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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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