What will be the pH of a 10^{-8} M HCl solu | vedclass

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(C) For a very dilute acid solution,the contribution of $H^+$ ions from the dissociation of water cannot be ignored.Total $[H^+] = [H^+]_{HCl} + [H^+]

Vedclass · Accepted Answer

$6.958$

Vedclass · Answer

(C) For a very dilute acid solution,the contribution of $H^+$ ions from the dissociation of water cannot be ignored.Total $[H^+] = [H^+]_{HCl} + [H^+]_{H_2O} = 10^{-8} + 10^{-7} \ M$.Total $[H^+] = 10^{-8} + 0.1 	imes 10^{-7} = 1.1 	imes 10^{-7} \ M$.$pH = -\log [H^+] = -\log (1.1 	imes 10^{-7}) = 7 - \log (1.1)$.Since $\log (1.1) \approx 0.0414$,$pH = 7 - 0.0414 = 6.9586 \approx 6.958$.

What will be the $pH$ of a $10^{-8} \ M \ HCl$ solution?

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