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JIPMER 2009
Exam
JIPMER
Year
2009
Subject
Chemistry
Chapter
Ionic Equilibrium

JIPMER 2009 · Chemistry · Ionic Equilibrium

A 0.1 aqueous solution of a weak acid is 2 \% ionised. If the ionic...

A 0.1 aqueous solution of a weak acid is 2%2 \% ionised. If the ionic product of water is 1×10−141 \times 10^{-14}, the [OH−]\left[\mathrm{OH}^{-}\right]is

Options

  1. A5×10−12M5 \times 10^{-12} \mathrm{M}
  2. B2×10−3M2 \times 10^{-3} \mathrm{M}
  3. C1×10−14M1 \times 10^{-14} \mathrm{M}
  4. DNone of these

Correct answer

A. 5×10−12M5 \times 10^{-12} \mathrm{M}

Step-by-step solution

Since, it is a weak acid, the equation to calculate [H+]\left[\mathrm{H}^{+}\right] is =C×α(α=% of  ionisation )=0.1×0.02=2×10−3MKw=[H+][OH−][OH−]=Kw[H+]=1×10−142×10−3=5×10−12M\begin{aligned} & =C \times \alpha(\alpha=\% \text{\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}
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