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 \(25^{\circ} \mathrm{C}\) ?

Short answer

The solubility product (Ksp) of strontium sulfate (SrSO₄) at 25°C is 3.15 × 10⁻⁷.

Step-by-step solution

Write the solubility reaction and Ksp expression

First, we need to write the solubility reaction of SrSO4 and the expression for its solubility product Ksp. The solubility reaction is as follows: SrSO₄ (s) ⇌ Sr²⁺ (aq) + SO₄²⁻ (aq) The solubility product Ksp is equal to the product of the concentrations of the dissolved ions: Ksp = [Sr²⁺] [SO₄²⁻]

Calculate the molar concentration from osmotic pressure

We are given the osmotic pressure (π) and need to find the concentration of the ions in the dissolved SrSO₄. Using the osmotic pressure formula, we can solve for the molar concentration C: π = nCRT Since SrSO₄ dissociates into 2 ions, the number of particles (n) is 2. The osmotic pressure is given as 21 torr. We need to convert this to the same unit as the ideal gas constant (R). Since 1 atm = 760 torr, we have: π = 21 torr × \(\frac{1\, \text{atm}}{760\, \text{torr}}\) = 0.02763 atm The temperature (T) is given in Celsius and will need to be converted to Kelvin for the purpose of this calculation: T = 25°C + 273.15 = 298.15 K The ideal gas constant R is given as 0.0821 L atm / K mol. Now we can plug in the values and solve for the concentration (C): 0.02763 atm = 2 × C × 0.0821 L atm / K mol × 298.15 K C = 0.02763 atm / (2 × 0.0821 L atm / K mol × 298.15 K) = 5.618 × 10⁻⁴ M

Calculate the solubility product (Ksp)

Now that we have the concentration of the ions, we can find the solubility product Ksp. Since the stoichiometry of the reaction is 1:1, the concentrations of Sr²⁺ and SO₄²⁻ are equal, and both are equal to the solubility of SrSO₄ in the dissolved state. Thus: Ksp = [Sr²⁺] [SO₄²⁻] = (5.618 × 10⁻⁴)² = 3.15 × 10⁻⁷ So, the solubility product (Ksp) of SrSO₄ at 25°C is 3.15 × 10⁻⁷.