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

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

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Slightly less than $7$

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(B) For a very dilute acid solution,the contribution of $H^+$ ions from the auto-ionization of water cannot be ignored.Total $[H^+] = [H^+]_{HCl} + [H^+]_{H_2O}$.Given $[H^+]_{HCl} = 10^{-8} \ M$.Let the concentration of $H^+$ from water be $x$.Then $[H^+]_{total} = 10^{-8} + x$.Since the water equilibrium is $[H^+][OH^-] = 10^{-14}$ and $[H^+] = [OH^-] + 10^{-8}$,we solve the quadratic equation $x^2 + 10^{-8}x - 10^{-14} = 0$.Solving this gives $[H^+] \approx 1.05 	imes 10^{-7} \ M$.$pH = -\log(1.05 	imes 10^{-7}) \approx 6.98$.Thus,the $pH$ is slightly less than $7$.

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

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