Solubility

Tutorial 10

    

Let $x$ $\frac{mol}{L}$ be the number of moles of calcium fluoride dissolved in $1$ L of a $0.01$ M solution of $NaF$ at $20^oC$. $x$ is the "solubility" of calcium fluoride in that solution. The dissolution of calcium fluoride $CaF_2(s)$ $Ca^{2+}(aq)$ $+$ $2F^-(aq)$ $K_s$ $=$ $1,7\cdot 10^{-10}\frac{mol}{L}$ produces per liter .......... mol $Ca^{2+}$ and .......... mol $F^-$ Sodium fluoride contributes per liter with .......... mol $F^-$, so that finally we have in the saturated solution : $[Ca^{2+}]$ = ..........$\frac{mol}{L}$ $[F^-]$ = ..........$\frac{mol}{L}$ The solubility product at $25^oC$ of $CaF_2$, expressed in terms of $x$, is therefore equal to $K_s=[Ca^{2+}][F^-]^2$ = .......... $\frac{mol^3}{L^3}$ - Find an equation for the solubility of calcium fluoride in $0.01$ M $NaF$ using those facts! - Solve this equation by means of an equation solver on your computer.

Calculation: ....... $x(2x$ $+$ $0,01)^2$ $=$ $K_s$ $x(2x$ $+$ $0.01)^2$ $=$ $1.7\cdot 10^{-10}$ ....... $x$ $=$ $1.7\cdot 10^{-6}$ ....... Solubility of $CaF_2$ in $ NaF \; 0.01 M$ $ =$ $ 1.7\cdot 10^{-6}\frac{mol}{L}$

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