1.

The solubility of BaSO_4 in water is 2.42xx10^(-3)g*L^(-1) at 298K. The value of its solubility product (K_(sp)) will be ( Given molar mass of BaSO_4=233g*mol^(-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 :`K_(sp)=[Ba^(2+)][SO_4^(2-)]=S^2`
[S=solubility of `BaSO_4`(in `mol*L^(-1)`)]
`S=(2.42xx10^(-3))/(233)mol*L^(-1)=0.0103xx10^(-3)mol*L^(-1)`
`:.K_(sp)=(0.0103xx10^(-3))^2=1.06xx10^(10)mol^2*L^(-2)`


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