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The degree of hydrolysis of ammonium acetate in its 0.1 M aqueous solution at 25 °C is about (Given: $K_{b} ({NH}_{4} OH)$ $= K_{a} (acetic acid) = 1.8 \times 10^{- 5} M)$

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0.552

$5.52 \times 10^{- 2}$

$5.52 \times 10^{- 3}$

$5.52 \times 10^{- 4}$

answer is C.

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Detailed Solution

$K_{h} = \frac{K_{w}}{K_{a} K_{b}} = \frac{(1.0 \times 10^{- 14} M^{2})}{{(1.8 \times 10^{- 5} M^{2})}^{2}}$

$\sqrt{K_{h}} = \frac{(1.0 \times 10^{- 7} M)}{(1.8 \times 10^{- 5} M)} = 5.56 \times 10^{- 3};$

$α = \frac{\sqrt{K_{h}}}{1 + \sqrt{K_{h}}} = \frac{5.56 \times 10^{- 3}}{1 + 5.56 \times 10^{- 3}} = 5.52 \times 10^{- 3}$

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