Weak acids and bases, Kw, pH and the solubility product
Answers are for parents and teachers.
1
An invented acid has Ka = 2.5 × 10⁻⁷ and C = 0.010 mol/L. Find [H⁺] and the pH (log 5 = 0.70).
[H⁺] = √(2.5 × 10⁻⁷ × 0.010) = √(25 × 10⁻¹⁰) = 5.0 × 10⁻⁵ mol/L. pH = 5 − log 5 = 5 − 0.70 = 4.30.
2
An invented weak base has Kb = 4.0 × 10⁻⁷ and C = 0.10 mol/L. Find [OH⁻], pOH and pH (log 2 = 0.30).
[OH⁻] = √(4.0 × 10⁻⁷ × 0.10) = √(4.0 × 10⁻⁸) = 2.0 × 10⁻⁴ mol/L. pOH = 4 − 0.30 = 3.70, so pH = 14 − 3.70 = 10.30.
3
20 mL of 0.050 mol/L HCl is diluted with water to a total volume of 1.0 L. Find the pH of the new solution.
n(HCl) = 0.050 × 0.020 = 1.0 × 10⁻³ mol, so the new concentration is 1.0 × 10⁻³ ÷ 1.0 = 1.0 × 10⁻³ mol/L, and pH = 3.
4
An invented salt AB₂ (M = 100 g/mol) has s = 5.0 × 10⁻⁵ mol/L. Find Ksp and the solubility in g/L.
Ksp = 4s³ = 4 × (5.0 × 10⁻⁵)³ = 4 × 1.25 × 10⁻¹³ = 5.0 × 10⁻¹³. Solubility = 5.0 × 10⁻⁵ × 100 = 5.0 × 10⁻³ g/L.
5
The salt AB (M = 150 g/mol) has Ksp = 2.5 × 10⁻¹¹. Find s in mol/L and in g/L.
s = √(2.5 × 10⁻¹¹) = √(25 × 10⁻¹²) = 5.0 × 10⁻⁶ mol/L. In g/L: 5.0 × 10⁻⁶ × 150 = 7.5 × 10⁻⁴ g/L.