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Chapter 17: Problem 56

A 1.00-L, solution saturated at \(25^{\circ} \mathrm{C}\) with lead(II) iodide contains \(0.54 \mathrm{~g}\) of \(\mathrm{Pbl}_{2}\). Calculate the solubility- product constant for this salt at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The solubility product constant (Ksp) for lead(II) iodide (PbI2) at 25掳C is approximately \(6.43 \times 10^{-9}\).

Step by step solution

01

Calculate the molar concentration of PbI2

We are given that a 1.00 L saturated solution contains 0.54 g of PbI2. To find the molar concentration, we will first convert the mass of PbI2 to moles and then divide by the volume of the solution in liters. Molecular weight of PbI2 = 207.2 (Pb) + 2 * 126.9 (I) = 460.0 g/mol Number of moles of PbI2 = \(\frac{0.54 \mathrm{~g}}{460.0 \mathrm{~g/mol}}\) = \(1.17 \times 10^{-3} \mathrm{mol}\) Molar concentration of PbI2 = \(\frac{1.17 \times 10^{-3} \mathrm{mol}}{1.00 \mathrm{~L}}\) = \(1.17 \times 10^{-3} \mathrm{M}\)
02

Write the balanced chemical equation and solubility product expression

For lead(II) iodide, the balanced chemical equation is: \[ \mathrm{PbI_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2I^{-} (aq)}\] The corresponding solubility product expression is: \[ K_{sp} = [\mathrm{Pb}^{2+}] [\mathrm{I}^{-}]^2\]
03

Determine the concentrations of Pb虏鈦� and I鈦� ions

Since the molar concentration of PbI2 is \(1.17 \times 10^{-3} \mathrm{M}\), the concentrations of the dissociated ions are as follows: - 1 mole of PbI2 gives 1 mole of Pb虏鈦� ions: \([\mathrm{Pb}^{2+}] = 1.17 \times 10^{-3} \mathrm{M}\) - 1 mole of PbI2 gives 2 moles of I鈦� ions: \([\mathrm{I}^{-}] = 2 \times 1.17 \times 10^{-3} \mathrm{M} = 2.34 \times 10^{-3} \mathrm{M}\)
04

Calculate the solubility product constant (Ksp)

Now, we will use the solubility product expression with the calculated ion concentrations to find Ksp: \[ K_{sp} = [\mathrm{Pb}^{2+}] [\mathrm{I}^{-}]^2 = (1.17 \times 10^{-3}) (2.34 \times 10^{-3})^2 = 6.43 \times 10^{-9} \] So, the solubility product constant for lead(II) iodide at 25掳C is approximately \(6.43 \times 10^{-9}\).

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Key Concepts

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

Lead(II) Iodide Solubility
The solubility of lead(II) iodide (PbI鈧�) refers to the amount of this compound that can dissolve in a solvent, creating a saturated solution. In a saturated solution, the maximum amount of solute is dissolved at a given temperature and pressure.
Lead(II) iodide is slightly soluble in water, meaning only a small amount dissolves before equilibrium is reached between the dissolved ions and the undissolved solid.
  • The dissolving process involves the dissociation of PbI鈧� into its constituent ions, piece by piece, in the water.
  • Because PbI鈧� is composed of lead (Pb虏鈦�) and iodide (I鈦�) ions, it dissolves by releasing these ions into the solution.
The solubility of a substance like lead(II) iodide can change with the temperature. Typically, increasing the temperature allows more PbI鈧� to dissolve, as more energy is available to break the ionic bonds. Understanding solubility is crucial since it underpins many reactions and processes in chemistry.
Molar Concentration Calculation
Calculating molar concentration helps to quantify the amount of a solute (like PbI鈧�) in a given volume of solution. The solution's concentration indicates how much solute is present, which is key for predicting the behavior of solutes in reactions and solutions.
In the problem, we start by converting the mass of PbI鈧� to moles since molarity is defined in terms of moles per liter.
  • First, you calculate the molar mass of PbI鈧� by adding the atomic masses of lead and iodide from the periodic table: 207.2 g/mol for lead and 126.9 g/mol for each iodide ion.

  • The total molar mass is 460.0 g/mol for PbI鈧�.

  • Next, divide the given mass by the molar mass: \( rac{0.54 ext{ g}}{460.0 ext{ g/mol}} = 1.17 imes 10^{-3} ext{ mol} \).

  • Finally, to find the molar concentration or molarity, divide the moles by the volume of the solution (in liters): \( rac{1.17 imes 10^{-3} ext{ mol}}{1.00 ext{ L}} = 1.17 imes 10^{-3} ext{ M} \).
Knowing the molar concentration is instrumental for determining the solubility product, helping to predict the solubility and behavior of ions in equilibrium.
Solubility Equilibrium
Solubility equilibrium is a dynamic state established when the rate of PbI鈧� dissolving equals the rate of Pb虏鈦� and I鈦� ions recombining into the solid. This balance allows us to define the solubility product constant, or Ksp.
In the case of lead(II) iodide, the equilibrium can be represented by the equation: \(\mathrm{PbI_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2I^{-} (aq)}\).
  • Ksp is a way to express the concentrations of the ions in a saturated solution at equilibrium directly through their concentrations.

  • For PbI鈧�, the expression is \( K_{sp} = [Pb^{2+}] [I^{-}]^2 \), showing how each ion contributes to the equilibrium constant.

  • The equilibrium expression indicates that a change in the concentration of either ion will impact the system until a new equilibrium is reached, maintaining the value of Ksp consistent under constant conditions.
By calculating Ksp, we can derive meaningful insights into the extent to which PbI鈧� can dissolve, a fundamental part of solubility and precipitation reactions.

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Most popular questions from this chapter

Assume that \(30.0 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) solution of a weak base B that accepts one proton is titrated with a \(0.10 \mathrm{M}\) solution of the monoprotic strong acid HA. (a) How many moles of HA have been added at the equivalence point? (b) What is the predominant form of B at the equivalence point? (c) Is the pH 7, less than 7, or more than 7 at the equivalence point? (d) Which indicator, phenolphthalein or methyl red, is likely to be the better choice for this titration?

Which of the following salts will be substantially more soluble in acidic solution than in pure water: (a) \(\mathrm{ZnCO}_{3^{*}}\) (b) \(\mathrm{ZnS}\), (c) \(\mathrm{Bil}_{3}\) (d) \(\mathrm{AgCN}_{4}\), (e) \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) ?

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(a) Calculate the percent ionization of \(0.125 \mathrm{M}\) lactic acid \(\left(K_{a}=1.4 \times 10^{-4}\right)\). (b) Calculate the percent ionization of \(0.125 \mathrm{M}\) lactic acid in a solution containing \(0.0075 \mathrm{M}\) sodium lactate. Buffers (Section 17.2)
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