molar solubility of `Cd(OH)_(2)(Ksp=2.5xx10^(-14))` in 0.1 M KOH solution is :-

Options

A. `2.5xx10^(-7)`M
B. `5xx10^(-8)`M
C. `5xx10^(-7)`M
D. `2.5xx10^(-12)`M

Answer

D

Explanation

`{:(Cd(OH)_(2)hArr,Cd^(2+),2OH^(-),,),(,S^('),(3s^(')+C),,),(KOhto,K^(+),+OH^(-),,),(,C,C,,),(Ksp=(s^('))(C)^(2),,,,):}`
`s^(')=s^(')=(K_(SP))/(C^(2))=(2.5xx10 ^(-14))/(10^(-2))=2.5xx10^(-12)`