Questions: Question 44 5 Points The Ksp of Ag2CO3(s) is 8.10 x 10^-12 a. Calculate the solubility of Ag2CO3(s) in water in mol / L. (The answer should have 3 sig figs)

Solution Steps

The dissolution of silver carbonate, \(\mathrm{Ag}_2\mathrm{CO}_3\), in water can be represented by the following equilibrium equation:

\[ \mathrm{Ag}_2\mathrm{CO}_3(s) \rightleftharpoons 2\mathrm{Ag}^+(aq) + \mathrm{CO}_3^{2-}(aq) \]

The solubility product constant, \(K_{sp}\), for this equilibrium is given by:

\[ K_{sp} = [\mathrm{Ag}^+]^2 [\mathrm{CO}_3^{2-}] \]

Let \(s\) be the solubility of \(\mathrm{Ag}_2\mathrm{CO}_3\) in mol/L. At equilibrium, the concentrations of the ions are:

• \([\mathrm{Ag}^+] = 2s\)
• \([\mathrm{CO}_3^{2-}] = s\)

Substitute the expressions for the ion concentrations into the \(K_{sp}\) expression:

\[ K_{sp} = (2s)^2 \cdot s = 4s^3 \]

Given that \(K_{sp} = 8.10 \times 10^{-12}\), we can solve for \(s\):

\[ 4s^3 = 8.10 \times 10^{-12} \]

\[ s^3 = \frac{8.10 \times 10^{-12}}{4} = 2.025 \times 10^{-12} \]

\[ s = \sqrt[3]{2.025 \times 10^{-12}} \]

Calculating the cube root:

\[ s \approx 1.27 \times 10^{-4} \, \text{mol/L} \]

Final Answer