Calculate the pH of a 0.095 M solution of HCN at 25 °C. [Ka for HCN (aq) = 4.9 x 10-10].?
回答 (1)
___________HCN(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CN⁻ ___ Ka = 4.9 × 10⁻¹⁰
Initial: _____ 0.095 M __________ 0 M ____ 0 M
Change: _____ -y M __________ +y M ___ +y M
At eqm: __ (0.095 - y) M ________ y M ____ y M
As Ka is very small, assume that 0.095 ≫ y and thus
[HCN] at eqm = (0.095 - y) M ≈ 0.095 M
Ka = [H₃O⁺] [CN⁻] / [HCN]
y² / 0.095 = 4.9 × 10⁻¹⁰
y = √(0.095 × 4.9 × 10⁻¹⁰) = 6.82 × 10⁻⁶ (The assumption that 0.095 ≫ y is correct.)
pH = -log[H₃O]⁺ = -log(6.82 × 10⁻⁶) = 5.2
收錄日期: 2021-04-18 17:22:11
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170725005012AAE7xNx
檢視 Wayback Machine 備份