1.

The solubility of BaSO_(4) in water 2.42 xx 10^(3) gL^(-1) at 298 K. The value of solubility product (K_(sp)) will be (Given molar mass of BASO_(4) = 233 g mol^(-1))

Answer»

`1.08 xx 10^(-10) Mol^(2) L^(-2)`
`1.08 xx 10^(-12) mol^(2) L^(-2)`
`1.08 xx 10^(-14) mol^(2) L^(-2)`
`1.08 xx 10^(14) mol^(2) L^(-2)`

SOLUTION :SOLUBILITY of `BaSO_(4) = 2.42 xx 10^(-3) gL^(-1)`
`:. s = (2.42 xx 10^(-3))/(233) = 1.038 xx 10^(-5) mol L^(-1)`
`K_(sp) = s^(2) = (1.038 xx 10^(-5))^(2)`
`= 1.08 xx 10^(-10) mol^(2) L^(-2)`


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