`pH` of a solution of `10 ml`. `1 N` sodium acetate and `50 ml 2N` acetic acid `(K_(a)=1.8xx10^(-5))` is approximately

Correct Answer - A
`pH=pK_(a)+log``([Sal t])/([Acid])`
`pH= -log(1.8xx10^(-5))+log``([10])/([100])`
`=-log 1.8+5+log 10^(-1)`
`= -0.2553+5-1=3.7447` or `=4`

`pH` of a solution of `10 ml`. `1 N` sodium acetate and `50 ml 2N` acetic acid `(K_(a)=1.8xx10^(-5))` is approximately

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