Q19.71P - Chemistry: Molecular Nature Of Matter And Change

Question

The solubility of zinc oxalate is$\mathbf{7}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{M}\mathbf{\text{ at 18 °C.}}$Calculate its${\mathit{K}}_{\mathbf{sp}}$.

Step-by-Step Solution

Answer

The ${\text{K}}_{\text{sp}}$ of zinc oxalate is $6.2\times {10}^{-5}.$ 

Concept Introduction

The  ${\mathit{Q}}_{\mathbf{s}\mathbf{p}}$-ion-product expression is obtained by multiplying the concentrations of ions generated by the dissolution of a chemical. When a solution is saturated, the ${\mathit{Q}}_{\mathbf{sp}}$ value is referred to as the ${\mathit{K}}_{\mathbf{s}\mathbf{p}}$ value (solubility-product constant).

 ${\mathbf{\text{MX}}}_{\mathbf{\text{2}}}{\mathbf{\text{ > > M}}}^{\mathbf{\text{2 + }}}{\mathbf{\text{ + 2X}}}^{\mathbf{\text{ - }}}$

We do not include the solid and liquid states in the ${\mathbf{K}}_{\mathbf{sp}}$ equations

 ${\mathbf{\text{K}}}_{\mathbf{s}\mathbf{p}}\mathbf{\text{= }}\left[{\text{M}}^{\text{2 + }}\right]{\left[{\text{X}}^{\text{ - }}\right]}^{\mathbf{\text{2}}}$.

$\mathbf{\text{(S)}}$ is equal to the concentration of one mol of the ion formed by dissolution of some compound.

information Provided

First, we can write the dissolution equation for ${\text{ZnC}}_{\text{2}}{\text{O}}_{\text{4}}$ :

$Zn{C}_2{O}_4\left(s\right)\iff Z{n}^{2+}\left(aq\right)+C_2{O}_4^{2-}\left(aq\right)S$ 

$$\begin{gathered} \left(Zn{C}_2{O}_4\right)=7.9\times {10}^{-3} \mathrm{M} \\ {K}_{sp}=\left[Z{n}^{2+}\right]\left[C_2{O}_4^{2-}\right] \\ \left[Z{n}^{2+}\right]=S \\ \left[Z{n}^{2+}\right]=7.9\times {10}^{-3}\mathrm{M} \end{gathered}$$ 

$$\begin{gathered} \left[C_2{O}_4^{2-}\right]=S \\ \left[C_2{O}_4^{2-}\right]=7.9\times {10}^{-3}M \\ {K}_{sp}=S\times S \\ {K}_{sp}=6.2\times {10}^{-5}. \end{gathered}$$

Therefore, the $K_{sp}$ of zinc oxalate is $6.2\times {10}^{-5}.$