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In a solution, $0.02 \mathrm{M}$ acetic acid is $4 \%$ dissociated. The $\left[\mathrm{OH}^{-}\right]$ in the solution is
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The correct answer is:
$1.25 \times 10^{-11}$
$\left[\mathrm{H}^{+}\right]=\sqrt{\alpha^{2} C^{2}}=\sqrt{(0.02)^{2} \cdot(0.04)^{2}}=8 \times 10^{-4}$
$\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14}$
$\therefore\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{8 \times 10^{-4}}=1.25 \times 10^{-11} \mathrm{M}$
$\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14}$
$\therefore\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{8 \times 10^{-4}}=1.25 \times 10^{-11} \mathrm{M}$
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