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Salt of Weak Acid and Weak Base

The degree of...

Question

The degree of hydrolysis of ammonium acetate is related to its hydrolysis constant by the expression

$a$ = $\frac{K h}{(1 + K h)}$

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$a$ = $\frac{K h}{(1 + K h^{2})}$

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$a$ = $\frac{\sqrt{K h}}{(1 + \sqrt{K h})}$

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$a$ = $\sqrt{\frac{K h}{(1 + K h)}}$

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Solution

The correct option is C
$a$ = $\frac{\sqrt{K h}}{(1 + \sqrt{K h})}$

We have

$N H_{4}^{+} + A c^{-} + H_{2} O$ = $N H_{4} O H + H A c$

$c (1 - α)$ $c (1 - α)$ $c α$ $c α$

$K_{h}$ = $\frac{[N H_{4} O H] [H A c]}{[N H_{4}^{+}] [A C^{-}]}$ = $\frac{(c α) (c α)}{(c (1 - α)) (c (1 - α))}$ = $\frac{α^{2}}{(1 - α)^{2}}$

or $\frac{α}{1 - α}$ = $\sqrt{K_{h}}$ or $α$ = $\frac{\sqrt{K_{h}}}{1 + \sqrt{K_{h}}}$

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Similar questions

The degree of hydrolysis of ammonium acetate

(i)

{N H}_{4} O H i s 4 \times 10^{- 5}

C H_{3} C O O H i s 1.8 \times 10^{- 5}

p H

p K_{a}

C H_{3} C O O H

p K_{b}

N H_{4} O H

4.76

4.75

(C H_{3} C O O N H_{4})

298

0.01

0.04

0.1 M

5.6 \times 10^{- 10}

25^{o} C

K_{h} = 3.175 \times 10^{- 5}

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