CHEM 122 - GENERAL CHEMISTRY II - Precipitation calculations Question Bank - with Complete Detailed Solutions 2025 - Set 3

CHEM 122- GENERAL
CHEMISTRY II- Precipitation
calculations Question Bank-
with Complete Detailed
Solutions 2025- Set 3
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
A student is conducting an experiment to determine the concentration of chloride
ions in a water sample. The student mixes the water sample with an excess of
silver nitrate (AgNO3) solution, which results in the formation of a white
precipitate of silver chloride (AgCl). The mass of the white precipitate obtained is
0.456 grams. If the molar mass of silver chloride is 143.32 g/mol, what is the
concentration of chloride ions in the original water sample in mol/L?

,Solution
Step 1: Write the balanced chemical equation for the precipitation reaction. The
balanced chemical equation for the precipitation of silver chloride (AgCl) from the
reaction of chloride ions (Cl−) and silver ions (Ag+) is:

Ag+ + Cl− → AgCl

Step 2: Determine the number of moles of silver chloride formed. Given that
the mass of silver chloride formed is 0.456 grams and the molar mass of silver
chloride is 143.32 g/mol, we can calculate the number of moles of silver chloride:

Mass 0.456g
Moles of AgCl = = Molar mass
143.32g/mol

Moles of AgCl ≈ 0.00318mol
Step 3: Determine the number of moles of chloride ions in the original water
sample. From the balanced chemical equation, we know that 1 mole of silver
chloride corresponds to 1 mole of chloride ions. Therefore, the number of moles
of chloride ions in the original water sample is also 0.00318 mol.
Step 4: Determine the volume of the original water sample. The concentration
of chloride ions is given by the formula:
Moles of solute
Concentration (mol/L) =
Volume of solution (L)
Since the number of moles of chloride ions is 0.00318 mol, we need to determine
the volume of the water sample.
Step 5: Calculate the concentration of chloride ions. The concentration of
chloride ions in the original water sample is given by:
0.00318mol
Concentration =
Volume of solution (L)
Therefore, to find the concentration of chloride ions in the original water
sample, we need the volume of the water sample in liters.


Question 2
Question
Calculate the concentration of chloride ions in a solution that results from mixing
200 mL of a 0.5 M sodium chloride solution with 500 mL of a 0.2 M calcium
chloride solution.




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,Solution
Step 1: Calculate the moles of chloride ions from each solution.

Moles of Cl− from sodium chloride = Volume × Molarity
= 0.2L × 0.5M
= 0.1moles


Moles of Cl− from calcium chloride = Volume × Molarity
= 0.5L × 0.2M
= 0.1moles
Step 2: Calculate the total moles of chloride ions.

Total moles of Cl− = 0.1moles + 0.1moles
= 0.2moles
Step 3: Calculate the total volume of the solution.
Total volume = 200mL + 500mL
= 0.2L + 0.5L
= 0.7L
Step 4: Calculate the concentration of chloride ions in the final solution.

− Total moles of Cl− Concentration of
Cl =
Total volume
0.2moles
=
0.7L
≈ 0.2857M
Therefore, the concentration of chloride ions in the final solution is
approximately 0.2857 M.


Question 3
Question
Calculate the concentration (in mol/L) of chloride ions in a solution prepared by
mixing 50 mL of 0.2 M barium chloride (BaCl 2) with 150 mL of 0.1 M sodium
chloride (NaCl).




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, Solution
Step 1: Calculate the moles of chloride ions from each compound. Step 2:
Determine the total volume of the solution. Step 3: Calculate the overall
concentration of chloride ions.
Step 1: The moles of chloride ions in 50 mL of 0.2 M BaCl 2 are:
50 mL × 0.2 mol/L = 0.01 mol
The moles of chloride ions in 150 mL of 0.1 M NaCl are:
150 mL × 0.1 mol/L = 0.015 mol
Step 2: The total volume of the solution is:
50 mL + 150 mL = 200 mL = 0.2 L
Step 3: Adding the moles of chloride ions from both compounds gives:
0.01 mol + 0.015 mol = 0.025 mol
Therefore, the concentration of chloride ions in the solution is:
0.025 mol
= 0.125 mol/L
0.2 L


Question 4
Question
Calculate the concentration of lead(II) iodide (PbI 2) that will precipitate when
200.0 mL of 0.200 M lead(II) nitrate (Pb(NO3)2) is mixed with 300.0 mL of
0.150 M sodium iodide (NaI). Assume the reaction goes to completion.
Solution
Step 1: Write the balanced chemical equation for the precipitation reaction
between lead(II) nitrate and sodium iodide.

Pb(NO3)2 + 2NaI → PbI2 + 2NaNO3

Step 2: Determine the limiting reactant by comparing the number of moles of
each reactant present. For lead(II) nitrate: - Moles of Pb(NO3)2 = 0.200 M × 0.200
L = 0.040 mol For sodium iodide: - Moles of NaI = 0.150 M × 0.300 L = 0.045 mol
Since lead(II) nitrate has fewer moles compared to sodium iodide, lead(II)
nitrate is the limiting reactant.
Step 3: Calculate the moles of lead(II) iodide formed using the limiting
reactant. - Moles of PbI2 = Moles of Pb(NO3)2 = 0.040 mol
Step 4: Calculate the concentration of lead(II) iodide in solution. - Total volume
of solution = 200.0 mL + 300.0 mL = 500.0 mL = 0.500 L - Concentration




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