Question

The solubility product of BaSO4 is 1.2 × 10 raise to the power -9 .Find out its solubility?? Anyone please help?? please

Answer 1

Ksp = [Ba+2] [SO4^-2]
1.2×10^-9 = s × s
s^2 = 1.2 ×10^-9
s = √ 1.2+×10^-9
s = 1.09×10^+3

Answer 2

The solubility of $BaSO_4$ with the given solubility product is $3.4\times10^{-5}$

Given:

Solubility product $(K_{sp})$ = $1.2\times 10^{-9}$

To find:

Solubility (S) = ?

Calculation:

$K_{sp} = [Ba^{2+}]+[SO_4^{2-}]$

$K_{sp} = S\times S$

$K_{sp} = S^2$

$S^2 = K_{sp}$

S = $\sqrt{K_{sp}}$

S = $\sqrt{1.2\times 10^{-9}}$

S = $3.4\times 10^{-5}$

Conclusion:

The solubility of $BaSO_4$ is calculated as $3.4\times 10^{-5}$.

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