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Q.

The hydrolysis constant of CH3COONa is given by

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a

large {K_h}, = ,frac{{{K_w}}}{{{K_a}}}

b

{K_h}, = ,frac{{{K_w}}}{{{K_b}}}

c

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

d

{K_h}, = ,{K_a}/{K_b}

answer is A.

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Detailed Solution

large C{H_3}COONaxrightarrow{{}}C{H_3}CO{O^ - } + N{a^ + }

CH3COO- is derivative of weak acid and so it will undergo hydrolysis.

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

large {K_H} = {left( {{K_b}} right)_{C{H_3}CO{O^ - }}} = frac{{{K_w}}}{{left( {{K_a}} right)}_{C{H_3}COOH}}

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