Practice Calculations Test - Solubility Equilibrium & Electrochemistry
Part A) Solubility Equilibrium
Section 1: Calculating Ksp from Molar Solubility
1. 1:1 Stoichiometry: The molar solubility of silver bromide (AgBr) is 8.8 × 10⁻⁷ mol/L. Calculate the Ksp.
2. 2:1 Stoichiometry: The solubility of Ag2CO3 is 1.3 × 10⁻⁴ mol/L at 25°C. Using the relationship Ksp =
(2x)²(x), calculate the Ksp.
3. Hydroxide Concentrations: In a saturated solution of Ca(OH)2, the concentration of hydroxide ions
[OH⁻] is found to be double the concentration of calcium ions [Ca²⁺]. If the molar solubility (x) is 0.011
mol/L, calculate the Ksp value using the expression Ksp = [Ca²⁺][OH⁻]².
4. Reverse Unit Calculation: A specific salt has a molar solubility of 1.5 × 10⁻⁵ mol/L. If this salt dissociates
into one cation and three anions (AB3), set up and solve for the Ksp using the appropriate algebraic
power.
5. Finding Ksp from Mass: A student finds that a maximum of 0.0012 moles of MgF2 dissolves in 1 L of
water. Use this information to calculate the Ksp for MgF2.
Section 2: Calculating Solubility from Ksp
1. Basic Square Root: The Ksp for CaCO3 is 4.8 × 10⁻⁹. Calculate its molar solubility in water.
2. Cube Root Calculation: Lead(II) iodide, PbI2, has a Ksp of 9.8 × 10⁻⁹ at 25°C. Solve for the molar
solubility (x) given the expression Ksp = 4x³.
, Practice Calculations Test - Solubility Equilibrium & Electrochemistry
3. Mass-Based Solubility: The Ksp for MgF2 is 6.4 × 10⁻⁹. Calculate the solubility in g/L (Molar mass of
MgF2 ≈ 62.3 g/mol).
4. Individual Ion Concentration: Using a Ksp of 3.2 × 10⁻¹¹ for CaF2, calculate the equilibrium
concentration of the fluoride ion [F⁻] specifically.
5. Comparison of Solubility: Given Ksp = 1.1 × 10⁻¹⁰ for Salt A (1:1 ratio) and Ksp = 1.1 × 10⁻¹⁰ for Salt B
(1:2 ratio), calculate the molar solubility for both and identify which is actually more soluble in water.
Section 3: The Common Ion Effect (Quantitative)
1. Solving with a Common Anion: Calculate the molar solubility of PbCrO4 (Ksp = 2.3 × 10⁻¹³) in a 0.10
mol/L solution of Na2CrO4.
2. The "500 Rule" Check: Perform the calculation to determine if x is negligible in the following scenario:
A salt with Ksp = 2.3 × 10⁻¹³ is added to a 0.10 M solution. Does the ratio (Initial Concentration / Ksp)
exceed 500?
3. Common Cation Effect: Calculate the solubility of BaCrO4 in a solution where 0.20 mol/L of BaCl2 has
already been dissolved.
4. Solubility Reduction Ratio: Calculate the solubility of AgCl (Ksp = 1.8 × 10⁻¹⁰) in pure water versus its
solubility in 0.50 M NaCl. By what factor did the solubility decrease?
5. High Concentration Common Ion: Determine the solubility of MgF2 (Ksp = 6.4 × 10⁻⁹) in a 0.15 M
solution of NaF, ensuring you account for the coefficient of the fluoride ion in your Ksp expression.
Section 4: Trial Ion Product (Qsp) and Mixing
1. Extreme Dilution: 0.050 mL of 6.0 mol/L AgNO3 is added to 1.0 L of 0.10 mol/L NaCl. Calculate the
new concentrations and determine if AgCl precipitates (Ksp = 1.8 × 10⁻¹⁰).
Part A) Solubility Equilibrium
Section 1: Calculating Ksp from Molar Solubility
1. 1:1 Stoichiometry: The molar solubility of silver bromide (AgBr) is 8.8 × 10⁻⁷ mol/L. Calculate the Ksp.
2. 2:1 Stoichiometry: The solubility of Ag2CO3 is 1.3 × 10⁻⁴ mol/L at 25°C. Using the relationship Ksp =
(2x)²(x), calculate the Ksp.
3. Hydroxide Concentrations: In a saturated solution of Ca(OH)2, the concentration of hydroxide ions
[OH⁻] is found to be double the concentration of calcium ions [Ca²⁺]. If the molar solubility (x) is 0.011
mol/L, calculate the Ksp value using the expression Ksp = [Ca²⁺][OH⁻]².
4. Reverse Unit Calculation: A specific salt has a molar solubility of 1.5 × 10⁻⁵ mol/L. If this salt dissociates
into one cation and three anions (AB3), set up and solve for the Ksp using the appropriate algebraic
power.
5. Finding Ksp from Mass: A student finds that a maximum of 0.0012 moles of MgF2 dissolves in 1 L of
water. Use this information to calculate the Ksp for MgF2.
Section 2: Calculating Solubility from Ksp
1. Basic Square Root: The Ksp for CaCO3 is 4.8 × 10⁻⁹. Calculate its molar solubility in water.
2. Cube Root Calculation: Lead(II) iodide, PbI2, has a Ksp of 9.8 × 10⁻⁹ at 25°C. Solve for the molar
solubility (x) given the expression Ksp = 4x³.
, Practice Calculations Test - Solubility Equilibrium & Electrochemistry
3. Mass-Based Solubility: The Ksp for MgF2 is 6.4 × 10⁻⁹. Calculate the solubility in g/L (Molar mass of
MgF2 ≈ 62.3 g/mol).
4. Individual Ion Concentration: Using a Ksp of 3.2 × 10⁻¹¹ for CaF2, calculate the equilibrium
concentration of the fluoride ion [F⁻] specifically.
5. Comparison of Solubility: Given Ksp = 1.1 × 10⁻¹⁰ for Salt A (1:1 ratio) and Ksp = 1.1 × 10⁻¹⁰ for Salt B
(1:2 ratio), calculate the molar solubility for both and identify which is actually more soluble in water.
Section 3: The Common Ion Effect (Quantitative)
1. Solving with a Common Anion: Calculate the molar solubility of PbCrO4 (Ksp = 2.3 × 10⁻¹³) in a 0.10
mol/L solution of Na2CrO4.
2. The "500 Rule" Check: Perform the calculation to determine if x is negligible in the following scenario:
A salt with Ksp = 2.3 × 10⁻¹³ is added to a 0.10 M solution. Does the ratio (Initial Concentration / Ksp)
exceed 500?
3. Common Cation Effect: Calculate the solubility of BaCrO4 in a solution where 0.20 mol/L of BaCl2 has
already been dissolved.
4. Solubility Reduction Ratio: Calculate the solubility of AgCl (Ksp = 1.8 × 10⁻¹⁰) in pure water versus its
solubility in 0.50 M NaCl. By what factor did the solubility decrease?
5. High Concentration Common Ion: Determine the solubility of MgF2 (Ksp = 6.4 × 10⁻⁹) in a 0.15 M
solution of NaF, ensuring you account for the coefficient of the fluoride ion in your Ksp expression.
Section 4: Trial Ion Product (Qsp) and Mixing
1. Extreme Dilution: 0.050 mL of 6.0 mol/L AgNO3 is added to 1.0 L of 0.10 mol/L NaCl. Calculate the
new concentrations and determine if AgCl precipitates (Ksp = 1.8 × 10⁻¹⁰).
