Q5 / 150 · MHT CET 2009 · Chemistry

The degree of dissociation of a $0.01 \mathrm{M}$ weak acid is $10^{-3}$. Its $\mathrm{pOH}$ is

  • A
    $\underline{5}$
  • B
    3
  • Correct answer
    C
    9
  • D
    11

Solution

$\left[\mathrm{H}^{+}\right]=\sqrt{K_{a} \cdot C}=139$ Or $=\sqrt{\alpha^{2} \cdot C^{2}} \quad\left[\therefore K_{a}=\alpha^{2} C\right]$ $$ \begin{aligned} =& \sqrt{\left(10^{-3}\right)^{2} \cdot(0.01)^{2}} \\ =& 10^{-5} \\ \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] &=-\log 10^{-5} \\ &=5 \end{aligned} $$ $\mathrm{pOH}=14-5=9$

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

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