(C) The ionic product of water $(K_w)$ at $25^{\circ}C$ is given by the expression: $K_w = [H^+][OH^-] = 10^{-14}$.Taking the negative logarithm on bo

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(C) The ionic product of water $(K_w)$ at $25^{\circ}C$ is given by the expression: $K_w = [H^+][OH^-] = 10^{-14}$.Taking the negative logarithm on both sides:$-\log(K_w) = -\log([H^+][OH^-]) = -\log(10^{-14})$.$pK_w = -(\log[H^+] + \log[OH^-]) = 14$.Since $pH = -\log[H^+]$ and $pOH = -\log[OH^-]$,we get:$pH + pOH = 14$.