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Chapter 17: Problem 116

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} ?$

Short Answer

Expert verified
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 by step solution

01

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.
02

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: \(P = 3.93 kPa * \frac{1 atm}{101.3 kPa} = 0.0388 atm\)
03

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: \[M = \frac{PV}{n_sRT} = \frac{0.0388 atm}{(2)(0.0821 L atm/(K mol))(298.15 K)}\] \[M = 8.03 × 10^{-5} mol/L\]
04

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: \[K_{sp} = (M)^2\]
05

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.

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Most popular questions from this chapter

A sample of \(500 \mathrm{mg}\) of an unknown monoprotic acid was dissolved in \(50.0 \mathrm{~mL}\) of water and titrated with \(0.200 \mathrm{M}\) KOH. The acid required \(20.60 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(10.30 \mathrm{~mL}\) of base had been added in the titration, the pH was found to be 4.20 . What is the \(\mathrm{p} K_{a}\) for the unknown acid?

For each of the following slightly soluble salts, write the net ionic equation, if any, for reaction with a strong acid: (a) MnS, (b) \(\mathrm{PbF}_{2}\), (c) \(\mathrm{AuCl}_{3}\) (d) \(\mathrm{Hg}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) (e) CuBr.

The value of \(K_{s p}\) for \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) is \(2.1 \times 10^{-20} .\) The \(\mathrm{AsO}_{4}^{3-}\) ion is derived from the weak acid \(\mathrm{H}_{3} \mathrm{AsO}_{4}\left(\mathrm{pK}_{a 1}=\right.\) \(\left.2.22 ; \mathrm{p} K_{a 2}=6.98 ; \mathrm{pK}_{a 3}=11.50\right)\) (a) Calculate the molar solubility of \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) in water. (b) Calculate the pH of a saturated solution of \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) in water.

A 10.0-mL sample of \(0.250 \mathrm{MHNO}_{3}\) solution is titrated with $0.100 M$ KOH solution. Calculate the pH of the solution after the following volumes of base have been added: (a) \(20.0 \mathrm{~mL}\), (b) \(24.9 \mathrm{~mL}\) (c) \(25.0 \mathrm{~mL}\) (d) \(25.1 \mathrm{~mL}\), (e) \(30.0 \mathrm{~mL}\).

A buffer, consisting of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-},\) helps control the pH of physiological fluids. Many carbonated soft drinks also use this buffer system. What is the \(\mathrm{pH}\) of a soft drink in which the major buffer ingredients are \(10.0 \mathrm{~g}\) of \(\mathrm{KH}_{2} \mathrm{PO}_{4}\) and \(10.0 \mathrm{~g}\) of $\mathrm{K}_{2} \mathrm{HPO}_{4}\( per \)0.500 \mathrm{~L}$ of solution?
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