(b) We know that, $pH = pK_{a} + \log\frac{\lbrack salt\rbrack}{\lbrack acid\rbrack}$ or $pH = - \log{10^{- 8} + \log\frac{1}{1}}$ ($\because\lbrack salt\rbrack = \lbrack acid\rbrack$) or $pH = 8$
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