Ammonia, NH3, is a weak base with a Kb = 1.8 x 10^-5. In a 0.8M solution of ammonia, what is the concentration of the ammonium ion, NH4+??
回答 (1)
2016-02-29 5:50 pm
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)
At eqm :
[NH₄⁺] = [OH⁻] = y M
[NH₃] = (0.8 - y) M ≈ 0.8 M, assuming that y ≪ 0.8
Kb = [NH₄⁺] [OH⁻] / [NH₃]
y² / 0.8 = 1.8 × 10⁻⁵
y² = 1.44 × 10⁻⁵
y = 3.8 × 10⁻³
(It fulfills the assumption that y ≪ 0.8)
At eqm, [NH₄⁺] = 3.8 × 10⁻³ M
At eqm :
[NH₄⁺] = [OH⁻] = y M
[NH₃] = (0.8 - y) M ≈ 0.8 M, assuming that y ≪ 0.8
Kb = [NH₄⁺] [OH⁻] / [NH₃]
y² / 0.8 = 1.8 × 10⁻⁵
y² = 1.44 × 10⁻⁵
y = 3.8 × 10⁻³
(It fulfills the assumption that y ≪ 0.8)
At eqm, [NH₄⁺] = 3.8 × 10⁻³ M
收錄日期: 2021-04-18 14:34:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160229093524AA3vFDA