Solubility

Tutorial 8

    

At $20^oC$, calcium fluoride ($CaF_2$) has a solubility product of $K_s=1,7\cdot10^{-10}\frac{mol^3}{L^3}$ Let $x$ $\frac{mol}{L}$ be the solubility of calcium fluoride. The dissolution of the pure salt is given by: $CaF_2(s)$ $Ca^{2+}(aq)$ $+$ $2F^{-}(aq)$ Therefore, expressed in terms of $x$, the molarity of fluoride ion in the saturated solution is equal to .......(2) The solubility product at $20^oC$ of $CaF_2$, expressed in terms of $x$, is therefore equal to $K_s$ $=$ $[Ca^{2+}][F^{-}]^2$ = ........(1) $\frac{mol^3}{L^3}$ Calculate the solubility using those facts!

Calculation: .......(2) $4x^3$ $=$ $K_s$ $4x^3$ $=$ $1,7\cdot 10^{-10}$ .......(3) $x$ $=$ $(\frac{1,7\cdot 10^{-10}}{4})^{\frac{1}{3}}$ $=$ $3,5\cdot10^{-4}$ .......(4) Solubility of calcium fluoride = $3,5\cdot10^{-4}\frac{mol}{L}$

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