Q67 / 160 · JIPMER 2009 · Chemistry

A 0.1 aqueous solution of a weak acid is $2 \%$ ionised. If the ionic product of water is $1 \times 10^{-14}$, the $\left[\mathrm{OH}^{-}\right]$is

  • Correct answer
    A
    $5 \times 10^{-12} \mathrm{M}$
  • B
    $2 \times 10^{-3} \mathrm{M}$
  • C
    $1 \times 10^{-14} \mathrm{M}$
  • D
    None of these

Solution

Since, it is a weak acid, the equation to calculate $\left[\mathrm{H}^{+}\right]$is $\begin{aligned} & =C \times \alpha(\alpha=\% \text { of ionisation }) \\ & =0.1 \times 0.02=2 \times 10^{-3} \mathrm{M} \\ K_w & =\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \\ {\left[\mathrm{OH}^{-}\right] } & =\frac{K_w}{\left[\mathrm{H}^{+}\right]}=\frac{1 \times 10^{-14}}{2 \times 10^{-3}} \\ & =5 \times 10^{-12} \mathrm{M} \end{aligned}$

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

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