The pH of 10^{-7} M NaOH is | vedclass

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(B) For a very dilute solution of a strong base like $10^{-7} \ M \ NaOH$,the contribution of $OH^-$ ions from the auto-ionization of water $(H_2O \ri

Vedclass · Accepted Answer

Between $7$ and $8$

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(B) For a very dilute solution of a strong base like $10^{-7} \ M \ NaOH$,the contribution of $OH^-$ ions from the auto-ionization of water $(H_2O ightleftharpoons H^+ + OH^-)$ cannot be neglected.Total $[OH^-] = [OH^-]_{	ext{base}} + [OH^-]_{	ext{water}}$.Let $[H^+] = x$. Then $[OH^-] = 10^{-7} + x$.Since $K_w = [H^+][OH^-] = 10^{-14}$,we have $x(10^{-7} + x) = 10^{-14}$.Solving the quadratic equation $x^2 + 10^{-7}x - 10^{-14} = 0$,we get $x \approx 0.618 	imes 10^{-7}$.$pH = -\log[H^+] = -\log(0.618 	imes 10^{-7}) = 7 - \log(0.618) \approx 7 - (-0.209) = 7.209$.Thus,the $pH$ lies between $7$ and $8$.

The $pH$ of $10^{-7} \ M \ NaOH$ is

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