Q19.70P - Chemistry: Molecular Nature Of Matter And Change

Question

The solubility of silver carbonate is $\mathbf{\text{0.032 M}}$ at $\mathbf{\text{20°C.}}$Calculate its$\mathbf{\text{Ksp.}}$

Step-by-Step Solution

Answer

The Ksp of silver carbonate is  $1.31\times {10}^{-4}$.

Concept Introduction

The Qsp -ion-product expression is obtained by multiplying the concentrations of ions generated by dissolution of a chemical. When a solution is saturated, the Qsp value is referred to as the Ksp 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 solid and liquid state in the Ksp equations

 $\mathbf{\text{Ksp = }}\left[{\text{M}}^{\text{2 + }}\right]{\left[{\text{X}}^{\text{ - }}\right]}^{\mathbf{\text{2}}}$.

Calculating the of silver carbonate

Let us calculate the Ksp of silver carbonate.

Given:

Solubility of silver carbonate$\text{ = 0.032 M at 20°C}$ 

To calculate the Ksp of silver carbonate, we first need to write the dissolution equation for ${\text{Ag}}_2{\text{CO}}_3:$

$$\begin{gathered} {\text{Ag}}_2{\text{CO}}_3\left(s\right)\iff 2{\text{Ag}}^{+}\left(aq\right)+{\text{CO}}_3^{2-}\left(aq\right) \\ \text{S}\left({\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}\right)\text{ = 0.032 M } \\ \text{Ksp = }{\left[{\text{Ag}}^{\text{ + }}\right]}^{\text{2}}\left[{\text{CO}}_{\text{3}}^{\text{2 - }}\right] \end{gathered}$$

We squared $\left[{\text{Ag}}^{+}\right]$ because we have two moles of ${\mathrm{Ag}}^{+}$ in the equation.

$$\begin{gathered} \left[{\text{Ag}}^{\text{ + }}\right]\text{ = 2S } \\ =2\times 0.032\mathrm{M} \\ =0.064\mathrm{M} \end{gathered}$$

$$\begin{gathered} \left[C{O}_3^{2-}\right]=S \\ \left[C{O}_3^{2-}\right] =0.032\mathrm{M} \\ Ksp=(2S{)}^2\times S \\ Ksp=1.31\times {10}^{-4} \end{gathered}$$

Therefore, the Ksp of silver carbonate is  $1.31\times {10}^{-4}$.