The pH of 10^{-7} N HCl is | vedclass

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(B) For a very dilute solution of a strong acid like $10^{-7} \ N \ HCl$,the contribution of $H^+$ ions from the dissociation of water cannot be ignor

Vedclass · Accepted Answer

$6.97$

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(B) For a very dilute solution of a strong acid like $10^{-7} \ N \ HCl$,the contribution of $H^+$ ions from the dissociation of water cannot be ignored.Total $[H^+] = [H^+]_{	ext{acid}} + [H^+]_{	ext{water}}$.$[H^+] = 10^{-7} + 10^{-7} = 2 	imes 10^{-7} \ M$.$pH = -\log[H^+] = -\log(2 	imes 10^{-7})$.$pH = 7 - \log(2) = 7 - 0.3010 = 6.699 \approx 6.97$.

The $pH$ of $10^{-7} \ N \ HCl$ is

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