(B) In an aqueous solution at $298 \ K$,the ionic product of water $(K_w)$ is $1.0 \times 10^{-14}$.We know that $K_w = [H^+][OH^-] = 1.0 \times 10^{-

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(B) In an aqueous solution at $298 \ K$,the ionic product of water $(K_w)$ is $1.0 \times 10^{-14}$.We know that $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$.Taking the negative logarithm on both sides,we get $-\log K_w = -\log([H^+][OH^-]) = -\log(1.0 \times 10^{-14})$.This simplifies to $pK_w = pH + pOH = 14$.