Q38 / 200 · WBJEE 2014 · Chemistry

The pH of $10^{-4} \mathrm{M}$ KOH solution will be

  • A
    4
  • B
    11
  • C
    10.5
  • Correct answer
    D
    10

Solution

$10^{-4} \mathrm{M} \mathrm{KOH}=10^{-4} \mathrm{M}\left[\mathrm{OH}^{-}\right]$ We know that $$ \left.\left[\mathrm{H}^{+}\right] \mathrm{IOH}^{-1}\right]=1 \times 10^{-14} $$ $$ \begin{array}{c} {\left[\mathrm{H}^{+}\right] \times 10^{-4}=1 \times 10^{-14}} \\ {\left[\mathrm{H}^{+}\right]=\frac{1 \times 10^{-14}}{1 \times 10^{-4}}=1 \times 10^{-10} \mathrm{M}} \\ \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(1 \times 10^{-10}\right) \\ =10 \end{array} $$

Hence, the correct answer is option: 'D'.

Relevant Questions