Question

the pH of aqueous solution of 1.0 molar solution of weak monoprotic Acid which is 1% ionised 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.......

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