The ionic product of water at 50^{circ} C is | vedclass

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(C) The ionic product of water is given by $K_w = [H^+][OH^-]$.For pure water,$[H^+] = [OH^-]$,so $K_w = [H^+]^2$.Given $K_w = 4 	imes 10^{-14}$ at $

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$6.7$ and neutral

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(C) The ionic product of water is given by $K_w = [H^+][OH^-]$.For pure water,$[H^+] = [OH^-]$,so $K_w = [H^+]^2$.Given $K_w = 4 	imes 10^{-14}$ at $50^{\circ} C$.Therefore,$[H^+] = \sqrt{4 	imes 10^{-14}} = 2 	imes 10^{-7} \ M$.The $pH$ is calculated as $pH = -\log[H^+] = -\log(2 	imes 10^{-7}) = -(\log 2 + \log 10^{-7}) = -(0.301 - 7) = 6.699 \approx 6.7$.Since the concentration of $H^+$ ions is equal to the concentration of $OH^-$ ions in pure water,the water remains neutral despite the $pH$ value being less than $7$.

The ionic product of water at $50^{\circ} C$ is $4 \times 10^{-14}$. The $pH$ and nature of water is :

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