Quick Chem Question, please someone help!?
2016-02-29 6:52 pm
I am confused on how to calculate pOH from Ka values? The question give you 0.15 M of HClO and a Ka value= 3.5x10^-8 and you have to find pOH. how would you do this??
回答 (1)
2016-02-29 7:07 pm
HClO(aq) + H₂O(l) ⇌ ClO⁻(aq) + H₃O⁺(aq) .... Kₐ
At eqm :
[ClO⁻] = [H₃O⁺] = y M
[HClO] = (0.15 - y) M
As Kₐ is small, assume that y ≪ 0.15.
Then, [HClO] = (0.15 - y) M ≈ 0.15 M
Kₐ = [ClO⁻] [H₃O⁺] / [HClO]
y² / 0.15 = 3.5 × 10⁻⁸
y = 7.25 × 10⁻⁵
pOH = 14 - pH = 14 - [-log(7.25 × 10⁻⁵)] = 9.9
OR :
pOH = -log[OH⁻] = -log(Kw/[H⁺]) = -log[(1 × 10⁻¹⁴)/(7.25 × 10⁻⁵)] = 9.9
At eqm :
[ClO⁻] = [H₃O⁺] = y M
[HClO] = (0.15 - y) M
As Kₐ is small, assume that y ≪ 0.15.
Then, [HClO] = (0.15 - y) M ≈ 0.15 M
Kₐ = [ClO⁻] [H₃O⁺] / [HClO]
y² / 0.15 = 3.5 × 10⁻⁸
y = 7.25 × 10⁻⁵
pOH = 14 - pH = 14 - [-log(7.25 × 10⁻⁵)] = 9.9
OR :
pOH = -log[OH⁻] = -log(Kw/[H⁺]) = -log[(1 × 10⁻¹⁴)/(7.25 × 10⁻⁵)] = 9.9
收錄日期: 2021-04-18 14:39:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160229105208AAO1BH7