✔ 最佳答案
As the pH of the solution is very close to 7, the autoprotolysis of water should be considered.
Let x M is the [H⁺] formed from the autoprotolysis of water, and y M be the [H⁺] formed from the dissociation of the acid, HA.
Then, total [H⁺] in the solution = (x + y) M = 10⁻⁶·⁷¹ M = 1.95 * 10⁻⁷ M
Consider the autoprotolysis of water :
____________ H₂O(l) __ ⇌ ___ OH⁻(aq) __ + __ H⁺(aq) ___ Kw = 1.00 * 10⁻¹⁴
Initial: ___________________ 0 M __________ 0 M
Change: _________________ +x M ________ +(x + y) M {y M of H⁺ formed from the acid}
At eqm: __________________ x M ____ (x + y) M = 1.95 * 10⁻⁷ M
At equilibrium : Kw = [OH⁻] [H⁺]
1.00 * 10⁻¹⁴ = x (1.95 * 10⁻⁷)
x = 5.13 * 10⁻⁸
y = (1.95 * 10⁻⁷) - (5.13 * 10⁻⁸) = 1.44 * 10⁻⁷
Consider the dissociation of the acid HA :
____________ HA(aq) __ ⇌ __ A⁻(aq) __ + __ H⁺(aq) ___ Ka
Initial: _____ 0.0173 M ______ 0 M ________ 0 M
Change: ______ -y M _______ +y M _____ +(x + y) M {x M of H⁺ formed from water}
At eqm: __ (0.0173 - y) M ____ y M ______ (x + y) M
At equilibrium :
[HA] = (0.0173 - y) M = (0.0173 - 1.44 * 10⁻⁷) M ≈ 0.0173 M
[A⁻] = y M = 1.44 * 10⁻⁷ M
[H⁺] = (x + y) M = 1.95 * 10⁻⁷ M
Ka = [A⁻] [H⁺] / [HA] = (1.44 × 10⁻⁷) * (1.95 × 10⁻⁷) / 0.0173 = 1.62 * 10⁻¹²
