In the equilibrium A^{-} + H_{2}O rightleftha | vedclass

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(A) Given: $K_{a} = 1.0 	imes 10^{-5}$ and concentration $C = 0.001 \ M = 10^{-3} \ M$.$K_{w} = 1.0 	imes 10^{-14}$.The hydrolysis constant $K_{h}$

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$10^{-3}$

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(A) Given: $K_{a} = 1.0 	imes 10^{-5}$ and concentration $C = 0.001 \ M = 10^{-3} \ M$.$K_{w} = 1.0 	imes 10^{-14}$.The hydrolysis constant $K_{h}$ is given by $K_{h} = \frac{K_{w}}{K_{a}} = \frac{10^{-14}}{10^{-5}} = 10^{-9}$.The degree of hydrolysis $(h)$ for a salt of a weak acid and strong base is given by $h = \sqrt{\frac{K_{h}}{C}}$.Substituting the values: $h = \sqrt{\frac{10^{-9}}{10^{-3}}} = \sqrt{10^{-6}} = 10^{-3}$.

In the equilibrium $A^{-} + H_{2}O \rightleftharpoons HA + OH^{-}$ $(K_{a} = 1.0 \times 10^{-5})$,the degree of hydrolysis of a $0.001 \ M$ solution of the salt is:

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