Question

Given the following solubility data, calculate the solubility product constant for each substance: (a) \(\mathrm{BaSO}_{4}, 3.9 \times 10^{-5} \mathrm{~mol} / \mathrm{L}\) (b) \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}, 7.8 \times 10^{-5} \mathrm{~mol} / \mathrm{L}\) (c) \(\mathrm{CaSO}_{4}, 0.67 \mathrm{~g} / \mathrm{L}\) (d) \(\mathrm{AgCl}, 0.0019 \mathrm{~g} / \mathrm{L}\)

Short answer

Ksp(\(\text{BaSO}_4\)) = \(1.52 \times 10^{-9}\), Ksp(\(\text{Ag}_2\text{CrO}_4\)) = \(1.9 \times 10^{-12}\), Ksp(\(\text{CaSO}_4\)) = \(2.42 \times 10^{-5}\), Ksp(\(\text{AgCl}\)) = \(1.77 \times 10^{-10}\).

Step-by-step solution

- Calculate Solubility for BaSO4

Given the solubility of \(\text{BaSO}_4\) as \(3.9 \times 10^{-5} \text{mol/L}\), the solubility product constant can be calculated using the equation \(K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]\). Since it dissociates into equal molar amounts of \( \text{Ba}^{2+} \) and \( \text{SO}_4^{2-} \), \(K_{sp} = (3.9 \times 10^{-5})^2 = 1.52 \times 10^{-9}\).

- Calculate Solubility for Ag2CrO4

Given the solubility of \(\text{Ag}_2\text{CrO}_4\) as \(7.8 \times 10^{-5} \text{mol/L}\), it dissociates into \( 2 \text{Ag}^+ \) and \( \text{CrO}_4^{2-} \). Therefore, \[K_{sp} = [2 \text{Ag}^+]^2[\text{CrO}_4^{2-}] = (2 \times 7.8 \times 10^{-5})^2 \times (7.8 \times 10^{-5}) = (1.56 \times 10^{-4})^2 \times 7.8 \times 10^{-5} = 1.9 \times 10^{-12}\].

- Convert Solubility of CaSO4 to mol/L

The solubility of \(\text{CaSO}_4\) is given in g/L. Convert it to mol/L using the molar mass \((\text{CaSO}_4) = 136.14 \text{g/mol}\). Thus, the molar solubility is \(\frac{0.67 \text{g/L}}{136.14 \text{g/mol}} = 4.92 \times 10^{-3} \text{mol/L}\).

- Calculate Ksp for CaSO4

Using the molar solubility \(\text{CaSO}_4\), \(K_{sp} = [\text{Ca}^{2+}][\text{SO}_4^{2-}] = (4.92 \times 10^{-3})^2 = 2.42 \times 10^{-5}\).

- Convert Solubility of AgCl to mol/L

The solubility of \(\text{AgCl}\) is given in g/L. Convert it to mol/L using the molar mass \((\text{AgCl}) = 143.32 \text{g/mol}\). Thus, the molar solubility is \(\frac{0.0019 \text{g/L}}{143.32 \text{g/mol}} = 1.33 \times 10^{-5} \text{mol/L}\).

- Calculate Ksp for AgCl

Using the molar solubility of \(\text{AgCl}\), \(K_{sp} = [\text{Ag}^+][\text{Cl}^-] = (1.33 \times 10^{-5})^2 = 1.77 \times 10^{-10}\).

Key concepts

Solubility

Solubility is the maximum amount of a substance that can dissolve in a solvent at a given temperature. It is often expressed in units like grams per liter (g/L) or moles per liter (mol/L). If the solubility is high, more of the substance can dissolve in the solvent. For example, the solubility of \(\text{BaSO}_4\) is given as \(3.9 \times 10^{-5} \text{mol/L}\), meaning that this amount of \(\text{BaSO}_4\) can dissolve per liter of water at the temperature specified.

Ion Dissociation

When ionic compounds dissolve in water, they dissociate into their constituent ions. This is a key concept in understanding solubility and chemical equilibria. For instance, \(\text{BaSO}_4\) dissociates into \( \text{Ba}^{2+} \) and \( \text{SO}_4^{2-} \) ions. Similarly, \(\text{AgCl}\) dissociates into \( \text{Ag}^+ \) and \( \text{Cl}^- \). Knowing how compounds dissociate helps in calculating the solubility product constant because it tells us the concentration of each ion in the solution.

Molar Mass Conversion

Molar mass conversion is essential when solubility data is given in grams per liter (g/L) instead of moles per liter (mol/L). To convert, you divide the given mass concentration by the molar mass of the compound. For example, the solubility of \(\text{CaSO}_4\) is given as 0.67 g/L. Using the molar mass of \( \text{CaSO}_4 (136.14 \text{g/mol})\), we convert it to mol/L by \( \frac{0.67\text{g/L}}{136.14\text{g/mol}} = 4.92 \times 10^{-3} \text{mol/L}\). This conversion is crucial for further calculations.

Chemical Equilibrium

Chemical equilibrium is when the rate of the forward reaction equals the rate of the reverse reaction in a closed system. This balance allows us to calculate the solubility product constant (\(K_{sp}\)). The \(K_{sp}\) value is the product of the molar concentrations of the dissociated ions raised to the power of their coefficients in the balanced equation. For example, for \(\text{Ag}_2\text{CrO}_4\), it dissociates as \(\text{Ag}_2\text{CrO}_4(s) \rightarrow 2 \text{Ag}^+(aq) + \text{CrO}_4^{2-}(aq)\). Hence, \(K_{sp} = [2 \text{Ag}^+]^2[\text{CrO}_4^{2-}]\). This method helps us understand how much of each ion is present in the solution at equilibrium.