What is the pH of a 10^{-8} M solution of HC | vedclass

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(D) Since $HCl$ is a strong acid,it dissociates completely in water.However,for very dilute solutions $(10^{-8} \ M)$,the contribution of $H^+$ ions f

Vedclass · Accepted Answer

$6.9$

Vedclass · Answer

(D) Since $HCl$ is a strong acid,it dissociates completely in water.However,for very dilute solutions $(10^{-8} \ M)$,the contribution of $H^+$ ions from the auto-ionization of water cannot be neglected.Let the concentration of $H^+$ ions from water be $x$.The total concentration of $H^+$ ions is $[H^+] = 10^{-8} + x$.The concentration of $OH^-$ ions from water is $[OH^-] = x$.Using the ionic product of water,$K_w = [H^+][OH^-] = 10^{-14}$.$(10^{-8} + x)(x) = 10^{-14} \implies x^2 + 10^{-8}x - 10^{-14} = 0$.Solving this quadratic equation for $x$ using the quadratic formula $x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{-10^{-8} + \sqrt{(10^{-8})^2 - 4(1)(-10^{-14})}}{2} = \frac{-10^{-8} + \sqrt{10^{-16} + 4 	imes 10^{-14}}}{2} \approx 0.95 	imes 10^{-8} \ M$.Total $[H^+] = 10^{-8} + 0.95 	imes 10^{-8} = 1.05 	imes 10^{-8} \ M$ (Correction: $1.05 	imes 10^{-7} \ M$ is the correct total concentration).$pH = -\log[H^+] = -\log(1.05 	imes 10^{-7}) \approx 6.98 \approx 6.9$.Thus,option $(D)$ is the correct answer.

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

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