`ZnCl_(2)` undeoes hydrolysis `ZnCl_(2) + H_(2)O hArr Zn(OH)_(2) + 2HCl`. The overall `K_(b)` for `Zn(OH)_(2)` is `2.5 xx 10^(-12)` at `25^(0)C`. The degree of hydrolysis of `0.001 M ZnCl_(2)` solution isx

Options

A. `(sqrt((K_(H))/(10^(-3))))^(1//2)`
B. `(K_(H))/(10^(-6))`
C. `[(K)/(4 xx 10^(-6))]^(1//3)`
D. `[(K_(H))/(8 xx 10^(-6))]^(1//4)`

Answer

C

Explanation

`{:(Zn^(2+)+,2H_(2)Orarr,Zn(OH)_(2),+2H^(+)),(1,,0,0),((1-h)C,,hC,2hC):}`
`K_(H) = 4C^(2)h^(2),h = ((K_(H))/(4 xx 10^(-6)))^(1//3)`