Question

The solubility of CuBr is `2 xx 10^(-4)` mol/at `25^(@)C`. The `K_(sp)` value for CuBr is
A. `4 xx 10^(-8) mol^(2) l^(-2)`
B. `4 xx 10^(-11) mol^(2) L^(-1)`
C. `4 xx 10^(-4) mol^(2) l^(-2)`
D. `4 xx 10^(-15) mol^(2) l^(-2)`

Answer 1

Correct Answer - A
`{:(CuBr,hArr,Cu^(+),+,Br^(-)),(K_(sp),,(S),,(S)):}`
`K_(sp) = S^(2) = (2 xx 10^(-4))^(2) = 4 xx 10^(-8)(mol^(2))/(l^(2))`