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Q.

The hydrolysis constant of ammonium acetate is given by

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a

large frac{{{K_w}}}{{{K_a}}}

b

frac{{{K_w}}}{{{K_b}}}

c

frac{{{K_w}}}{{{K_a}.,{K_b}}}

d

K_a. K_b

answer is C.

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Detailed Solution

large C{H_3}COON{H_4}xrightarrow{{}}C{H_3}CO{O^ - }left( {aq} right) + NH_4^ + left( {aq} right)

Both the acid radical and the basic radical are derivative of weak electrolytes, so both of them will undergo hydrolysis.

large C{H_3}CO{O^ - } + {H_2}O rightleftharpoons C{H_3}COOH + O{H^-}

large NH_4^ + + {H_2}O rightleftharpoons N{H_4}OH + {H^ + }

large OR

large C{H_3}CO{O^ - } + NH_4^ + + {H_2}O rightleftharpoons C{H_3}COOH + N{H_4}OH

For the above hydrolysis reaction, we need to find the hydrolysis cosntant, K_h

Consider the following equilibria,

large NH_4^ + + O{H^ - } rightleftharpoons N{H_4}OH;frac{1}{{{K_b}}} to left( 2 right)

large C{H_3}CO{O^ - } + {H^ + } rightleftharpoons C{H_3}COOH;frac{1}{{Ka}} to left( 3 right)

large {H_2}O rightleftharpoons {H^ + } + O{H^ - };{K_w} to left( 4 right)

large Adding{text{ 2, 3 and 4,}}

large C{H_3}CO{O^ - } + NH_4^ + + {H_2}O rightleftharpoons N{H_4}OH + C{H_3}COOH

large {K_{eq}} = {K_w} times frac{1}{{{K_a}}} times frac{1}{{{K_b}}}

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