The pH of a 2.0M solution of a weak acid HA is 5, what is the Ka for this acid?
回答 (1)
2016-05-31 12:22 am
HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq) ...... Kₐ
At equilibrium :
[H₃O⁺] = 10^(-pH) = 1 × 10⁻⁵ M
[A⁻] = [H₃O⁺] = 1 × 10⁻⁵ M
[HA] = (2.0 - 1 × 10⁻⁵) M ≈ 2.0 M
Kₐ = [H₃O⁺] [A⁻] / [HA] = (1 × 10⁻⁵)² / 2.0 = 5 × 10⁻¹¹ (M)
At equilibrium :
[H₃O⁺] = 10^(-pH) = 1 × 10⁻⁵ M
[A⁻] = [H₃O⁺] = 1 × 10⁻⁵ M
[HA] = (2.0 - 1 × 10⁻⁵) M ≈ 2.0 M
Kₐ = [H₃O⁺] [A⁻] / [HA] = (1 × 10⁻⁵)² / 2.0 = 5 × 10⁻¹¹ (M)
收錄日期: 2021-04-18 14:57:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160530161605AAcDxVd