The pH of a 10^{-8} M HCl aqueous 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 neglected.Total $[H^+] = [H^+]_{HCl} +

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Between $6$ and $7$

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(B) For a very dilute acid solution,the contribution of $H^+$ ions from the auto-ionization of water cannot be neglected.Total $[H^+] = [H^+]_{HCl} + [H^+]_{H_2O}$.Given $[H^+]_{HCl} = 10^{-8} \ M$.Let the concentration of $H^+$ from water be $x$.Then,$[H^+]_{total} = 10^{-8} + x$ and $[OH^-] = x$.Since $K_w = [H^+][OH^-] = 10^{-14}$,we have $(10^{-8} + x)(x) = 10^{-14}$.$x^2 + 10^{-8}x - 10^{-14} = 0$.Solving this quadratic equation using the formula $x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{-10^{-8} + \sqrt{(10^{-8})^2 - 4(1)(-10^{-14})}}{2} = \frac{-10^{-8} + \sqrt{10^{-16} + 4 	imes 10^{-14}}}{2} \approx 0.95 	imes 10^{-7} \ M$.Total $[H^+] = 10^{-8} + 0.95 	imes 10^{-7} = 1.05 	imes 10^{-7} \ M$.$pH = -\log[H^+] = -\log(1.05 	imes 10^{-7}) = 7 - \log(1.05) \approx 6.98$.Thus,the $pH$ lies between $6$ and $7$.

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

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