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The percentage hydrolysis of $0.15 \mathrm{M}$ solution of ammonium acetate, $\mathrm{K}_{\mathrm{a}}$ for $\mathrm{CH}_{3} \mathrm{COOH}$ is $1.8 \times 10^{-5}$
and $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_{3}$ is $1.8 \times 10^{-5}$
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and $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_{3}$ is $1.8 \times 10^{-5}$
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Verified Answer
The correct answer is:
0.556
$\alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}} \times \mathrm{K}_{\mathrm{b}}}}=\sqrt{\frac{1 \times 10^{-14}}{1.8 \times 10^{-5} \times 1.8 \times 10^{-5}}}=0.55$
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