Question

Question:Suppose 100.0 mL of a $\mathbf{0}.\mathbf{0010}\mathbf{M}{\mathbf{CaCl}}_{\mathbf{2}}$solution is added to 50.0 mL of a$\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{MNaF}$ solution at 25°C. Determine whether${\mathbf{CaF}}_{\mathbf{2}}\left(\mathbf{s}\right)$ (role="math" localid="1661513203101" ${\mathbf{K}}_{\mathbf{sp}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}$) tends to precipitate from this mixture.

Short answer

Answer

A precipitate of CaF₂ is not formed during this reaction.

Step-by-step solution

The relation between reaction quotient and solubility product  

When

$$\begin{gathered} {\mathbf{Q}}_{\mathbf{sp}}\mathbf{<}{\mathbf{K}}_{\mathbf{sp}}\mathbf{,} \mathbf{precipitate} \mathbf{is} \mathbf{not} \mathbf{formed}\mathbf{.} \\ {\mathbf{Q}}_{\mathbf{sp}}\mathbf{=}{\mathbf{K}}_{\mathbf{sp}}\mathbf{,} \mathbf{equilibrium} \mathbf{position}\mathbf{.} \\ {\mathbf{Q}}_{\mathbf{sp}}\mathbf{>}{\mathbf{K}}_{\mathbf{sp}}\mathbf{,} \mathbf{precipitate} \mathbf{is} \mathbf{formed}\mathbf{.} \end{gathered}$$

Where,Q_sp is the reaction quotient, k_sp is the solubility product

The value of reaction quotient

The balanced chemical equation for the given chemical reaction is shown below:

$CaC{l}_2\left(aq\right)+2NaF\left(aq\right)\to Ca{F}_2\left(s\right)+2NaCl\left(aq\right)$

The value of $Q_{sp}$can be calculated as shown below:

$$\begin{gathered} Q_{sp}=\left[C{a}^{2+}\right]{\left[F^{-}\right]}^2 \\ \left[C{a}^{2+}\right]=\frac{0.0010 M\times 0.100 L}{\left(0.100+0.050\right) L} \\ =6.66\times {10}^{-4} M \\ \left[F^{-}\right]=\frac{6.0\times {10}^{-5} M\times 0.050 L}{\left(0.100+0.050\right) L} \\ =2.0\times {10}^{-5} M \\ {Q}_{sp}=\left[C{a}^{2+}\right]{\left[F^{-}\right]}^2 \\ =\left(6.66\times {10}^{-4}\right){\left(2.0\times {10}^{-5}\right)}^2 \\ =2.66\times {10}^{-13} \end{gathered}$$

Explanation

Given the value of $K_{sp}=3.9\times {10}^{-11}$.

Hence, $Q_{sp}<K_{sp}$. No precipitate of CaF₂ is formed during this reaction.