The solubility of BaSO_{4} in water is 2.42 t | vedclass

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Question

(A) The solubility $(s)$ in $mol \ L^{-1}$ is calculated by dividing the solubility in $g \ L^{-1}$ by the molar mass of $BaSO_{4}$.$s = \frac{2.42

Vedclass · Accepted Answer

$1.08 	imes 10^{-10} \ mol^{2} \ L^{-2}$

Vedclass · Answer

(A) The solubility $(s)$ in $mol \ L^{-1}$ is calculated by dividing the solubility in $g \ L^{-1}$ by the molar mass of $BaSO_{4}$.$s = \frac{2.42 	imes 10^{-3} \ g \ L^{-1}}{233 \ g \ mol^{-1}} \approx 1.0386 	imes 10^{-5} \ mol \ L^{-1}$.For a salt like $BaSO_{4}$,the dissociation is $BaSO_{4}(s) ightleftharpoons Ba^{2+}(aq) + SO_{4}^{2-}(aq)$.The solubility product $K_{sp} = [Ba^{2+}][SO_{4}^{2-}] = s 	imes s = s^{2}$.$K_{sp} = (1.0386 	imes 10^{-5})^{2} \approx 1.078 	imes 10^{-10} \ mol^{2} \ L^{-2}$.Rounding to two decimal places,we get $1.08 	imes 10^{-10} \ mol^{2} \ L^{-2}$.

The solubility of $BaSO_{4}$ in water is $2.42 \times 10^{-3} \ g \ L^{-1}$ at $298 \ K$. The value of solubility product $(K_{sp})$ will be (Given molar mass of $BaSO_{4} = 233 \ g \ mol^{-1}$)

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