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(A) For the hydrolysis of a salt of a weak acid and a strong base,the equilibrium reaction is $A^{-} + H_{2}O \rightleftharpoons HA + OH^{-}$.The equi

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$\frac{K_{w}}{K_{a}}$

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(A) For the hydrolysis of a salt of a weak acid and a strong base,the equilibrium reaction is $A^{-} + H_{2}O \rightleftharpoons HA + OH^{-}$.The equilibrium constant for this reaction is the hydrolysis constant,$K_{h}$.$K_{h} = \frac{[HA][OH^{-}]}{[A^{-}]}$.Multiplying the numerator and denominator by $[H^{+}]$,we get $K_{h} = \frac{[HA][OH^{-}][H^{+}]}{[A^{-}][H^{+}]}$.Since $K_{w} = [H^{+}][OH^{-}]$ and $K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}$,it follows that $K_{h} = \frac{K_{w}}{K_{a}}$.

In the hydrolysis of a salt of a weak acid and a strong base,$A^{-} + H_{2}O \rightleftharpoons HA + OH^{-}$,the hydrolysis constant $(K_{h})$ is equal to...

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