At 25^{circ} C, pH of a 10^{-8} M aqueous KO | vedclass

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Question

(B) For a very dilute solution of a strong base, the contribution of $OH^{-}$ ions from the dissociation of water cannot be neglected. $KOH \longright

Vedclass · Accepted Answer

$7.02$

Vedclass · Answer

(B) For a very dilute solution of a strong base, the contribution of $OH^{-}$ ions from the dissociation of water cannot be neglected. $KOH \longrightarrow K^{+} + OH^{-}$ $[OH^{-}]_{KOH} = 10^{-8} \ M$ $[OH^{-}]_{water} = 10^{-7} \ M$ Total $[OH^{-}] = 10^{-8} + 10^{-7} = 10^{-7}(0.1 + 1) = 1.1 	imes 10^{-7} \ M$ $pOH = -\log(1.1 	imes 10^{-7}) = 7 - \log(1.1) \approx 7 - 0.0414 = 6.9586$ Since $pH + pOH = 14$ at $25^{\circ} C$,$pH = 14 - 6.9586 = 7.0414 \approx 7.04$. Given the options, the closest value is $7.02$.

At $25^{\circ} C$, $pH$ of a $10^{-8} \ M$ aqueous $KOH$ solution will be

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