1.

The solubility of BaSO_(4) in water is 2.33 xx 10^(-3) gm/litre. Its solubility product will be (molecular weight of BaSO_(4) = 233)

Answer»

`1 XX 10^(-5)`
`1 xx 10^(-10)`
`1 xx 10^(-15)`
`1 xx 10^(-20)`

Solution :The solubility of `BaSO_(4)` in g/litre is given `2.33 xx 10^(-3)` `because` in mole/litre. `N = (W)/(m.wt) = 1 xx 10^(-5) = (2.33 xx 10^(-3))/(233)`
Because `BaSO_(4)` is a compound
`K_(sp) = S^(2) = [1 xx 10^(-5)]^(2) = 1 xx 10^(-10)`.


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