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Chapter 16: Problem 83

On a hot day, a \(200.0-\mathrm{mL}\) sample of a saturated solution of \(\mathrm{Pb} \mathrm{I}_{2}\) was allowed to evaporate until dry. If \(240 \mathrm{mg}\) of solid \(\mathrm{PbI}_{2}\) was collected after evaporation was complete, calculate the \(K_{\text {sp }}\) value for \(\mathrm{Pb} \mathrm{I}_{2}\) on this hot day.

Short Answer

Expert verified
The solubility product constant (Ksp) for PbI2 on this hot day is approximately \(7.0 \times 10^{-9}\).

Step by step solution

01

Write down the balanced chemical equation

The balanced chemical equation for the dissociation of PbI2 in solution is: \(PbI_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2I^- (aq)\)
02

Calculate the moles of PbI2

First, we need to calculate the moles of PbI2. The problem states that 240 mg of solid PbI2 was collected after evaporation. - Convert the mass of PbI2 to grams by dividing 240 mg by 1000: \(240\,mg\, PbI_2 \times \frac{1\,g}{1000\,mg} = 0.24\,g\, PbI_2\) Next, we need the molar mass of PbI2. Using the periodic table, we find the molar mass of each constituent: - Lead (Pb): 207.2 g/mol - Iodine (I): 126.9 g/mol Summing these up for PbI2, we have: \(Molar\,mass\,of\,PbI_2 = 207.2\,g/mol + 2 \times 126.9\,g/mol = 460.0\,g/mol\) Now we can determine the moles of PbI2 by dividing the grams of PbI2 by its molar mass: \(\frac{0.24\,g\, PbI_2}{460.0\, g/mol} = 5.2 \times 10^{-4}\, mol\, PbI_2\)
03

Calculate the concentration of Pb2+ and I- ions

Now that we have the moles of PbI2, we can find the concentrations of the Pb2+ and I- ions in the saturated solution. Since the initial volume of the solution was 200.0 mL, we can convert it to liters: \(200.0\,mL = 0.200\,L\) Using the balanced chemical equation provided, we see that 1 mole of PbI2 produces 1 mole of Pb2+ ions and 2 moles of I- ions in solution. Therefore, the concentrations of Pb2+ and I- ions are: - \(Pb^{2+}\): \(\frac{5.2 \times 10^{-4}\, mol\, PbI_2}{0.200\,L} = 2.6 \times 10^{-3}\, M\, Pb^{2+}\) - \(I^-\): Since 1 mole of PbI2 produces 2 moles of I- ions, multiply the PbI2 moles by 2: \(2 \times 5.2 \times 10^{-4}\, mol = 10.4 \times 10^{-4}\, mol\, I^-\) Now, calculate the concentration of I- ions: \(\frac{10.4 \times 10^{-4}\, mol}{0.200\,L} = 5.2 \times 10^{-3}\, M\, I^-\)
04

Calculate the Ksp value for PbI2

Now that we have the concentrations of Pb2+ and I- ions in the saturated solution, we can calculate the Ksp value for PbI2 using the formula: \(K_{sp} = [Pb^{2+}] [I^-]^2\) Plug in the values we calculated in the previous step: \(K_{sp} = (2.6 \times 10^{-3}\, M)(5.2 \times 10^{-3}\, M)^2\) Calculate the Ksp value: \(K_{sp} = 7.0 \times 10^{-9}\) The solubility product constant (Ksp) for PbI2 on this hot day is approximately \(7.0 \times 10^{-9}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Chemical equilibrium occurs when the rate at which a chemical reaction proceeds forward equals the rate going in reverse. This results in the concentration of reactants and products remaining constant over time. In the case of the dissociation of lead(II) iodide (PbI鈧�) in solution, the equilibrium is represented by the equation:
  • Lead iodide dissociates into lead ( Pb^{2+} ) and iodide ( I^- ) ions.
  • The reactions in both directions occurring at the same rate maintain the equilibrium.
  • The concentrations of Pb^{2+} and I^- ions do not change under stable conditions.
Recognizing the state of equilibrium is crucial when calculating the solubility product constant (K_{sp}). Understanding this balance helps in determining how solubility is influenced by various factors.
Molar Mass Calculation
Calculating the molar mass of a compound is essential for stoichiometry and chemical calculations. The molar mass is the mass of one mole of a substance, typically measured in grams per mole (g/mol).
For the compound PbI鈧�:
  • Lead (Pb) has a molar mass of 207.2 g/mol.
  • Iodine (I) has a molar mass of 126.9 g/mol.
  • Since there are two iodine atoms in PbI鈧�, the total molar mass is:
  • \[207.2 \, g/mol + 2 \times 126.9 \, g/mol = 460.0 \, g/mol\]
This calculation allows you to convert between mass and moles, which is fundamental in solving chemistry problems involving solubility and reaction equations.
Concentration Calculation
Calculating concentration involves determining how much solute is present in a given volume of solution. Concentration is usually expressed in molarity, defined as moles per liter (M). In this context:
  • Start with the number of moles of PbI鈧� obtained from its mass, using the molar mass.
  • Convert the solution volume from milliliters to liters.
  • For Pb^{2+}, divide the moles of PbI鈧� by the volume in liters:\[(5.2 \times 10^{-4} \, mol \, PbI_2) / 0.200 \, L = 2.6 \times 10^{-3} \, M\]
  • For I^- ions, since they are double the moles of Pb^{2+}, multiply and then divide by the volume:\[(10.4 \times 10^{-4} \, mol \, I^-) / 0.200 \, L = 5.2 \times 10^{-3} \, M\]
These concentrations are vital for calculating the solubility product constant, which provides insights into the solubility of the salt.
Balanced Chemical Equation
Writing balanced chemical equations is essential for representing chemical reactions accurately. A balanced equation indicates the relative amounts of reactants and products and their forms. For PbI鈧� in a solution:
  • The dissociation is expressed as:\[PbI_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2I^- (aq)\]
  • This shows that one mole of PbI鈧� yields one mole of Pb^{2+} ions and two moles of I^- ions.
  • Balanced equations allow you to determine stoichiometric relationships, such as the ratio of ions formed.
Understanding how to balance and interpret these equations is crucial for any further calculations involving chemical equilibrium and solubility.

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Most popular questions from this chapter

The solubility of \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) in a \(0.20-\mathrm{M} \mathrm{KIO}_{3}\) solution is \(4.4 \times 10^{-8} \mathrm{~mol} / \mathrm{L}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3} .\)

Write balanced equations for the dissolution reactions and the corresponding solubility product expressions for each of the following solids. a. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) b. \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) c. \(\mathrm{BaF}_{2}\)

The \(K_{\text {sp }}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(2 \times 10^{-32}\). At what \(\mathrm{pH}\) will a \(0.2-M \mathrm{Al}^{3+}\) solution begin to show precipitation of \(\mathrm{Al}(\mathrm{OH})_{3}\) ?

Consider \(1.0 \mathrm{~L}\) of an aqueous solution that contains \(0.10 \mathrm{M}\) sulfuric acid to which \(0.30\) mole of barium nitrate is added. Assuming no change in volume of the solution, determine the \(\mathrm{pH}\), the concentration of barium ions in the final solution, and the mass of solid formed.

A solution is prepared by mixing \(100.0 \mathrm{~mL}\) of \(1.0 \times 10^{-2} M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(100.0 \mathrm{~mL}\) of \(1.0 \times 10^{-3} M \mathrm{NaF}\). Will \(\mathrm{PbF}_{2}(s)\) \(\left(K_{\text {sp }}=4 \times 10^{-8}\right)\) precipitate?
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