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

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 

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