A 0.10 m solution of hf is 8.0 ionised. What is ka
Answers
Answered by
0
Answer:
Explanation:
HA(aq) <=> H+(aq) + A-(aq)
Percent ionization = ([H+]/[HA]) x 100%
1 = ([H+]/0.1 M) x 100
(1) (0.1)/100 = [H+]
0.001 M = [H+]
Answered by
34
Explanation:
HF(aq) <=> H+(aq) + F-(aq)
Percent ionization = ([H+]/[HF]) x 100%
8= ([H+]/0.1 M) x 100
(8) (0.1)/100 = [H+]
0.008 M = [H+]
pH = -log[H+] = -log (0.008) = 2.09
Similar questions
Hindi,
6 months ago
Geography,
6 months ago
Chemistry,
6 months ago
Business Studies,
1 year ago
Science,
1 year ago