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 at \(25^{\circ} \mathrm{C}\) is \(1.25 \times 10^{-6}\).

Step-by-step solution

Write down the osmotic pressure equation

The osmotic pressure equation is given by: \(Π = cRT\) where Π is the osmotic pressure, c is the molar concentration of the solute, R is the ideal gas constant, and T is the absolute temperature in Kelvin.

Write down the Van't Hoff equation relating osmotic pressure and molar concentration

The Van't Hoff equation is given by: \(Π = c_i × R × T\) where c_i is the effective concentration of the solute particles.

Write down the balanced chemical equation for strontium sulfate dissolution

The balanced chemical equation for the dissolution of strontium sulfate is: \(SrSO_4(s) ↔ Sr^{2+}(aq) + SO^{2-}_4(aq)\)

Write down the solubility product (Ksp) equation for strontium sulfate

The solubility product (Ksp) equation for strontium sulfate is: \(K_{sp} = [Sr^{2+}] × [SO^{2-}_4]\)

Convert the given temperature to Kelvin and determine the ideal gas constant value

To convert the given temperature, 25°C, to Kelvin, add 273.15: \(25°C + 273.15 = 298.15K\) The value of the ideal gas constant, R, is 0.0821 L atm/mol K.

Calculate the molar concentration of the solute using the osmotic pressure equation

Given the osmotic pressure (21 torr), convert the pressure to atm: \(1 Torr = \frac{1}{760} atm\) \(21 Torr = \frac{21}{760} atm = 0.02763 atm\) Now, we can calculate the molar concentration (c) using the osmotic pressure equation: Π = cRT where Π = 0.02763 atm, R = 0.0821 L atm/mol K, and T = 298.15 K. Solve for c: \(c = \frac{Π}{RT} = \frac{0.02763}{(0.0821)(298.15)} = 1.12 × 10^{-3} M\)

Calculate the solubility product (Ksp) using the molar concentration

Since the dissolution results in a 1:1 ratio of Sr²⁺ and SO₄²⁻ ions, the molar concentration for each ion is also 1.12 × 10^{-3} M. Using the solubility product equation: \(K_{sp} = [Sr^{2+}] × [SO^{2-}_4] = (1.12 × 10^{-3})(1.12 × 10^{-3}) = 1.25 × 10^{-6}\) The solubility product of strontium sulfate at 25°C is 1.25 × 10^{-6}.