Problem 116 - Chemistry: The Central Science in SI Units, Global Edition

Question

The osmotic pressure of a saturated solution of lead(II) sulfate \(\left(\mathrm{PbSO}_{4}\right)\) at \(25^{\circ} \mathrm{C}\) is \(3.93 \mathrm{kPa}\). What is the solubility product of this salt at \(25^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The solubility product of lead(II) sulfate at 25°C can be calculated using the osmotic pressure formula and the ideal gas law. Convert temperature and pressure to Kelvin and atm, and calculate the molar concentration of the solution. Write the solubility product expression for PbSO₄ and substitute the molar concentration value to find the solubility product. The solubility product of lead(II) sulfate at 25°C is \(K_{sp} = 6.45 × 10^{-9}\).

Step 1: Write down the osmotic pressure formula and the ideal gas law

The osmotic pressure formula is: \[Π = n_sMRT\] where Π is the osmotic pressure, n_s is the number of solute particles, M is the molarity of the solution, R is the gas constant (0.0821 L atm/(K mol)), and T is the temperature in Kelvin. The ideal gas law is: \[PV = nRT\] where P is the pressure, V is the volume, n is the moles of the gas, R is the gas constant, and T is the temperature in Kelvin.

Step 2: Convert the given temperature and pressure to Kelvin and atm

We are given the temperature in Celsius and the pressure in kPa. We need to convert them to the appropriate units for the equations mentioned above. Temperature: \(T = 25°C + 273.15 = 298.15 K\) Pressure: \(P = 3.93 kPa * \frac{1 atm}{101.3 kPa} = 0.0388 atm\)

Step 3: Determine the molar concentration of the solution

Now we can use the osmotic pressure formula and the ideal gas law to find the molar concentration of the solution. We write the ideal gas law in terms of moles, n, and solve for the molarity, M. From the osmotic pressure formula and the ideal gas law combined, we have: \[\frac{PV}{RT} = n_sM\] where n_s= number of ions in solution = 2 (1 for Pb²⁺ and 1 for SO₄²⁻) Now, solve for M: \[M = \frac{PV}{n_sRT} = \frac{0.0388 atm}{(2)(0.0821 L atm/(K mol))(298.15 K)}\] \[M = 8.03 × 10^{-5} mol/L\]

Step 4: Write down the solubility product expression for PbSO₄

The solubility product expression for lead(II) sulfate is: \[K_{sp} = [Pb^{2+}][SO_4^{2-}]\] Since the molar concentration of both ions is the same, we can write the expression as: \[K_{sp} = (M)^2\]

Step 5: Calculate the solubility product for PbSO₄ at 25°C

Substitute the value of M from Step 3 into the solubility product equation, and solve for K_sp: \[K_{sp} = (8.03 × 10^{-5} mol/L)^2\] \[K_{sp} = 6.45 × 10^{-9}\] The solubility product of lead(II) sulfate at 25°C is 6.45 × 10⁻⁹.

Key Concepts

Osmotic Pressure

Osmotic pressure is a crucial concept, particularly in solutions. It refers to the pressure that must be applied to a solution to prevent the inward flow of water through a semipermeable membrane. In essence, it's the measure of a solution's potential to "pull" in water, influencing its concentration balance with neighboring compartments.
This phenomenon arises because solutions have a tendency to reach equilibrium and balance the concentration of solutes across a membrane. In our exercise, we used osmotic pressure to determine the concentration of dissolved lead(II) sulfate, using the formula:

• \(Π = n_sMRT\)

Here, \(Π\) is the osmotic pressure, \(n_s\) represents the number of solute particles released in the solution, \(M\) is the molarity, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin, ensuring the units are consistent across calculations. Calculating osmotic pressure requires converting the temperature to Kelvin and the pressure to atmospheres, as these are the standard units for these kinds of calculations in chemistry. This conversion is crucial for accuracy in the mathematical calculations that follow.

Ideal Gas Law

The ideal gas law is one of the most fundamental equations in chemistry, allowing us to relate the properties of gases. It’s mathematically expressed as:

• \[PV = nRT\]

In this equation, \(P\) represents the pressure, \(V\) is the volume, \(n\) is the number of moles of gas present, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin. The ideal gas law assumes a perfect gas scenario where the gas particles themselves occupy no space and experience no intermolecular forces. This assumption is crucial when relating the behavior of gases under various conditions related to pressure, volume, and temperature.
In the context of the exercise, the ideal gas law is used to derive the formula for the osmotic pressure and find the molar concentration of the solution. This is because the behavior of the dissolved ions in a solution can often be analogized to that of a gas in terms of pressure and concentration relations.

Molar Concentration

Molar concentration, also known as molarity, refers to the number of moles of solute per liter of solution. It's a critical concept in solution chemistry and is represented by the symbol \(M\). Molarity helps in understanding how concentrated a solution is, and in calculating reaction yields or other properties.
The formula is:

• \[M = rac{n_sPV}{RT}\]

Where \(n_s\) is the number of solute particles, \(P\) denotes pressure, \(V\) is volume, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin. In the practice exercise, we found that the molar concentration for the solution of lead(II) sulfate was 8.03 × 10⁻⁵ mol/L. This value provides insight into how much of the lead(II) sulfate has dissolved in water, allowing us to further calculate its solubility product, an indicator of its solubility in water.

Lead(II) Sulfate

Lead(II) sulfate, represented by the formula \( ext{PbSO}_4\), is a chemical compound known for its low solubility in water. This characteristic makes it an interesting subject for studying solubility products. The solubility product constant \(K_{sp}\) for lead(II) sulfate is a value that indicates the extent to which \( ext{PbSO}_4\) can dissolve in water to form its constituent ions: lead ions \(( ext{Pb}^{2+})\) and sulfate ions \(( ext{SO}_4^{2-})\).
The solubility product expression can be written as:

• \[K_{sp} = [ ext{Pb}^{2+}][ ext{SO}_4^{2-}]\]

From the exercise, the concentration of both ions was equal, simplifying the equation to \((M)^2\). Using the molarity calculated earlier, the solubility product \(K_{sp}\) was determined to be 6.45 × 10⁻⁹. This low \(K_{sp}\) value confirms the limited solubility of lead(II) sulfate in water, explaining why it tends to precipitate rather than dissolve fully.