Chemistry, asked by birjesh6796, 1 year ago

A 0.10 m solution of hf is 8.0 ionised. What is ka

Answers

Answered by 7siddharth0
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 itzsakshii
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

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