Question

If the solubility of \(\mathrm{BaSO}_{4}\) (mol. wt. 233) is \(2.33 \times 10^{-4}\) \(\mathrm{g} / 100 \mathrm{~mL}\) then the solubility product of \(\mathrm{BaSO}_{4}\) is: (a) \(1 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\) (b) \(1 \times 10^{-10} \mathrm{~mol} \mathrm{~L}^{-1}\) (c) \(1 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\) (d) \(1 \times 10^{-8} \mathrm{~mol} \mathrm{~L}^{-1}\)

Short answer

The solubility product of \( \text{BaSO}_4 \) is \( 1 \times 10^{-10} \text{ mol}^2/\text{L}^2 \).

Step-by-step solution

Understand the Concept

We need to determine the solubility product constant, or K_{sp}, for \( \text{BaSO}_4 \). The solubility product is the product of the molar concentrations of the constituent ions raised to the power of their stoichiometric coefficients in the dissolution equation.

Write the Dissolution Equation

The dissolution of \( \text{BaSO}_4 \) in water is represented by the equation: \[ \text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]This shows that one mole of \( \text{BaSO}_4 \) produces one mole each of \( \text{Ba}^{2+} \) and \( \text{SO}_4^{2-} \) ions.

Calculate Molar Solubility

The given solubility of \( \text{BaSO}_4 \) is \( 2.33 \times 10^{-4} \text{ g/100 mL} \). First, convert grams to moles using the formula:\[ \text{Molarity} = \frac{\text{Grams/Liter}}{\text{Molecular Weight}} \]Convert \( 100 \text{ mL} = 0.1 \text{ L} \), so the solubility in g/L is:\[ 2.33 \times 10^{-4} \text{ g/100 mL} \times 10 = 2.33 \times 10^{-3} \text{ g/L} \]Convert to moles:\[ \frac{2.33 \times 10^{-3} \text{ g/L}}{233 \text{ g/mol}} = 1 \times 10^{-5} \text{ mol/L} \]

Calculate Solubility Product

Since each \( \text{BaSO}_4 \) molecule produces one \( \text{Ba}^{2+} \) and one \( \text{SO}_4^{2-} \), we know that:\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = (1 \times 10^{-5} \text{ mol/L}) \times (1 \times 10^{-5} \text{ mol/L}) \]Thus, the solubility product \( K_{sp} = 1 \times 10^{-10} \text{ mol}^2/\text{L}^2 \).

Key concepts

Molar Solubility

Molar solubility is a measure of how much of a substance can dissolve in a solvent to form a solution at a specific temperature. It is expressed in terms of moles per liter (mol/L). To find the molar solubility of a compound, you need to convert the compound's solubility from mass terms (grams per liter) to moles per liter.
For example, the exercise provides the solubility of \(\mathrm{BaSO}_4\) as \(2.33 \times 10^{-4}\) g/100 mL. First, you convert mL to L, obtaining \(2.33 \times 10^{-3}\) g/L.
Next, use the formula:

• \[\text{Molarity} = \frac{\text{Grams/Liter}}{\text{Molecular Weight}}\]
• \(\text{Molarity} = \frac{2.33 \times 10^{-3} \text{ g/L}}{233 \text{ g/mol}} = 1 \times 10^{-5} \text{ mol/L}\)

This calculation shows that the molar solubility of \(\mathrm{BaSO}_4\) is \(1 \times 10^{-5}\) mol/L.

Dissolution Equation

The dissolution equation reveals how a solid compound dissolves in water to form ions. The key to understanding this process is to recognize how many ions form from one unit of the compound. This equation is crucial in finding the solubility product (\(K_{sp}\)) since it shows the stoichiometry, or molar ratio, of ions produced.
For \(\mathrm{BaSO}_4\), the dissolution process is represented by:

• \[\text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq)\]

This equation indicates that one mole of \(\text{BaSO}_4\) breaks into one mole of \(\text{Ba}^{2+}\) ions and one mole of \(\text{SO}_4^{2-}\) ions in solution. This 1:1 stoichiometric ratio is crucial for calculating \(K_{sp}\).
Understanding the dissolution equation helps determine how changes in concentration affect the solubility and how to establish equilibrium conditions for the dissolving species.

Ionic Concentrations

Ionic concentrations are the amounts of individual ions present in a solution, expressed in moles per liter (mol/L). Calculating ionic concentrations is essential for determining the solubility product constant (\(K_{sp}\)).
In the solution, when \(\mathrm{BaSO}_4\) dissolves completely, the concentration of \(\text{Ba}^{2+}\) ions is equal to the concentration of \(\text{SO}_4^{2-}\) ions because the dissolution equation, \[\text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq)\], shows a 1:1 ratio of these ions.
If the molar solubility of \(\mathrm{BaSO}_4\) is \(1 \times 10^{-5} \text{ mol/L}\), both ions are present at this concentration, \( [\text{Ba}^{2+}] = 1 \times 10^{-5} \text{ mol/L}\) and \( [\text{SO}_4^{2-}] = 1 \times 10^{-5} \text{ mol/L}\).
To find the \(K_{sp}\), use the formula:

• \[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]
• \[ K_{sp} = (1 \times 10^{-5})(1 \times 10^{-5}) = 1 \times 10^{-10} \text{ mol}^2/\text{L}^2 \]

Thus, knowing ionic concentrations allows for precise calculation of solubility product, which in turn indicates the solubility of a compound in a solution at equilibrium.