Question

Question:A solution is prepared by mixing 250.0 mL of $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{M}\mathbf{}\mathbf{Ce}{\left({\mathrm{NO}}_3\right)}_{\mathbf{3}}$ and 150.0 mL of $\mathbf{10}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{M}\mathbf{,}{\mathbf{KIO}}_{\mathbf{3}}$ at 25°C. Determine whether $\mathbf{Ce}{\left({\mathbf{IO}}_{\mathbf{3}}\right)}_{\mathbf{3}}\left(\mathbf{s}\right)$ ( ${\mathbf{K}}_{\mathbf{sp}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}$) tends to precipitate from this mixture.

Short answer

Answer

A precipitate of $Ce{\left(I{O}_3\right)}_3$ is formed during this reaction.

Step-by-step solution

The relation between reaction quotient and solubility product

When

Where, Q_spis the reaction quotient, K_sp is the solubility product.

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

The value of reaction quotient

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

$Ce{\left(N{O}_3\right)}_3\left(aq\right)+3KI{O}_3\left(aq\right)\to Ce{\left(I{O}_3\right)}_3\left(s\right)+3KN{O}_3\left(aq\right)$

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

$$\begin{gathered} Q_{sp}=\left[C{e}^{3+}\right]{\left[I{O_3}^{-}\right]}^3 \\ \left[C{e}^{3+}\right]=\frac{2.0\times {10}^{-3} M\times 0.250 L}{\left(0.250+0.150\right) L} \\ \left[I{O_3}^{-}\right]=\frac{10\times {10}^{-2} M\times 0.150 L}{\left(0.250+0.150\right) L} \\ =3.75\times {10}^{-2} M \\ {Q}_{sp}=\left[C{e}^{3+}\right]{\left[I{O_3}^{-}\right]}^3 \\ =\left(1.25\times {10}^{-3}\right){\left(3.75\times {10}^{-2}\right)}^3 \\ =6.59\times {10}^{-8} \end{gathered}$$

Explanation

Given the value of $K_{sp}=1.9\times {10}^{-10}$.

Hence, $Q_{sp}>K_{sp}$. A precipitate of $Ce{\left(I{O}_3\right)}_3$ is formed during this reaction.