Question

The solubility of zinc oxalate is$\mathbf{7}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{M}\mathbf{\text{at18°C.}}$Calculate its${\mathit{K}}_{\mathbf{sp}}$.

Short answer

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

Step-by-step solution

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 ${\text{ZnC}}_{\text{2}}{\text{O}}_{\text{4}}$:

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

role="math" localid="1663309639326" $$\begin{gathered} \left(Zn{C}_2{O}_4\right)=7.9\times {10}^{-3}\mathrm{M} \\ {K}_{sp}=\left[Z{n}^{2+}\right]\left[C_2{O}_4^{2-}\right] \\ \left[Z{n}^{2+}\right]=S \\ \left[Z{n}^{2+}\right]=7.9\times {10}^{-3}\mathrm{M} \end{gathered}$$

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

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