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The solubility of BaSO4 in water is 2.4...

The solubility of `BaSO_4` in water is ` 2.42 xx 10^(-3) gL^(-1) ` at 298 K. The value of its solubility product `(K_(sp)) ` will be (Given molar mass of `BaSO_4= 233 g " mol"^(-1))`

A

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

B

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

C

`1.08 xx 10^(-10) "mol"^(2)L^(-2)`

D

`1.08 xx 10^(-8) "mol"^(2) L^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
C
Solubility of `BaSO_(4), s=(2.42xx10^(-3))/(233) ("mol" L^(-1))`
`=1.04xx10^(-5) ("mol" L^(-1))`
`BaSO_(4)(s) hArr underset(s)(Ba^(2))(aq) + underset(s)(SO_(4)^(2-))(aq)`
`K_(sp)=[Ba^(2+)][SO_(4)^(2-)]=s^(2)`
`=(1.04xx 10^(-5))^(2)`
`=1.08xx10^(-10) "mol"^(2)L^(-2)`
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