At 50,^oC,the pH of pure water will be ...... | vedclass

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(D) For pure water,the dissociation equilibrium is $H_2O \rightleftharpoons H^+ + OH^-$. At any temperature,the ionic product of water is $K_w = [H^+]

Vedclass · Accepted Answer

$6.63$

Vedclass · Answer

(D) For pure water,the dissociation equilibrium is $H_2O \rightleftharpoons H^+ + OH^-$. At any temperature,the ionic product of water is $K_w = [H^+][OH^-]$. Since $[H^+] = [OH^-]$ in pure water,$K_w = [H^+]^2$. Taking the negative logarithm on both sides,we get $pK_w = -\log(K_w) = -\log([H^+]^2) = -2\log[H^+] = 2pH$. Therefore,$pH = \frac{pK_w}{2}$. Given $pK_w = 13.26$ at $50\,^oC$,we have $pH = \frac{13.26}{2} = 6.63$.

At $50\,^oC$,the $pH$ of pure water will be ........... (Given: $pK_w = 13.26$ at $50\,^oC$)

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