Question

The ph of an aqueous solution of a 0.1 m solution of a weak monoprotic acid, which is 1%, ionized is

Answer 1

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+]

pH = -log[H+] = -log (0.001) = 3

@skb

Answer 2

HIII.......

YOUR ANSWER☺✔

THANKS....❗❗