Solubility

Tutorial 7

    

Barium sulfate ($BaSO_4$) has the solubility product $K_s=9,9\cdot10^{-11}\frac{mol^2}{L^2}$. Let $x$ $\frac{mol}{L}$ be the solubility of baryum sulfate. The dissolution of the pure salt is given by: $BaSO_4(s)$ $Ba^{2+}(aq)$ + $SO_4^{2-}(aq)$ The solubility product at $25^oC$ of $BaSO_4$, expressed in terms of $x$, is therefore equal to $K_s$ $=$ $[Ba^{2+}][SO_4^{2-}]$ = ........(1) $\frac{mol^2}{L^2}$ Calculate the solubility using these facts!

Calculation: .......(2) $x^2$ $=$ $K_s$ $x^2$ $=$ $9.9\cdot 10^{-11}$ .......(3) $x$ $=$ $\sqrt{9.9\cdot 10^{-11}}$ $=$ $9.95\cdot10^{-6}$ .......(4) Solubility of barium sulfate = $9.95\cdot10^{-6}\frac{mol}{L}$

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