Problem 31 - Chemistry in 30 days

Question

The solubility of \(\mathrm{BaSO}_{4}\) in water is \(2.33 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}\). Its solubility product will be (molecular weight of \(\mathrm{BaSO}_{4}=233\) ). (a) \(1 \times 10^{-5}\) (b) \(1 \times 10^{-10}\) (c) \(1 \times 10^{-15}\) (i) \(1 \times 10^{-20}\)

Step-by-Step Solution

Answer

The solubility product is \(1 \times 10^{-10}\), option (b).

Step 1: Determine the Moles of BaSO4 Dissolved

The solubility of BaSO4 is given as \(2.33 \times 10^{-3}\ \text{g L}^{-1}\). To find the moles of \(\text{BaSO}_4\) per liter, we use its molar mass (233 g/mol): \[\frac{2.33 \times 10^{-3} \text{ g L}^{-1}}{233 \text{ g mol}^{-1}} = 1 \times 10^{-5} \text{ mol L}^{-1}.\] This value represents the concentration of \(\text{Ba}^{2+}\) and \(\text{SO}_4^{2-}\) ions, since one unit of \(\text{BaSO}_4\) dissociates to produce one solution of each ion.

Step 2: Write the Solubility Product Expression

The solubility product \(K_{sp}\) for \(\text{BaSO}_4\) is given by the expression: \[K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}].\] Since the dissolution of \(\text{BaSO}_4\) can be represented by the equation: \[\text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq),\] the concentrations of \(\text{Ba}^{2+}\) and \(\text{SO}_4^{2-}\) arising from the solid are both \(1 \times 10^{-5}\ \text{mol L}^{-1}.\)

Step 3: Calculate the Solubility Product Ksp

Substitute the ion concentrations into the \(K_{sp}\) expression: \[K_{sp} = (1 \times 10^{-5})(1 \times 10^{-5}) = 1 \times 10^{-10}.\] Thus, the solubility product \(K_{sp}\) for \(\text{BaSO}_4\) is \(1 \times 10^{-10}.\)

Step 4: Match to Given Options

Compare the calculated \(K_{sp}\) value with the options provided: (a) \(1 \times 10^{-5}\), (b) \(1 \times 10^{-10}\), (c) \(1 \times 10^{-15}\), (i) \(1 \times 10^{-20}\). The option closest to the calculated value is (b) \(1 \times 10^{-10}\).

Key Concepts

BaSO4 Solubility

Understanding the solubility of barium sulfate, ( BaSO_4), in water is essential for comprehending its behavior in various chemical contexts. Solubility refers to the maximum amount of a substance that can dissolve in a solvent, at a specified temperature and pressure, to form a saturated solution.
BaSO4 is known for its low solubility in water. This means it only dissolves in very small amounts. In our example, we know that BaSO4's solubility in water is approximately \(2.33 \times 10^{-3}\) ext{g L}^{-1}. This translates to a delicate balance where only limited amounts of BaSO_4 can dissociate to interact with water molecules effectively.
This low solubility is due to the strong ionic bonds between the Ba^{2+} and SO_4^{2-} ions within the BaSO_4 crystal lattice. The energy required to break these interactions is higher than the energy to dissolve the substance fully in water.

Ksp Calculation

The solubility product constant, known as K_{sp}, is a valuable way to quantify the extent of solubility of a sparingly soluble ionic compound like BaSO_4. K_{sp} is a specific equilibrium constant that applies to solubility reactions:
When BaSO4 dissolves in water, it separates into its ions as shown:

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

The K_{sp} expression for the above reaction is:

• \( K_{sp} = [ ext{Ba}^{2+}][ ext{SO}_4^{2-}] \)

Since both Ba^{2+} and SO_4^{2-} ions are produced in equal molar amounts during dissolution, their concentrations are the same, calculated to be \(1 \times 10^{-5} ext{ mol L}^{-1} \). Plugging these values into the K_{sp}expression, we calculate the value to be:

• \( K_{sp} = (1 \times 10^{-5})(1 \times 10^{-5}) = 1 \times 10^{-10} \)

This shows us that the product of the concentrations of ions in a saturated solution equals the K_{sp} value, indicating equilibrium.

Dissolution Equilibrium

Dissolution equilibrium occurs when a solid solute establishes a dynamic balance with its ions in solution. In the context of BaSO_4, when it dissolves in water, it reaches a point where the rate of dissolution (and formation of Ba^{2+} and SO_4^{2-} ions) is equal to the rate of precipitation (for the recombination of these ions into the solid form).
Achieving this balance doesn't imply that no more dissolving occurs, but rather, the amount of dissolved ions becomes constant, indicating both processes are happening at the same rate.
The dissolution equilibrium is crucial for understanding how substances like BaSO4 dissolve. In practice, knowing this equilibrium concept helps predict the behavior of sparingly soluble salts under different conditions. Let's keep in mind that equilibrium can shift with changes in temperature, pressure, and the presence of other ions or compounds.