1.

The solubility of CuBr is 2 xx 10^(-4) mol/at 25^(@)C. The K_(sp) value for CuBr is

Answer»

`4 xx 10^(-8) mol^(2) L^(-2)`
`4 xx 10^(-11) mol^(2) L^(-1)`
`4 xx 10^(-4) mol^(2) l^(-2)`
`4 xx 10^(-15) mol^(2) l^(-2)`

Solution :`{:(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))`


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