calculate the K _sp of sparingly soluble salt XY₂ (s) at 25 ^oC , if the vapour pressure of its saturated solution in water is 31.78 mm of Hg and vapour pressure of pure water is 31.82 mm of Hg?

1)0.0233 mol³ L^-3

2)5.42 x 10^-4 mol² L^-2

3)4.37 x 10 ^-5 mol ³ L^-3

4)5.05 x 10 ^-5 mol ² L^-2

To calculate the solubility product constant (Ksp) of the sparingly soluble salt XY₂ at 25 °C, we can use the given vapor pressures of the saturated solution and pure water. This approach utilizes Raoult's Law, which relates the vapor pressure of a solution to the concentration of solute particles in it.

Understanding Vapor Pressure and Raoult's Law

Raoult's Law states that the vapor pressure of a solvent in a solution (P_solution) is equal to the vapor pressure of the pure solvent (P_pure) multiplied by the mole fraction of the solvent in the solution (X_solvent):

P_solution = P_pure × X_solvent

Step 1: Calculate the Mole Fraction of Water

First, we need to find the mole fraction of water in the saturated solution. We can rearrange Raoult's Law to find X_solvent:

X_solvent = P_solution / P_pure

Substituting the given values:

• P_solution = 31.78 mm Hg
• P_pure = 31.82 mm Hg

X_solvent = 31.78 mm Hg / 31.82 mm Hg ≈ 0.9987

Step 2: Calculate the Mole Fraction of the Solute

The mole fraction of the solute (X_solute) can be found using the relationship:

X_solute = 1 - X_solvent ≈ 1 - 0.9987 = 0.0013

Step 3: Relate Mole Fraction to Concentration

For a saturated solution of XY₂, we can express the mole fraction of the solute in terms of its molarity (C) and the total number of moles in the solution. Assuming the volume of the solution is 1 L, the total number of moles of solvent (water) is approximately 55.5 moles (since the density of water is about 1 g/mL).

Using the formula for mole fraction:

X_solute = n_solute / (n_solute + n_solvent)

Let n_solute = C (the molarity of XY₂), then:

0.0013 = C / (C + 55.5)

Step 4: Solve for C

Rearranging the equation gives:

0.0013(C + 55.5) = C

0.0013C + 0.07215 = C

0.07215 = C - 0.0013C

0.07215 = 0.9987C

C ≈ 0.07215 / 0.9987 ≈ 0.0723 mol/L

Step 5: Calculate Ksp

The dissociation of XY₂ in water can be represented as:

XY₂ (s) ⇌ X²⁺ (aq) + 2Y^- (aq)

From this equation, we see that for every 1 mole of XY₂ that dissolves, it produces 1 mole of X²⁺ and 2 moles of Y^-. Therefore, if the solubility (C) of XY₂ is 0.0723 mol/L, then:

• [X²⁺] = C = 0.0723 mol/L
• [Y^-] = 2C = 2 × 0.0723 = 0.1446 mol/L

Now, we can express Ksp as:

Ksp = [X²⁺][Y^-]² = (0.0723)(0.1446)²

Ksp = (0.0723)(0.0209) ≈ 0.00151

Final Calculation

To express Ksp in the correct units, we can convert it to mol³L^-3:

Ksp ≈ 1.51 × 10^-3 mol³L^-3

Identifying the Correct Answer

Looking at the provided options, the closest value to our calculated Ksp is:

• 1) 0.0233 mol³L^-3
• 2) 5.42 × 10^-4 mol²L^-2
• 3) 4.37 × 10^-5 mol³L^-3
• 4) 5.05 × 10^-5 mol²L^-2

Thus, the correct answer is approximately 5.42 × 10^-4 mol²L^-2, which corresponds to option 2.