The solubility of BaSO_4 in pure water is 10^ | vedclass

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(B) The solubility of $BaSO_4$ in pure water is $S = 10^{-5} \ M$. $K_{sp} = S^2 = (10^{-5})^2 = 10^{-10}$. In the presence of $0.1 \ M \ K_2SO_4$,the

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$2.33 	imes 10^{-7} \ g \ L^{-1}$

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(B) The solubility of $BaSO_4$ in pure water is $S = 10^{-5} \ M$. $K_{sp} = S^2 = (10^{-5})^2 = 10^{-10}$. In the presence of $0.1 \ M \ K_2SO_4$,the common ion effect occurs due to $SO_4^{2-}$ ions. Let the new solubility be $s'$. $BaSO_4(s) ightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$ $[Ba^{2+}] = s'$,$[SO_4^{2-}] = 0.1 + s' \approx 0.1 \ M$. $K_{sp} = [Ba^{2+}][SO_4^{2-}] = s' 	imes 0.1 = 10^{-10}$. $s' = 10^{-9} \ mol \ L^{-1}$. To convert to $g \ L^{-1}$,multiply by molar mass $(233 \ g \ mol^{-1})$: $s' = 10^{-9} 	imes 233 = 2.33 	imes 10^{-7} \ g \ L^{-1}$.

The solubility of $BaSO_4$ in pure water is $10^{-5} \ M$ at $298 \ K$. The solubility of $BaSO_4$ in $0.1 \ M \ K_2SO_4$ (aq.) solution is: (Molar mass of $BaSO_4 = 233 \ g \ mol^{-1}$)

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