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Chapter 17: Problem 114

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

Expert verified
The solubility product constant (K鈧涒倸) of strontium sulfate at 25掳C is approximately 1.23 脳 10鈦烩伓.

Step by step solution

01

Write down the expression for osmotic pressure

The osmotic pressure (蟺) of a solution can be determined using the formula: 蟺 = (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鈦宦� K鈦宦�) and T is the temperature in Kelvin.
02

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 鈮� 0.02763 atm
03

Calculate the molar concentration of the solution

We know 蟺 = (n/V)RT. Therefore, n/V (molar concentration) can be calculated as: n/V = 蟺 / RT We are given the temperature 25掳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鈦宦� K鈦宦� 脳 298.15 K) 鈮� 0.00111 mol/L
04

Write the solubility product expression for strontium sulfate

Strontium sulfate (SrSO鈧�) dissociates in water as follows: SrSO鈧�(s) 鈬� Sr虏鈦�(aq) + SO鈧劼测伝(aq) The solubility product constant (K鈧涒倸) can be written as: K鈧涒倸 = [Sr虏鈦篯[SO鈧劼测伝]
05

Calculate the solubility product constant (K鈧涒倸)

Since the saturated solution is at equilibrium, the concentrations of Sr虏鈦� and SO鈧劼测伝 are equal (0.00111 mol/L). Therefore, we can calculate the solubility product constant: K鈧涒倸 = (0.00111 mol/L)(0.00111 mol/L) 鈮� 1.23 脳 10鈦烩伓 So, the solubility product of strontium sulfate at 25掳C is approximately 1.23 脳 10鈦烩伓.

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