what is the Ka of a 0.0981 M solution of hydrocyanic acid with a pH of 6.00?
回答 (2)
2017-08-04 2:45 pm
pH at equilibrium = 6.00
Then, [H⁺] at equilibrium = 10⁻⁶·⁰⁰ M = 1 × 10⁻⁶ M
____________HCN(aq) + __ H₂O(l) ⇌ __H₃O⁺(aq) __ + __ CN⁻(aq) __ Ka
Initial: ______ 0.0981 M ______________ 0 M ___________ 0 M
At eqm: _ (0.0981 - 1×10⁻⁶) M ________ 1×10⁻⁶ M ______ 1×10⁻⁶ M
Ka = [H₃O⁺] [CN⁻] / [HCN] = (1 × 10⁻⁶)² / (0.0981 - 1 × 10⁻⁶) = 1.02 × 10⁻¹¹
Then, [H⁺] at equilibrium = 10⁻⁶·⁰⁰ M = 1 × 10⁻⁶ M
____________HCN(aq) + __ H₂O(l) ⇌ __H₃O⁺(aq) __ + __ CN⁻(aq) __ Ka
Initial: ______ 0.0981 M ______________ 0 M ___________ 0 M
At eqm: _ (0.0981 - 1×10⁻⁶) M ________ 1×10⁻⁶ M ______ 1×10⁻⁶ M
Ka = [H₃O⁺] [CN⁻] / [HCN] = (1 × 10⁻⁶)² / (0.0981 - 1 × 10⁻⁶) = 1.02 × 10⁻¹¹
2017-08-04 4:17 pm
easier:
pH = 0.5(pKa - log(0.0981)) = 6
pKa = 10.99
Ka = 1.02*10^-11 M
pH = 0.5(pKa - log(0.0981)) = 6
pKa = 10.99
Ka = 1.02*10^-11 M
收錄日期: 2021-05-01 13:12:59
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170804062832AA7Gpyt