(B) Given,$[OH^{-}] = 1 \times 10^{-5} \ M$. First,calculate the $pOH$ of the solution using the formula: $pOH = - \log [OH^{-}] = - \log (1 \times 10

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(B) Given,$[OH^{-}] = 1 \times 10^{-5} \ M$. First,calculate the $pOH$ of the solution using the formula: $pOH = - \log [OH^{-}] = - \log (1 \times 10^{-5}) = 5$. At $25^{\circ}C$,the relationship between $pH$ and $pOH$ is given by: $pH + pOH = 14$. Substituting the value of $pOH$: $pH = 14 - 5 = 9$. Therefore,the correct option is $(B)$.