Chapter 18, Problem 61E

Interpretation Introduction

Interpretation:

The solubility product constant for silver carbonate is to be stated.

Concept introduction:

Solubility product is the product of concentrations of ions raised to the power of their stoichiometric coefficients at equilibrium. It is represented by ${\text{K}}_{\text{sp}}$. It is applicable for saturated solutions.

Answer to Problem 61E

The solubility product constant for silver carbonate is $8.001\times {10}^{-12}$.

Explanation of Solution

The equilibrium equation for the magnesium fluoride is shown below.

${\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}\rightleftarrows 2{\text{Ag}}^{+}+{\text{CO}}_{\text{3}}{}^{2-}$

The solubility product constant for magnesium fluoride is calculated by the formula shown below.

${\text{k}}_{\text{sp}}={\left[{\text{Ag}}^{+}\right]}^2\left[{\text{CO}}_3{}^{2-}\right]$…(1)

The molar concentration of silver carbonate $\left[{\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}\right]$ is calculated as formula shown below.

$\left[{\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}\right]=\frac{1000\times {\text{mass}}_{{\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}}}{{\left(\text{Molar}\text{mass}\right)}_{{\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}}\times {\text{volume}}_{\text{mL}}}$…(2)

The value of ${\text{mass}}_{{\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}}$, ${\left(\text{Molar}\text{mass}\right)}_{{\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}}$, and ${\text{volume}}_{\text{mL}}$ are $\text{8}\text{.75}\text{mg}$, $275.\text{8}\text{g}/\text{mol}$ and $2.50\times {10}^{\text{2}}\text{mL}$ respectively.

Substitute the value of ${\text{mass}}_{{\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}}$, ${\left(\text{Molar}\text{mass}\right)}_{{\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}}$, and ${\text{volume}}_{\text{mL}}$ in equation (2)

$\begin{array}{rll}\left[{\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}\right]& =& \frac{1000\times \text{8}\text{.7}\text{mg}\times \left(\frac{\text{1}\text{g}}{\text{1000}\text{mg}}\right)}{275.\text{8}\text{g}/\text{mol}\times \text{250}\text{mL}}\\ & =& \frac{\text{8}\text{.7}}{68950}\text{mol}/\text{mL}\\ & =& 1.26\times {10}^{-4}\text{M}\end{array}$

The value of $\left[{\text{Ag}}^{+}\right]$ is calculated as shown below.

$\left[{\text{Ag}}^{+}\right]=2\left[{\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}\right]$…(3)

Substitute the value of ${\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}$ in equation (3).

$\begin{array}{rll}\left[{\text{Ag}}^{+}\right]& =& 2\left[{\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}\right]\\ & =& 2\times 1.26\times {10}^{-4}\text{M}\\ & =& 2.52\times {10}^{-4}\text{M}\end{array}$

The value of $\left[{\text{CO}}_3{}^{2-}\right]$ is calculated as shown below.

$\left[{\text{CO}}_3{}^{2-}\right]=\left[{\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}\right]$…(4)

Substitute the value of ${\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}$ in equation (4).

$\begin{array}{rll}\left[{\text{CO}}_3{}^{2-}\right]& =& \left[{\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}\right]\\ & =& 1.26\times {10}^{-4}\text{M}\end{array}$

Substitute the value of $\left[{\text{Ag}}^{+}\right]$ and $\left[{\text{CO}}_3{}^{2-}\right]$ in equation (1).

$\begin{array}{rll}{\text{k}}_{\text{sp}}& =& {\left[{\text{Ag}}^{+}\right]}^2\left[{\text{CO}}_3{}^{2-}\right]\\ & =& {\left(2.52\times {10}^{-4}\text{M}\right)}^2\left(1.26\times {10}^{-4}\text{M}\right)\\ & =& 8.001\times {10}^{-12}\end{array}$

Conclusion

The solubility product constant for silver carbonate is $8.001\times {10}^{-12}$.