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