Questions: Calculate the solubility of Ag2CO3 in water at 25°C. You'll find Ksp data in the ALEKS Data tab. Round your answer to 2 significant digits.

Solution Steps

To calculate the solubility of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) in water at \(25^{\circ} \mathrm{C}\), we need to use the solubility product constant (\(K_{sp}\)) for \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\). Let's assume the \(K_{sp}\) value is provided as \(8.46 \times 10^{-12}\).

The dissolution of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) in water can be represented by the following equation: \[ \mathrm{Ag}_{2} \mathrm{CO}_{3} (s) \rightleftharpoons 2 \mathrm{Ag}^{+} (aq) + \mathrm{CO}_{3}^{2-} (aq) \]

The solubility product expression for this equilibrium is: \[ K_{sp} = [\mathrm{Ag}^{+}]^2 [\mathrm{CO}_{3}^{2-}] \]

Let \(s\) be the solubility of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) in mol/L. Then, at equilibrium:

• The concentration of \(\mathrm{Ag}^{+}\) ions is \(2s\).
• The concentration of \(\mathrm{CO}_{3}^{2-}\) ions is \(s\).

Substitute these into the \(K_{sp}\) expression: \[ K_{sp} = (2s)^2 \cdot s = 4s^3 \]

Substitute the given \(K_{sp}\) value: \[ 8.46 \times 10^{-12} = 4s^3 \] \[ s^3 = \frac{8.46 \times 10^{-12}}{4} = 2.115 \times 10^{-12} \] \[ s = \sqrt[3]{2.115 \times 10^{-12}} \approx 1.287 \times 10^{-4} \, \text{mol/L} \]

The molar mass of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) is approximately \(275.75 \, \mathrm{g/mol}\). Therefore, the solubility in \(\mathrm{g/L}\) is: \[ \text{Solubility} = 1.287 \times 10^{-4} \, \text{mol/L} \times 275.75 \, \mathrm{g/mol} \approx 0.0355 \, \mathrm{g/L} \]

Final Answer