The pH of a 10^{-10} M NaOH solution is nea | vedclass

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Question

(B) For a very dilute solution of a strong base like $NaOH$,the contribution of $OH^-$ ions from water cannot be neglected.Total $[OH^-] = [OH^-]_{NaO

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) For a very dilute solution of a strong base like $NaOH$,the contribution of $OH^-$ ions from water cannot be neglected.Total $[OH^-] = [OH^-]_{NaOH} + [OH^-]_{H_2O} = 10^{-10} + 10^{-7} \approx 10^{-7} \ M$.$pOH = -\log[OH^-] = -\log(10^{-7}) = 7$.Since $pH + pOH = 14$ at $298 \ K$,$pH = 14 - 7 = 7$.Therefore,the $pH$ is nearest to $7$.

The $pH$ of a $10^{-10} \ M \ NaOH$ solution is nearest to

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