Given that K_w for water is 10^{-13} M^2 at | vedclass

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(D) The ionic product of water is defined as $K_w = [H^+][OH^-]$.By definition,$pK_w = -\log(K_w)$.Given $K_w = 10^{-13} \ M^2$,we have $pK_w = -\log(

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$13$

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(D) The ionic product of water is defined as $K_w = [H^+][OH^-]$.By definition,$pK_w = -\log(K_w)$.Given $K_w = 10^{-13} \ M^2$,we have $pK_w = -\log(10^{-13}) = 13$.Since $pH = -\log[H^+]$ and $pOH = -\log[OH^-]$,the relationship between $pH$,$pOH$,and $pK_w$ is given by $pH + pOH = pK_w$.Therefore,at $62 \ ^\circ C$,the sum $pH + pOH = 13$.

Given that $K_w$ for water is $10^{-13} \ M^2$ at $62 \ ^\circ C$,compute the sum of $pOH$ and $pH$ for a neutral aqueous solution at $62 \ ^\circ C$.

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