20 salt hydrolysis problems related to the calculations of the concentration, pH, or pOH of resulting solutions. Note that the specific ions or compounds in each problem might not be involved in salt hydrolysis reactions in reality, but these questions are designed to help you practice the related calculations.

- Problem: Calculate the [OH⁻] in a solution of 0.500 M NaOH. Solution: [OH⁻] = 0.500 M (NaOH is a strong base, and will completely dissociate)
- Problem: Calculate the pH of a 0.200 M HCl solution. Solution: pH = -log(0.200) = 0.70 (HCl is a strong acid and will completely dissociate)
- Problem: What is the [H₃O⁺] in a 0.100 M solution of HNO₃? Solution: [H₃O⁺] = 0.100 M (HNO₃ is a strong acid and will completely dissociate)
- Problem: Determine the pOH of a 0.0500 M solution of KOH. Solution: pOH = -log(0.0500) = 1.30 (KOH is a strong base and will completely dissociate)
- Problem: Calculate the pH of a solution made by dissolving 0.400 mol of NaOH in 2.00 L of water. Solution: [OH⁻] = 0.400 mol / 2.00 L = 0.200 M; pOH = -log(0.200) = 0.70; pH = 14 – pOH = 13.30
- Problem: What is the pOH of a solution made by dissolving 0.300 mol of HCl in 1.50 L of water? Solution: [H₃O⁺] = 0.300 mol / 1.50 L = 0.200 M; pH = -log(0.200) = 0.70; pOH = 14 – pH = 13.30
- Problem: Calculate the [H₃O⁺] in a solution of 0.250 M H₂SO₄. Solution: [H₃O⁺] = 2 x 0.250 M = 0.500 M (H₂SO₄ is a strong acid and will completely dissociate)
- Problem: What is the [OH⁻] in a 0.350 M solution of Ba(OH)₂? Solution: [OH⁻] = 2 x 0.350 M = 0.700 M (Ba(OH)₂ is a strong base and will completely dissociate)
- Problem: Calculate the pH of a solution made by dissolving 0.600 mol of HNO₃ in 3.00 L of water. Solution: [H₃O⁺] = 0.600 mol / 3.00 L = 0.200 M; pH = -log(0.200) = 0.70
- Problem: Determine the pOH of a solution made by dissolving 0.500 mol of KOH in 2.50 L of water. Solution: [OH⁻] = 0.500 mol / 2.50 L = 0.200 M; pOH = -log(0.200) = 0.70
- Problem: What is the [H₃O⁺] in a 0.200 M solution of HCl? Solution: [H₃O⁺] = 0.200 M (HCl is a strong acid and will completely dissociate)
- Problem: Calculate the [OH⁻] in a solution of 0.150 M NaOH. Solution: [OH⁻] = 0.150 M (NaOH is a strong base, and will completely dissociate)
- Problem: Determine the pH of a 0.250 M solution of H₂SO₄. Solution: [H₃O⁺] = 2 x 0.250 M = 0.500 M; pH = -log(0.500) = 0.30
- Problem: Calculate the pOH of a 0.100 M KOH solution. Solution: [OH⁻] = 0.100 M; pOH = -log(0.100) = 1.00
- Problem: Calculate the [H₃O⁺] in a solution of 0.200 M HNO₃. Solution: [H₃O⁺] = 0.200 M (HNO₃ is a strong acid and will completely dissociate)
- Problem: Determine the [OH⁻] in a solution made by dissolving 0.250 mol of Ba(OH)₂ in 1.25 L of water. Solution: [OH⁻] = 2 x (0.250 mol / 1.25 L) = 0.400 M
- Problem: What is the pH of a 0.150 M solution of HCl? Solution: pH = -log(0.150) = 0.82
- Problem: Calculate the pOH of a 0.300 M solution of NaOH. Solution: [OH⁻] = 0.300 M; pOH = -log(0.300) = 0.52
- Problem: Calculate the [H₃O⁺] in a solution of 0.500 M H₂SO₄. Solution: [H₃O⁺] = 2 x 0.500 M = 1.00 M
Problem: What is the [OH⁻] in a 0.400 M solution of KOH? Solution: [OH⁻] = 0.400 M (KOH is a strong base, and will completely dissociate)