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The hydrolysis constant of CH₃COONa is given by

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

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

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

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

answer is A.

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Detailed Solution

$\large C{H_3}COONa\xrightarrow{{}}C{H_3}CO{O^ - } + N{a^ + }$

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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