 M barium chloride (BaCl2) and 0.15 M sodium sulfate
(Na2SO4). Calculate the concentration of the sulfate ion after a precipitation
reaction, where all sulfate reacts with barium to form insoluble barium sulfate
(BaSO4). The solubility product (Ksp) of barium sulfate is 1.1×10−10.
Solution
Step 1: Write the balanced equation for the precipitation reaction between
barium chloride and sodium sulfate:
BaCl2+ Na2SO4→BaSO4+ 2NaCl
Step 2: Determine the limiting reactant. Since barium sulfate is insoluble, it
will precipitate out until one of the ions runs out. Calculate the moles of each ion
in the solution: - Moles of sulfate ion from sodium sulfate: 0.15 M ×2 mol/L =
0.30 mol/L - Moles of barium ion from barium chloride: 0.10 M ×1 mol/L =
0.10 mol/L
Since there are more moles of sulfate ion (limiting reactant), barium will be
completely consumed.
Step 3: Calculate the concentration of sulfate ion after the reaction. Since all
sulfate reacts with barium to form insoluble barium sulfate, the concentration
of sulfate ion after the reaction will be zero.
Therefore, the concentration of the sulfate ion after the precipitation reaction
is 0 M .
Question 24
Question
A solution contains 0.15 M of silver ion (Ag+) and 0.13 M of sulfate ion (SO2−
4).
What is the minimum concentration of sodium chloride (NaCl) that needs to be
added to the solution in order to precipitate all of the silver ion as silver chloride
(AgCl) in a 1.0 L solution? (Hint: The Ksp of silver chloride is 1.77 ×10−10)
Solution
Step 1: Write the balanced chemical equation for the precipitation of silver
chloride:
Ag++ Cl−→AgCl
Step 2: Write the equation for the equilibrium constant (Ksp):
Ksp = [Ag+][Cl−]
20
Step 3: Substitute the initial concentrations into the Ksp expression. Since
Cl−is coming from NaCl, we assume that the concentration of NaCl is equal
to that of Cl−:
Ksp = (0.15)(x)
1.77 ×10−10 = 0.15x
Step 4: Solve for xto find the minimum concentration of NaCl needed:
x=1.77 ×10−10
0.15 = 1.18 ×10−9M
Therefore, the minimum concentration of NaCl that needs to be added to
the solution is 1.18 ×10−9M.
Question 25
Question
At a weather station, the annual precipitation can be modeled by the function
P(t)=5t2−30t+ 40, where trepresents the month of the year with January
as t= 1 and December as t= 12. Calculate the total precipitation for the year.
Solution
Step 1: To calculate the total precipitation for the year, we need to find the
integral of the precipitation function P(t) over the interval [1,12].
Step 2: The integral of the precipitation function P(t)=5t2−30t+ 40 is
given by
Z12
1
(5t2−30t+ 40) dt
Step 3: To find the integral, we first calculate the antiderivative of each
term:
Z5t2dt =5
3t3+C, Z−30t dt =−15t2+C, Z40 dt = 40t+C
Step 4: Applying the antiderivative to the integral, we have
Z12
1
(5t2−30t+ 40) dt =5
3t3−15t2+ 40t12
1
Step 5: Evaluate the antiderivative at the upper and lower limits:
=5
3(12)3−15(12)2+ 40(12)−5
3(1)3−15(1)2+ 40(1)
Step 6: Simplifying further, we get
= (960 −2160 + 480) −5
3−15 + 40
21
Step 7: Finally, compute the total precipitation for the year:
= 280 −5
3+ 15 −40 = 254.33 units
Therefore, the total precipitation for the year is 254.33 units.
Question 26
Question
A solution contains 20 grams of calcium chloride (CaCl2) in 200 mL of water.
If sodium carbonate (Na2CO3) solution is added to this solution, how many
grams of CaCO3will precipitate out?
Solution
Step 1: Write the balanced chemical equation for the reaction between calcium
chloride and sodium carbonate.
CaCl2(aq) + Na2CO3(aq)−→ CaCO3(s)+2NaCl(aq)
Step 2: Calculate the number of moles of calcium chloride present in the
solution. Given: Mass of CaCl2: 20 grams Molar mass of CaCl2: 40.08 (Ca) +
2(35.45) (Cl) = 110.98 g/mol
Number of moles of CaCl2=20 g
110.98 g/mol ≈0.18 mol
Step 3: Calculate the number of moles of CaCO3that can be formed. From
the balanced chemical equation, 1 mole of CaCl2produces 1 mole of CaCO3.
Therefore, number of moles of CaCO3= number of moles of CaCl2= 0.18
mol
Step 4: Calculate the mass of CaCO3formed. Molar mass of CaCO3: 40.08
(Ca) + 12.01 (C) + 3(16.00) (O) = 100.09 g/mol
Mass of CaCO3= number of moles of CaCO3×molar mass of CaCO3Mass
of CaCO3= 0.18 mol ×100.09 g/mol = 18.02 grams
Therefore, 18.02 grams of CaCO3will precipitate out.
Question 27
Question
A solution contains 0.15 M calcium chloride (CaCl2) and 0.20 M silver nitrate
(AgNO3). Determine if a precipitate forms when these two solutions are mixed.
If so, calculate the mass of precipitate formed when 500.0 mL of each solution
are combined.
22
Solution
Step 1: Write the chemical equation for the possible precipitation reaction be-
tween calcium chloride and silver nitrate:
CaCl2+ 2AgNO3−→ Ca(NO3)2+ 2AgCl
Step 2: Determine the possible products of the reaction: - The cation from
CaCl2is calcium (Ca2+). - The anion from AgNO3is nitrate (NO−
3). - The
possible precipitate is silver chloride (AgCl).
Step 3: Determine the net ionic equation for the reaction:
Ca2+ + 2NO−
3+ 2Ag++ 2Cl−−→ Ca2+ + 2NO−
3+ 2AgCl
Step 4: Determine if a precipitate will form by examining the solubility rules.
Silver chloride is insoluble in water, so a white precipitate of silver chloride will
form when these solutions are mixed.
Step 5: Calculate the moles of each reactant: - For calcium chloride:
moles of CaCl2= 0.15 M ×0.5 L = 0.075 mol
- For silver nitrate:
moles of AgNO3= 0.20 M ×0.5 L = 0.100 mol
Step 6: Determine the limiting reactant by looking at the stoichiometry of
the reaction. Since the reaction consumes 2 moles of silver nitrate for every 1
mole of calcium chloride, calcium chloride is the limiting reactant.
Step 7: Calculate the mass of precipitate formed (silver chloride):
moles of AgCl = 0.075 mol ×2 mol AgCl
1 mol CaCl2×143.32 g/mol = 21.41 g
Therefore, when 500.0 mL of 0.15 M calcium chloride and 0.20 M silver
nitrate are mixed, a white precipitate of silver chloride weighing 21.41 g will
form.
Question 28
Question
Calculate the concentration of sulfate ions (SO2−
4) in a solution prepared by
mixing 100.0 mL of 0.200 M sodium sulfate (Na2SO4) with 150.0 mL of 0.400
M barium chloride (BaCl2). Assume complete precipitation of barium sulfate
(BaSO4) occurs.
23
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between sodium sulfate and barium chloride:
Na2SO4(aq) + BaCl2(aq)→BaSO4(s) + 2NaCl(aq)
Step 2: Determine the limiting reactant. Since Na2SO4and BaCl2are in a
1:1 molar ratio, we need to find out which reactant will run out first.
Calculate the moles of Na2SO4:
Moles of Na2SO4= Volume of Na2SO4×Molarity of Na2SO4
= 0.100 L ×0.200 mol/L = 0.020 mol
Calculate the moles of BaCl2:
Moles of BaCl2= Volume of BaCl2×Molarity of BaCl2
= 0.150 L ×0.400 mol/L = 0.060 mol
Since 0.020 mol of Na2SO4is less than 0.060 mol of BaCl2, Na2SO4is the
limiting reactant.
Step 3: Calculate the moles of barium sulfate formed using the limiting
reactant:
Moles of BaSO4= Moles of Na2SO4= 0.020 mol
Step 4: Calculate the concentration of sulfate ions present in the solution
after precipitation:
Volume of final solution = 0.100L+ 0.150L= 0.250L
Concentration of sulfate ions = Moles of sulfate ions
Volume of final solution
=0.020 mol
0.250 L = 0.080 mol/L
Therefore, the concentration of sulfate ions in the final solution is 0.080
mol/L.
Question 29
Question
Calculate the concentration of sulfate ions in a solution after adding excess
barium chloride to 100.0 mL of a 0.200 M sodium sulfate solution. Assume
complete precipitation of barium sulfate.
24
Solution
Step 1: Write the balanced chemical equation for the reaction between sodium
sulfate and barium chloride.
Na2SO4+ BaCl2→BaSO4+ 2NaCl
Step 2: Determine the limiting reactant. Since barium chloride is added in
excess, the limiting reactant is sodium sulfate.
Step 3: Use stoichiometry to find the moles of sulfate ions in the solution.
Moles of Na2SO4= Volume ×Molarity = 0.100 L ×0.200 mol/L = 0.020 mol
Step 4: Since each mole of Na2SO4produces one mole of SO2−
4ions, the
concentration of sulfate ions in the solution is the same as the concentration of
sodium sulfate.
Concentration of SO2−
4= 0.200 M
Question 30
Question
A chemist is tasked with determining the amount of precipitate formed when
100.0 mL of 0.200 M silver nitrate solution is added to 150.0 mL of 0.100 M
sodium chloride solution. Assuming that a precipitate forms according to the
reaction:
AgN O3(aq) + N aCl(aq)→AgCl(s) + NaNO3(aq)
Calculate the mass of silver chloride (AgCl) that will precipitate out.
(Hint: First, determine the limiting reagent in the reaction to find the
amount of precipitate formed.)
Solution
Step 1: Write the balanced chemical equation for the reaction and determine
the limiting reagent. The balanced chemical equation is:
AgN O3(aq) + N aCl(aq)→AgCl(s) + NaNO3(aq)
From the balanced equation, the molar ratio between silver nitrate (AgNO3)
and silver chloride (AgCl) is 1:1, and between sodium chloride (NaCl) and silver
chloride (AgCl) is 1:1.
Calculate the moles of silver nitrate and sodium chloride:
Moles of AgN O3= (0.100 mol/L) * (0.100 L) = 0.020 mol
Moles of NaCl = (0.150 mol/L) * (0.200 L) = 0.030 mol
Since the molar ratios are 1:1, silver nitrate is the limiting reagent because
it produces fewer moles of silver chloride.
Step 2: Calculate the mass of silver chloride precipitate formed. The molar
mass of AgCl is 143.32 g/mol.
25
Calculate the mass of AgCl precipitated out using the moles of AgNO3:
Mass of AgCl = moles of AgN O3* molar mass of AgCl
Mass of AgCl = 0.020 mol * 143.32 g/mol = 2.87 g
Therefore, the mass of silver chloride that will precipitate out is 2.87 grams.
Question 31
Question
A chemist is conducting an experiment that involves mixing two solutions to
form a precipitate. The chemist mixes 150.0 mL of a 0.200 M solution of calcium
chloride with 200.0 mL of a 0.150 M solution of sodium sulfate. Calculate the
mass of calcium sulfate (CaSO4) that will precipitate out of solution.
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction:
CaCl2+N a2SO4→CaSO4+ 2N aCl
Step 2: Determine the limiting reactant by calculating the moles of each
reactant.
Moles of calcium chloride (CaCl2):
0.150 mol/L ×0.150 L = 0.0300 mol
Moles of sodium sulfate (Na2SO4):
0.200 mol/L ×0.150 L = 0.0300 mol
Both reactants have the same number of moles, so CaCl2is the limiting
reactant.
Step 3: Calculate the moles of CaSO4formed using the mole ratio from the
balanced chemical equation.
Moles of CaSO4= 0.0300 mol ×1 mol CaSO4
1 mol CaCl2
= 0.0300 mol
Step 4: Convert moles of CaSO4to grams.
Molar mass of CaSO4:
40.08 g/mol + 32.06 g/mol + 4(16.00 g/mol) = 136.06 g/mol
Mass of CaSO4:
0.0300 mol ×136.06 g/mol = 4.08 g
Therefore, the mass of calcium sulfate precipitated out of solution is 4.08
grams.
26
Question 32
Question
Calculate the molarity of chloride ions in a solution that contains 0.250 moles
of calcium chloride, CaCl2, dissolved in 500.0 mL of solution.
Solution
Step 1: Determine the number of moles of chloride ions present in calcium
chloride. Given that calcium chloride, CaCl2, dissociates into three ions when
dissolved in water, the number of moles of chloride ions is 3 times the moles
of calcium chloride. Number of moles of Cl−ions = 3 ×0.250 moles = 0.750
moles
Step 2: Calculate the total volume of the solution in liters. The volume
provided is 500.0 mL, which is equivalent to 0.500 L.
Step 3: Calculate the molarity of chloride ions. Molarity (M) is defined
as the moles of solute divided by the volume of solution in liters. Therefore,
molarity of chloride ions = 0.750 moles
0.500 L = 1.50 M
Therefore, the molarity of chloride ions in the solution is 1.50 M.
Question 33
Question
Calculate the concentration of sulfate ions in a solution when 100.0 mL of 0.200
M silver sulfate solution is mixed with 150.0 mL of 0.500 M potassium chloride
solution. Assume complete precipitation of silver sulfate.
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between silver sulfate and potassium chloride:
Ag2SO4+ 2KCl −→ 2AgCl +K2SO4
Step 2: Determine the limiting reactant. Since the reaction goes to comple-
tion and all the silver sulfate precipitates, we need to find out how much of the
sulfate ions comes from the silver sulfate:
Moles of Ag2SO4= 100.0 mL ×0.200 mol/L = 0.0200 mol
Moles of SO2−
4= 0.0200 mol ×1 mol SO2−
4
1 mol Ag2SO4
= 0.0200 mol
Step 3: Calculate the final volume of the solution:
Vfinal = 100.0 mL + 150.0 mL = 250.0 mL = 0.250 L
27
Step 4: Calculate the concentration of the sulfate ions in the final solution:
Concentration of SO2−
4=0.0200 mol
0.250 L = 0.0800 M
Therefore, the concentration of sulfate ions in the solution is 0.0800 M.
Question 34
Question
A student is conducting an experiment to determine the concentration of chlo-
ride ions in a water sample by precipitation with silver ions. The student mixes
50.0 mL of the water sample with an excess of silver nitrate solution. The re-
sulting precipitate of silver chloride is then filtered, dried, and found to have a
mass of 0.527 g.
Calculate the concentration of chloride ions (in mol/L) in the original water
sample.
(Given: molar mass of AgCl = 143.32 g/mol)
Solution
Step 1: Calculate the moles of silver chloride precipitate formed.
Moles of AgCl = Mass of AgCl
Molar mass of AgCl
Moles of AgCl = 0.527 g
143.32 g/mol = 0.003675 mol
Step 2: Since the reaction between chloride ions and silver ions is a 1:1 ratio,
the moles of chloride ions is equal to the moles of silver chloride formed.
Moles of chloride ions = 0.003675 mol
Step 3: Calculate the volume of the water sample in liters.
Volume of water sample (L) = 50.0 mL
1000 mL/L = 0.0500 L
Step 4: Calculate the concentration of chloride ions in the original water
sample.
Concentration of chloride ions (mol/L) = Moles of chloride ions
Volume of water sample (L)
Concentration of chloride ions (mol/L) = 0.003675 mol
0.0500 L = 0.0735 mol/L
Therefore, the concentration of chloride ions in the original water sample is
0.0735 mol/L.
28
Question 35
Question
Given that the solubility product constant (Ksp) of lead(II) chloride (PbCl2) is
1.6×10−5at 25◦C, calculate the molar solubility of lead(II) chloride in a 0.050
M hydrochloric acid solution. Assume that the dissolution of PbCl2does not
significantly impact the concentration of Cl−ions in the solution.
Solution
Step 1: Write the equilibrium equation for the dissolution of lead(II) chloride.
The equilibrium equation for the dissolution of PbCl2is:
PbCl2⇌Pb2+ + 2Cl−
Step 2: Write the expression for the solubility product constant (Ksp). The
solubility product constant can be expressed as:
Ksp = [Pb2+][Cl−]2
Step 3: Let xbe the molar solubility of lead(II) chloride in the solution.
After dissociation, we have:
[Pb2+] = x
[Cl−]=2x
Step 4: Substitute the values into the Ksp expression. Plugging in the values,
we get:
1.6×10−5= (x)(2x)2
Step 5: Solve for x.
1.6×10−5= 4x3
x=3
r1.6×10−5
4
x= 0.020 M
Therefore, the molar solubility of lead(II) chloride in a 0.050 M hydrochloric
acid solution is 0.020 M.
29
