The solubility product

Barium sulfate ($BaSO_4$) has the following solubility product: $K_s$ $=$ $9,9\cdot10^{-11}\frac{mol^2}{L^2}$ Let $x$ $\frac{mol}{L}$ be the solubility of barium sulfate. Dissolution of this pure salt is given by: $BaSO_4(s)$ $Ba^{2+}(aq)$ $+$ $SO_4^{2-}(aq)$ The product of solubility at $25^oC$ of $BaSO_4$, expressed as a function of $x$, is therefore worth $K_s=[Ba^{2+}][SO_4^{2-}]$ = ........(1) $\frac{mol^2}{L^2}$ Calculate solubility using these data!

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 = $9,95\cdot10^{-6}\frac{mol}{L}$