What is the ph of a 0.023 m solution of hcn (ka = 4.9 x 10-10) ?

Question

Isiah

Chemistry

19 November, 01:08

What is the ph of a 0.023 m solution of hcn (ka = 4.9 x 10-10) ?

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Answers (1)

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Cesar Diaz · Answer 1

The ionic equation is simply:

HCN - - > H + + CN-

Therefore the formula for Ka is:

Ka = [H+] [CN-] / [HCN]

[H+] = [CN-] = x

4.9 x 10^-10 = x^2 / 0.023

x^2 = 1.127 x 10^-11

x = 3.357 x 10^-6 = [H+] = [CN-]

pH = - log [H+]

pH = - log 3.357 x 10^-6

pH = 5.47