1.

The equilibrium `A^(-) + H_(2)O hArr HA + OH^(-) (K_(a) = 1.0 xx 10^(-5))`. The degree of hydrolysis of 0.001 M solution of the salt isA. `10^(-3)`B. `10^(-4)`C. `10^(-5)`D. `10^(-6)`

Answer» Correct Answer - A
`K_(a) = 1.0 xx 10^(-5)`
`K_(h)` = hydrolysis constant
`K_(h) = (K_(w))/(K_(a)) = (10^(-14))/(10^(-5)) = 10^(-9)`
Degree of hydrolysis (h) `= sqrt((K_(h))/(C)) = sqrt((10^(-9))/(0.001))`
`= sqrt(10^(-6)) = 10^(-3) rArr h = 10^(-3)`.


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