Question

Given that the dissociation constant for $ H_{2}O $ is $ K_{w} =1\times 10^{-14} $ what is the pH of a $ M \, KOH $ solution?

• A. $ 11 $
• B. $ 3 $
• C. $ 14 $
• D. $ 10^{-11} $

Answer 1

Answer: (A)

Given, $K_{w}=1 \times 10^{-14} mol ^{2} L ^{-2}$
Molarity of $KOH =$ normality of $K O H $
$N=1 \times 10^{-3}$ or $\left[O H^{-}\right]=1 \times 10^{-3}\left[H^{+}\right]$
$=\frac{1 \times 10^{-14}}{\left[O H^{-}\right]}=\frac{1 \times 10^{-14}}{1 \times 10^{-3}}=1 \times 10^{-11}$
$ p H=-\log \left[H^{+}\right]$
$p H=-\log 10^{-11} p H=11$