Question

0.10 M solution of HF is 8.0 % associated with what is ka

Answer 1

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