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  4. At 25°C, pH of a 10-8 M aqueous KOH solution will be

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

WBJEEWBJEE 2013Equilibrium

Solution:

$NaOH \rightleftharpoons Na ^{+}+ OH ^{-} \left[ OH ^{-}\right]=10^{-8} M$
$H _{2} O \rightleftharpoons H ^{+}+ OH ^{-} \left[ OH ^{-}\right]=10^{-7} M$
$\therefore \left[ OH ^{-}\right]_{\text {total }}=\left(10^{-8}+10^{-7}\right) M$
$=10^{-7}(1.1)$
$=1.1 \times 10^{-7}$
$\therefore pOH =\log 1.1 \times 10^{-7} \simeq 6.98$
$\therefore pH =14-6.98$
$=7.02$