1.

The solubility of BaSO_(4) in water is 2.42xx10^(-3)g L^(-1) at 298 K. The value of its solubility product (K_(sp)) will be (Given molar mass of BaSO_(4)=233 g L^(-1))

Answer»

`1.08xx10^(-10) "mol"^(2)L^(-2)`
`1.08xx10^(-12) "mol"^(2)L^(-2)`
`1.08xx10^(-14) "mol"^(2)L^(-2)`
`1.08xx10^(-8) "mol"^(2)L^(-2)`

Solution :Solubility (S) `=2.42xx10^(-3) g L^(-1)`
`=(2.42xx10^(-3))/(233) "mol" L^(-1)`
`=0.0104xx10^(-3) "mol" L^(-1)`
`=1.04xx10^(-5) "mol" L^(-1)`
`{:(BaSO_(4) ,hArr,Ba^(2+),+,SO_(4)^(2-)),(,,S,,S):}`
`K_(sp)=[Ba^(2+)][SO_(4)^(2-)]=SxxS=S^(2)`
`=(1.04xx10^(-5) "mol" L^(-1))^(2)`
`=1.08xx10^(-10) "mol"^(2)L^(-2)`


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