The pH of a 10^{-8} M solution of HCl in wat | vedclass

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(C) For a very dilute acid solution,the contribution of $H^+$ ions from the auto-ionization of water cannot be neglected. The total concentration of $

Vedclass · Accepted Answer

between $6$ and $7$

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(C) For a very dilute acid solution,the contribution of $H^+$ ions from the auto-ionization of water cannot be neglected. The total concentration of $H^+$ ions is given by: $[H^+]_{total} = [H^+]_{HCl} + [H^+]_{water} = 10^{-8} \ M + 10^{-7} \ M$. $[H^+]_{total} = (0.1 	imes 10^{-7} + 1 	imes 10^{-7}) \ M = 1.1 	imes 10^{-7} \ M$. Now,calculate the $pH$: $pH = -\log[H^+]_{total} = -\log(1.1 	imes 10^{-7})$. $pH = 7 - \log(1.1) \approx 7 - 0.0414 = 6.9586$. Thus,the $pH$ lies between $6$ and $7$.

The $pH$ of a $10^{-8} \ M$ solution of $HCl$ in water is

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