pK_b ofCN^(-)is 4.7Calculate (a) hydrolysis constant and (b) degree of hydrolysis of 0.5M aqueous NaCN solution.

pK_b ofCN^(-)is 4.7Calculate (a) hydrolysis constant and (b) degree of

1.	pK_b ofCN^(-)is 4.7Calculate (a) hydrolysis constant and (b) degree of hydrolysis of 0.5M aqueous NaCN solution.
Answer» Solution :`pk_b `of ` CN^(-) = 4.7 ` therefore `k_b `of ` CN^(-) = 10^(-4.7) = 2 xx 10^(-5)` Hydrolysis constant (K) is given in TERMS of ionic PRODUCT of WATER (`K_w`) as `k_h = (k_w)/(k_a) =k_w xx (k_b)/(k_w)=k_b = 2 xx 10^(-5)` Degreeof hydrolysis(H )is givenas , h`= sqrt( k_h //c) = sqrt(2 xx 10^(-5) //0.5 ) = 6.3 xx 10^(-3)`