CHEM 122 - GENERAL CHEMISTRY II - Precipitation calculations Question Bank - with Complete Detailed Solutions 2025 - Set 1

CHEM 122- GENERAL
CHEMISTRY II- Precipitation
calculations Question Bank-
with Complete Detailed
Solutions 2025- Set 1
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 chemistry student is conducting an experiment where a silver nitrate solution
reacts with a sodium chloride solution to produce silver chloride precipitate. The
student mixes 100.0 mL of 0.200 M silver nitrate solution with 150.0 mL of 0.150
M sodium chloride solution. What mass of silver chloride precipitate will form?
(Hint: Silver chloride is insoluble in water and will precipitate out of solution).

,Solution
Step 1: Write the balanced chemical equation for the reaction between silver
nitrate and sodium chloride:

AgNO3 + NaCl → AgCl + NaNO3

Step 2: Determine the limiting reactant by comparing the number of moles of
each reactant. The number of moles of silver nitrate:

moles of AgNO3 = Molarity × Volume (L) moles of

AgNO3 = 0.200mol/L × 0.100L = 0.0200mol

The number of moles of sodium chloride:

moles of NaCl = Molarity × Volume (L) moles of NaCl =

0.150mol/L × 0.150L = 0.0225mol

Since silver nitrate and sodium chloride react in the ratio of 1:1, the limiting
reactant is silver nitrate because it produces fewer moles of product.
Step 3: Calculate the mass of silver chloride precipitate that will form. The
molar mass of silver chloride (AgCl) is approximately 143.32 g/mol.

moles of AgCl = moles of AgNO3 mass of AgCl
= moles of AgCl × molar mass of AgCl mass of AgCl =
0.0200mol × 143.32g/mol = 2.8664g

Therefore, the mass of silver chloride precipitate that will form is 2.8664
grams.


Question 2
Question
A solution contains 250 mL of 0.2 M silver nitrate (AgNO 3) and 300 mL of 0.3
M sodium chloride (NaCl). Calculate the mass of silver chloride (AgCl) that will
precipitate when the two solutions are mixed.
(Hint: AgCl precipitates according to the following reaction: AgNO 3 + NaCl →
AgCl ↓ +NaNO3)


Solution
Step 1: Calculate the moles of AgNO3 and NaCl using the formula n = M×V.




2

,For AgNO3: nAgNO3 = 0.2mol/L × 0.250L = 0.05mol
For NaCl: nNaCl = 0.3mol/L × 0.300L = 0.09mol
Step 2: Determine the limiting reactant by comparing the moles of each
reactant. According to the balanced chemical equation, 1 mole of AgNO 3 reacts
with 1 mole of NaCl to form 1 mole of AgCl. Since the mole ratio is 1:1, the limiting
reactant is the one that produces the least amount of product. In this case, AgNO 3
is the limiting reactant because it produces 0.05 moles of AgCl compared to 0.09
moles from NaCl.
Step 3: Calculate the mass of AgCl precipitated using the molar mass of AgCl
(143.32 g/mol).

Mass of AgCl = nAgCl×Molar mass of AgCl = 0.05mol×143.32g/mol = 7.166g

Therefore, when 250 mL of 0.2 M AgNO3 is added to 300 mL of 0.3 M NaCl,
7.166 g of AgCl will precipitate.


Question 3
Question
Calculate the solubility of lead(II) iodide (PbI2) at 25 C. The Ksp of PbI2 is 7.1 × 10−9.


Solution
Step 1: Write the solubility equilibrium expression for lead(II) iodide:

PbI2 ⇌ Pb2+ + 2I−

The equilibrium constant expression for this reaction is:

Ksp = [Pb2+][I−]2
Step 2: Let’s assume that the solubility of PbI2 is x. This means the equilibrium
concentrations of Pb2+ and I− will be x and 2x respectively.
Step 3: Substitute the equilibrium concentrations into the Ksp expression:

7.1 × 10−9 = (x)(2x)2

Step 4: Solve for x:




3

, Step 5: Calculate the solubility of lead(II) iodide:

x ≈ 5.72 × 10−3 M

Therefore, the solubility of lead(II) iodide at 25 C is approximately 5.72 × 10−3
M.



Question 4
Question
Calculate the molarity of barium chloride (BaCl 2) solution when 50.0 mL of a 0.200
M solution of silver nitrate (AgNO3) is added to 75.0 mL of a solution containing
0.150 M of sodium sulfate (Na2SO4). Assume the reaction goes to completion and
that barium sulfate (BaSO4) precipitates.


Solution
Step 1: Write the balanced chemical equation for the reaction between silver
nitrate and sodium sulfate to find the net ionic equation.
AgNO3 + Na2SO4 → Ag2SO4 + NaNO3
The net ionic equation is:
Ag
Step 2: Determine the limiting reagent to find the moles of barium sulfate
formed.
From the net ionic equation, it is clear that 1 mole of barium sulfate is formed
from 1 mole of silver nitrate.
Moles of silver nitrate = 0.0500 L × 0.200 mol/L = 0.0100 mol
Since the reaction ratio is 1:1, moles of barium sulfate formed = 0.0100 mol
Step 3: Calculate the moles of barium chloride formed using the mole ratio
from the balanced equation.
From the balanced equation for the precipitation reaction:
BaCl2 + Na2SO4 → BaSO4 + NaCl
We have a 1:1 mole ratio between barium chloride and barium sulfate.
Therefore, moles of barium chloride = moles of barium sulfate = 0.0100 mol
Step 4: Calculate the molarity of the barium chloride solution.
Volume of barium chloride solution = 75.0 mL + 50.0 mL = 125.0 mL = 0.1250
L
Molarity of barium chloride solution = 0800 M
Therefore, the molarity of the barium chloride solution is 0.0800 M.


Question 5




4