Strong and weak electrolytes, Ka and Ostwald’s law
Answers are for parents and teachers.
1
Say whether each is a strong or a weak electrolyte and write its ionisation: (a) KNO₃, (b) HF.
(a) strong, ionic: KNO₃ → K⁺ + NO₃⁻; (b) weak acid: HF ⇌ H⁺ + F⁻.
2
For an invented acid HA at equilibrium, [H₃O⁺] = [A⁻] = 3.0 × 10⁻⁴ mol/L and [HA] = 0.020 mol/L. Find Ka.
Ka = (3.0 × 10⁻⁴)² ÷ 0.020 = 9.0 × 10⁻⁸ ÷ 0.020 = 4.5 × 10⁻⁶.
3
An invented acid has Ka = 1.0 × 10⁻⁷. Find α at C = 0.0010 mol/L, and then after the solution is diluted to 0.00025 mol/L.
α = √(1.0 × 10⁻⁷ ÷ 0.0010) = √(1.0 × 10⁻⁴) = 0.010. After dilution α = √(1.0 × 10⁻⁷ ÷ 0.00025) = √(4.0 × 10⁻⁴) = 0.020: it doubles because C is four times smaller.
4
Two invented acids at C = 0.010 mol/L have Ka = 4.0 × 10⁻⁶ (acid X) and Ka = 4.0 × 10⁻⁸ (acid Y). Find α of each and say how many times larger the first is.
α(X) = √(4.0 × 10⁻⁶ ÷ 0.010) = √(4.0 × 10⁻⁴) = 0.020. α(Y) = √(4.0 × 10⁻⁸ ÷ 0.010) = √(4.0 × 10⁻⁶) = 0.0020. The first is 0.020 ÷ 0.0020 = 10 times larger.
5
Why does the concentration of H⁺ in a weak acid fall when the acid is diluted, although α rises? Answer in two sentences.
[H⁺] = αC. When the solution is diluted C falls faster than α rises (α only grows with the square root of the dilution), so [H⁺] becomes smaller.