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Calculate the pH of 0.01 M solution of N...

Calculate the pH of 0.01 M solution of `NH_(4)CN.` The dissociation constants `K_(a)` for `HCN=6.2xx10^(-10)and K_(b)` for `NH_(3)=1.6xx10^(-5).`

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The correct Answer is:
`9.21`
`NH_(4)CN` is a salt of weak acid, weak base.
Calculate `pK_(a) and pK_(b)`. We get `pK_(a) = 9.21 , pK_(b) = 4.80`. Then `pH = 7 + (1)/(2) [pK_(a) - pK_(b)]`
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