How many grams of CaC_2O_4 (molecular weight | vedclass

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(A) The dissociation of $CaC_2O_4$ is given by: $CaC_2O_4(s) \rightleftharpoons Ca^{2+}(aq) + C_2O_4^{2-}(aq)$.Let the solubility be $S \ mol \ L^{-1}

Vedclass · Accepted Answer

$0.0064$

Vedclass · Answer

(A) The dissociation of $CaC_2O_4$ is given by: $CaC_2O_4(s) ightleftharpoons Ca^{2+}(aq) + C_2O_4^{2-}(aq)$.Let the solubility be $S \ mol \ L^{-1}$.Then,$K_{sp} = [Ca^{2+}][C_2O_4^{2-}] = S 	imes S = S^2$.Given $K_{sp} = 2.5 	imes 10^{-9}$,so $S = \sqrt{2.5 	imes 10^{-9}} = \sqrt{25 	imes 10^{-10}} = 5 	imes 10^{-5} \ mol \ L^{-1}$.The mass of $CaC_2O_4$ required for $1 \ L$ solution is: $	ext{Mass} = S 	imes 	ext{Molecular Weight} = 5 	imes 10^{-5} \ mol \ L^{-1} 	imes 128 \ g \ mol^{-1} = 640 	imes 10^{-5} \ g = 0.0064 \ g$.

How many grams of $CaC_2O_4$ (molecular weight $= 128$) on dissolving in $1 \ L$ of distilled water will give a saturated solution? $[K_{sp}(CaC_2O_4) = 2.5 \times 10^{-9} \ mol^2 L^{-2}]$

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