CHEM 122 Exam 2 - GENERAL CHEMISTRY
II - Precipitation calculations Question
Bank - with Complete Detailed
Solutions 2025 - Set 8
This Exam typically involve mixing solutions, identifying
potential precipitates using solubility rules, calculating moles
of reactants, determining the limiting reactant, and then
finding the concentration (molarity) of ions or the amount of
precipitate formed after the reaction, often using
stoichiometry and <<< Ksp >> concepts, as seen in examples
with Ba(NO₃)₂ and Na₂SO₄ forming BaSO (The Ultimate Guide
for YR. 2026/2027)
Liberty University
Question 1
Question
Calculate the concentration of barium sulfate (BaSO 4) in a solution if 50.0 mL of a
0.0250 M barium nitrate (Ba(NO 3)2) solution is mixed with 25.0 mL of a 0.0200
M sodium sulfate (Na2SO4) solution. The reaction between barium nitrate and
sodium sulfate forms barium sulfate precipitate according to the following
balanced chemical equation:
Ba(NO3)2 + Na2SO4 → BaSO4 ↓ +2NaNO3
,Solution
Step 1: Calculate the moles of Ba 2+ and SO ions in each solution. Moles of
Ba2+ from barium nitrate solution:
Moles of Ba2+ = Volume×Concentration = 50.0mL×0.0250mol/L = 0.00125mol Moles of
SO from sodium sulfate solution:
Moles of SO = Volume×Concentration = 25.0mL×0.0200mol/L = 0.0005mol
Step 2: Determine the limiting reactant in the reaction by comparing the moles of
2+
Ba and SO ions.
Since the balanced chemical equation indicates a 1:1 ratio of Ba(NO 3)2 and
Na2SO4 needed to form BaSO4, the limiting reactant will be the one that produces
the least amount of BaSO4. In this case, SO from sodium sulfate is the limiting
reactant because it produces 0.0005 mol of BaSO 4 compared to 0.00125 mol of
Ba2+ from barium nitrate.
Step 3: Calculate the concentration of BaSO4 ions in the final solution.
The volume of the final solution is 50.0 mL + 25.0 mL = 75.0 mL = 0.0750 L.
Since 0.0005 mol of BaSO4 is produced in this volume, the concentration of BaSO 4
is:
0.0005mol
Concentration of BaSO4 = = 0.00667M
0.0750L
Therefore, the concentration of barium sulfate in the final solution is 0.00667 M.
Question 2
Question
A solution contains 0.3 M calcium nitrate and 0.6 M sodium sulfate. Will a
precipitate form when these two solutions are combined? If so, what mass of the
precipitate is formed?
Solution
Step 1: Write the balanced equation for the precipitation reaction between calcium
nitrate and sodium sulfate:
Ca(NO3)2 + Na2SO4 → CaSO4 + 2NaNO3
2
, Step 2: Determine the possible products of the reaction. Calcium sulfate is
insoluble and will precipitate out of solution, while sodium nitrate will stay in
solution.
Step 3: Determine the net ionic equation for the reaction:
Step 4: Calculate the concentrations of the ions in the combined solution.
Since both solutions are dissolved in water, assume that the volumes are
additive.
For calcium ions: [Ca2+] = 0.3M
For sulfate ions: [
Step 5: Use the concentrations of the ions to determine the reaction quotient, Q,
for the precipitation reaction.
Step 6: Compare the reaction quotient to the solubility product constant, Ksp, for
calcium sulfate. If Q > Ksp, a precipitate will form.
The Ksp for calcium sulfate is 2.4×10−5. Since Q > Ksp (0.18 ¿ 2.4×10−5), a
precipitate of calcium sulfate will form.
Step 7: Calculate the mass of the precipitate formed using the stoichiometry
of the balanced equation. Since 1 mole of calcium sulfate is formed for every 1
mole of calcium nitrate reacted,
Moles of CaSO4 = Molarity of Ca(NO3)2 × Volume of Ca(NO3)2
= 0.3mol/L × volume of calcium nitrate solution in L
Then, use the molar mass of calcium sulfate to calculate the mass of the precipitate
formed.
Question 3
Question
Calculate the concentration of sulfate ions in a solution that results from mixing
250.0 mL of 0.200 M barium chloride and 300.0 mL of 0.150 M sodium sulfate.
Assume complete reaction and that the final volume of the solution is 500.0 mL.
Solution
Step 1: Write the balanced chemical equation for the reaction between barium chloride
(BaCl2) and sodium sulfate (Na2SO4) to determine the mole ratio.
BaCl2 + Na2SO4 → BaSO4 ↓ +2NaCl
3
, Step 2: Calculate the moles of barium chloride and sodium sulfate, respectively,
using the given concentrations and volumes.
Moles of BaCl2 = 0.200M × 0.2500L = 0.0500mol
Moles of Na2SO4 = 0.150M × 0.3000L = 0.0450mol
Step 3: Determine the limiting reactant by comparing the mole ratios from
the balanced chemical equation. Since the mole ratio of BaCl 2 to Na2SO4 is 1:1,
BaCl2 is the limiting reactant.
Step 4: Use the limiting reactant to calculate the moles of barium sulfate formed.
Moles of BaSO4 = 0.0500mol
Step 5: Calculate the concentration of sulfate ions in the final solution. Since 1
mole of barium sulfate produces 1 mole of sulfate ions, the moles of sulfate ions
in the final solution is also 0.0500 mol. Now, we need to find the final volume of
the solution, which is 500.0 mL.
Step 6: Calculate the final concentration of sulfate ions in the solution.
0.0500mol
Concentration of sulfate ions = = 0.100M
0.5000L
Therefore, the concentration of sulfate ions in the final solution is 0.100 M.
Question 4
Question
A solution contains 0.15 M BaCl2 and 0.20 M Na2SO4. Will a precipitation reaction
occur when these solutions are mixed? If so, what mass of BaSO 4 will be formed?
(Given: Ksp = 1.1 × 10−10 for BaSO4)
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction:
BaCl2 (aq) + Na2SO4 (aq) −−→ BaSO4 (s) + 2NaCl (aq)
Step 2: Determine the ions that will be present in solution after mixing: Ba 2+ from
BaCl2 - SO from Na2SO4
Step 3: Calculate the ion product (Qsp) to determine if precipitation will occur:
Qsp = [Ba2+][SO42−] = (0.15)(0.20) = 0.03
Step 4: Compare Qsp to Ksp to determine if precipitation will occur: Since Qsp = 0.03
> Ksp = 1.1 × 10−10, a precipitation reaction will occur.
Step 5: Calculate the mass of BaSO 4 formed: - Let’s assume the volume of the
resulting solution is 1 L for easier calculations. - Convert moles of BaSO 4 to mass
using its molar mass (233.4 g/mol):
4
II - Precipitation calculations Question
Bank - with Complete Detailed
Solutions 2025 - Set 8
This Exam typically involve mixing solutions, identifying
potential precipitates using solubility rules, calculating moles
of reactants, determining the limiting reactant, and then
finding the concentration (molarity) of ions or the amount of
precipitate formed after the reaction, often using
stoichiometry and <<< Ksp >> concepts, as seen in examples
with Ba(NO₃)₂ and Na₂SO₄ forming BaSO (The Ultimate Guide
for YR. 2026/2027)
Liberty University
Question 1
Question
Calculate the concentration of barium sulfate (BaSO 4) in a solution if 50.0 mL of a
0.0250 M barium nitrate (Ba(NO 3)2) solution is mixed with 25.0 mL of a 0.0200
M sodium sulfate (Na2SO4) solution. The reaction between barium nitrate and
sodium sulfate forms barium sulfate precipitate according to the following
balanced chemical equation:
Ba(NO3)2 + Na2SO4 → BaSO4 ↓ +2NaNO3
,Solution
Step 1: Calculate the moles of Ba 2+ and SO ions in each solution. Moles of
Ba2+ from barium nitrate solution:
Moles of Ba2+ = Volume×Concentration = 50.0mL×0.0250mol/L = 0.00125mol Moles of
SO from sodium sulfate solution:
Moles of SO = Volume×Concentration = 25.0mL×0.0200mol/L = 0.0005mol
Step 2: Determine the limiting reactant in the reaction by comparing the moles of
2+
Ba and SO ions.
Since the balanced chemical equation indicates a 1:1 ratio of Ba(NO 3)2 and
Na2SO4 needed to form BaSO4, the limiting reactant will be the one that produces
the least amount of BaSO4. In this case, SO from sodium sulfate is the limiting
reactant because it produces 0.0005 mol of BaSO 4 compared to 0.00125 mol of
Ba2+ from barium nitrate.
Step 3: Calculate the concentration of BaSO4 ions in the final solution.
The volume of the final solution is 50.0 mL + 25.0 mL = 75.0 mL = 0.0750 L.
Since 0.0005 mol of BaSO4 is produced in this volume, the concentration of BaSO 4
is:
0.0005mol
Concentration of BaSO4 = = 0.00667M
0.0750L
Therefore, the concentration of barium sulfate in the final solution is 0.00667 M.
Question 2
Question
A solution contains 0.3 M calcium nitrate and 0.6 M sodium sulfate. Will a
precipitate form when these two solutions are combined? If so, what mass of the
precipitate is formed?
Solution
Step 1: Write the balanced equation for the precipitation reaction between calcium
nitrate and sodium sulfate:
Ca(NO3)2 + Na2SO4 → CaSO4 + 2NaNO3
2
, Step 2: Determine the possible products of the reaction. Calcium sulfate is
insoluble and will precipitate out of solution, while sodium nitrate will stay in
solution.
Step 3: Determine the net ionic equation for the reaction:
Step 4: Calculate the concentrations of the ions in the combined solution.
Since both solutions are dissolved in water, assume that the volumes are
additive.
For calcium ions: [Ca2+] = 0.3M
For sulfate ions: [
Step 5: Use the concentrations of the ions to determine the reaction quotient, Q,
for the precipitation reaction.
Step 6: Compare the reaction quotient to the solubility product constant, Ksp, for
calcium sulfate. If Q > Ksp, a precipitate will form.
The Ksp for calcium sulfate is 2.4×10−5. Since Q > Ksp (0.18 ¿ 2.4×10−5), a
precipitate of calcium sulfate will form.
Step 7: Calculate the mass of the precipitate formed using the stoichiometry
of the balanced equation. Since 1 mole of calcium sulfate is formed for every 1
mole of calcium nitrate reacted,
Moles of CaSO4 = Molarity of Ca(NO3)2 × Volume of Ca(NO3)2
= 0.3mol/L × volume of calcium nitrate solution in L
Then, use the molar mass of calcium sulfate to calculate the mass of the precipitate
formed.
Question 3
Question
Calculate the concentration of sulfate ions in a solution that results from mixing
250.0 mL of 0.200 M barium chloride and 300.0 mL of 0.150 M sodium sulfate.
Assume complete reaction and that the final volume of the solution is 500.0 mL.
Solution
Step 1: Write the balanced chemical equation for the reaction between barium chloride
(BaCl2) and sodium sulfate (Na2SO4) to determine the mole ratio.
BaCl2 + Na2SO4 → BaSO4 ↓ +2NaCl
3
, Step 2: Calculate the moles of barium chloride and sodium sulfate, respectively,
using the given concentrations and volumes.
Moles of BaCl2 = 0.200M × 0.2500L = 0.0500mol
Moles of Na2SO4 = 0.150M × 0.3000L = 0.0450mol
Step 3: Determine the limiting reactant by comparing the mole ratios from
the balanced chemical equation. Since the mole ratio of BaCl 2 to Na2SO4 is 1:1,
BaCl2 is the limiting reactant.
Step 4: Use the limiting reactant to calculate the moles of barium sulfate formed.
Moles of BaSO4 = 0.0500mol
Step 5: Calculate the concentration of sulfate ions in the final solution. Since 1
mole of barium sulfate produces 1 mole of sulfate ions, the moles of sulfate ions
in the final solution is also 0.0500 mol. Now, we need to find the final volume of
the solution, which is 500.0 mL.
Step 6: Calculate the final concentration of sulfate ions in the solution.
0.0500mol
Concentration of sulfate ions = = 0.100M
0.5000L
Therefore, the concentration of sulfate ions in the final solution is 0.100 M.
Question 4
Question
A solution contains 0.15 M BaCl2 and 0.20 M Na2SO4. Will a precipitation reaction
occur when these solutions are mixed? If so, what mass of BaSO 4 will be formed?
(Given: Ksp = 1.1 × 10−10 for BaSO4)
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction:
BaCl2 (aq) + Na2SO4 (aq) −−→ BaSO4 (s) + 2NaCl (aq)
Step 2: Determine the ions that will be present in solution after mixing: Ba 2+ from
BaCl2 - SO from Na2SO4
Step 3: Calculate the ion product (Qsp) to determine if precipitation will occur:
Qsp = [Ba2+][SO42−] = (0.15)(0.20) = 0.03
Step 4: Compare Qsp to Ksp to determine if precipitation will occur: Since Qsp = 0.03
> Ksp = 1.1 × 10−10, a precipitation reaction will occur.
Step 5: Calculate the mass of BaSO 4 formed: - Let’s assume the volume of the
resulting solution is 1 L for easier calculations. - Convert moles of BaSO 4 to mass
using its molar mass (233.4 g/mol):
4
