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Which of the following is correct for buffer [Salt = S; acid = A] ?

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$\large p{K_a}\, = \,pH\, + \,\log \frac{{[S]}}{{[A]}}$

$pH\, = \,p{K_a}\, + \,\log \frac{{[S]}}{{[A]}}$

$[{H^ + }]\, = \,{10^{ - p{K_a}}}$

$pH\, = \, - \log \frac{{[S]}}{{[A]}}$

answer is B.

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Detailed Solution

\mathop {C{H_3}COOH}\limits_{a - x} \,\,\,\,\,\, \rightleftharpoons \,\,\,\,\mathop {C{H_3}CO{O^ - }}\limits_{x + s} + \mathop {{H^ + }}\limits_x

\mathop {C{H_3}COONa}\limits_{s - s} \,\,\,\,\,\, \rightleftharpoons \,\,\,\,\mathop {C{H_3}CO{O^ - }}\limits_{s + x} + \mathop {{H^ + }}\limits_s

K_{a} = \frac{(x + s) (H^{+})}{(a - x)} [H^{+}] = K_{a} \frac{(acid)}{(salt)}; neglecting x

$p^{H} = p^{K_{a}} + \log \frac{(salt)}{(acid)}$ .

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