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What is $\left[\mathrm{NH}_4^{+}\right]$in a solution that is $0.02 \mathrm{M} \mathrm{NH}_3$ $0.01 \mathrm{M} \mathrm{KOH} ?\left[K_b\left(\mathrm{NH}_3\right)=1.8 \times 10^{-5}\right]$
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The correct answer is:
$3.6 \times 10^{-5} \mathrm{M}$
$3.6 \times 10^{-5} \mathrm{M}$
$\mathrm{NH}_4 \mathrm{OH} \rightleftharpoons \mathrm{NH}_4^{+}+\mathrm{OH}^{-}$
$K_b=\frac{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_4 \mathrm{OH}\right]}$
$1.8 \times 10^{-5}=\frac{\left[\mathrm{NH}_4^{+}\right][0.01]}{[0.02]}$
$\therefore \quad\left[\mathrm{NH}_4^{+}\right]=3.6 \times 10^{-5} \mathrm{M}$
$K_b=\frac{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_4 \mathrm{OH}\right]}$
$1.8 \times 10^{-5}=\frac{\left[\mathrm{NH}_4^{+}\right][0.01]}{[0.02]}$
$\therefore \quad\left[\mathrm{NH}_4^{+}\right]=3.6 \times 10^{-5} \mathrm{M}$
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