Problem 114 - Chemistry The Central Science

Question

The osmotic pressure of a saturated solution of strontium sulfate at \(25^{\circ} \mathrm{C}\) is 21 torr. What is the solubility product of this salt at \(25^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The solubility product constant (Kₛₚ) of strontium sulfate at 25°C is approximately 1.23 × 10⁻⁶.

Step 1: Write down the expression for osmotic pressure

The osmotic pressure (π) of a solution can be determined using the formula: π = (n/V)RT where n is the amount of solute in moles, V is the volume of the solution in liters, R is the ideal gas constant (0.08206 L atm mol⁻¹ K⁻¹) and T is the temperature in Kelvin.

Step 2: Convert the given osmotic pressure to atm

We are given the osmotic pressure of 21 torr. Since the gas constant R is given in atm, we need to convert the pressure to atm: 1 atm = 760 torr So, 21 torr = (21/760) atm ≈ 0.02763 atm

Step 3: Calculate the molar concentration of the solution

We know π = (n/V)RT. Therefore, n/V (molar concentration) can be calculated as: n/V = π / RT We are given the temperature 25°C, which is 298.15 K in Kelvin. Therefore, n/V can be calculated as: n/V = 0.02763 atm / (0.08206 L atm mol⁻¹ K⁻¹ × 298.15 K) ≈ 0.00111 mol/L

Step 4: Write the solubility product expression for strontium sulfate

Strontium sulfate (SrSO₄) dissociates in water as follows: SrSO₄(s) ⇌ Sr²⁺(aq) + SO₄²⁻(aq) The solubility product constant (Kₛₚ) can be written as: Kₛₚ = [Sr²⁺][SO₄²⁻]

Step 5: Calculate the solubility product constant (Kₛₚ)

Since the saturated solution is at equilibrium, the concentrations of Sr²⁺ and SO₄²⁻ are equal (0.00111 mol/L). Therefore, we can calculate the solubility product constant: Kₛₚ = (0.00111 mol/L)(0.00111 mol/L) ≈ 1.23 × 10⁻⁶ So, the solubility product of strontium sulfate at 25°C is approximately 1.23 × 10⁻⁶.