Question

The precipitate of ${CaF}_2$ ( $K_{sp}=1.7\times {10}^{-10}$) is obtained when equal volumes of the following are mixed
(1) ${10}^{-4}M{Ca}^{2+}+{10}^{-4}{MF}^{-}$
(2)${10}^{-2}M{Ca}^{2+}+{10}^{-3}{MF}^{-}$

Answer 1

Equation

$\begin{array}{c}Ca{F}_2\left(g\right)\to C{a}_2\left(aq\right)+2F\left(aq\right)\\ K_{SP}=\left(C{a}_2^{+}\right)\left(F^{-2}\right)=1.7\times {10}^{-10}\end{array}$

When combining the two solutions, the concentration of each in will be halved from its initial concentration.

So, in solutions the final concentration of the $C{a}^{2+}$ will be $5\times {10}^{-5}$ of the concentration of $F^{-}$ will also be $5\times {10}^{-5}$, so

$\left(5\times {10}^{-5}\right){\left(5\times {10}^{-5}\right)}^2=1.25\times {10}^{-13}$

Since this is below $K_{Sp}$,no precipitate will from,

From, $\left(2\right)$

$\left(5\times {10}^{-3}\right){\left(5\times {10}^{-4}\right)}^2=1.25\times {10}^{-19}$

Since this is larger than $K_{Sp}$, a precipitate will form.

Hence the precipitate of $Ca{F}_2\left(K_{Sp}=1.7\times {10}^{-10}\right)$ is obtained when equal volume are mixed.