Question

A 0.1 m solution of hf is 8% dissociated thn ka for hf is

Answer 1

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Please complete your question

Answer 2

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