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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, \( K_{sp} \), for lead(II) iodide at \(25^{\circ}\)C is approximately \(6.42 \times 10^{-9}\).

Step by step solution

01

Understanding the Dissolution Equation

The dissolution of lead(II) iodide in water can be represented by the equation: \[ \text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq) \]This means that when lead(II) iodide dissolves, it produces one lead ion and two iodide ions.
02

Calculate the Moles of Lead(II) Iodide

We begin by calculating the molar mass of \( \text{PbI}_2 \). The atomic masses are approximately: \( \text{Pb} = 207.2 \) and \( \text{I} = 126.9 \). Thus, \[ \text{Molar mass of PbI}_2 = 207.2 + 2(126.9) = 461 \text{ g/mol} \]Now, calculate the moles of \( \text{PbI}_2 \) in the solution:\[ \text{Moles of PbI}_2 = \frac{0.54 \text{ g}}{461 \text{ g/mol}} \approx 0.00117 \text{ mol} \]
03

Determine Ion Concentrations

From the dissolution equation, each mole of \( \text{PbI}_2 \) dissolves into 1 mole of \( \text{Pb}^{2+} \) and 2 moles of \( \text{I}^- \). Therefore:- \( \text{[Pb}^{2+}] = 0.00117 \text{ mol/L}\)- \( \text{[I}^-] = 2 \times 0.00117 \text{ mol/L} = 0.00234 \text{ mol/L}\)
04

Calculate the Solubility-Product Constant

The solubility-product constant \( K_{sp} \) for \( \text{PbI}_2 \) is given by:\[ K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 \]Substitute the ion concentrations:\[ K_{sp} = (0.00117)(0.00234)^2 \]Performing the calculation:\[ K_{sp} = 0.00117 \times 0.00234 \times 0.00234 \approx 6.42 \times 10^{-9} \]

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Key Concepts

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

Solubility-Product Constant (Ksp)
The solubility-product constant, often abbreviated as Ksp, is a vital concept in chemistry, particularly when studying solubility equilibria. It helps us understand the extent to which a compound can dissolve in water. The Ksp is an equilibrium constant that applies to a sparingly soluble ionic compound. This means that for a dissolved ionic compound in a saturated solution, the multiplication of the concentrations of the ions each raised to the power of their coefficients in the balanced dissolution equation remains constant at a given temperature. For example, when dealing with lead(II) iodide, PbI₂, the Ksp reflects the product of the molarity of lead ions and the square of the molarity of iodide ions in the solution. The Ksp can help predict how much of the compound will dissolve in the water under standard conditions. It’s crucial for anticipating the behavior of ions in diverse chemical environments, from laboratory settings to environmental systems.
Dissolution Equation
In chemistry, the dissolution equation provides a clear picture of what happens when a solid solute dissolves in a solvent, typically water. For lead(II) iodide, the dissolution reaction can be written as: \[ \text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq) \] This equation shows that one molecule of lead(II) iodide, a solid, separates into one lead ion and two iodide ions when it enters the solution. Understanding this equation is crucial because it tells us the stoichiometric relationships between the reactants and the products. This dissolution process dynamically reaches equilibrium, meaning while some of the solid dissolves into ions, some ions will recombine into the solid, maintaining a steady state at equilibrium. This equilibrium is what allows us to use Ksp to determine the solubility characteristics of compounds.
Ion Concentrations in Solution
Calculating ion concentrations in a solution is a step that follows the dissolution equation. The process begins after determining the number of moles of the solute present. For lead(II) iodide, after finding the moles, you can deduce the concentrations of dissolved ions. One mole of PbIâ‚‚ yields one mole of \( \text{Pb}^{2+} \) ions and two moles of \( \text{I}^- \) ions. Therefore, if you have 0.00117 moles of PbIâ‚‚ dissolved in a liter of solution, you have 0.00117 M of \( \text{Pb}^{2+} \) and 0.00234 M of \( \text{I}^- \). This calculation is essential for finding Ksp because ion concentrations precisely determine the saturation point of the compound in the solution. These quantitative insights provide valuable information on which conditions could lead to precipitation or continued dissolution, aiding in both academic and practical chemical problem-solving.

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Most popular questions from this chapter

Predict whether the equivalence point of each of the following titrations is below, above, or at pH 7: (a) benzoic acid titrated with \(\mathrm{KOH},(\mathbf{b})\) ammonia titrated with iodic acid, (c) hydroxylamine with hydrochloric acid.

A 1.50-L solution saturated at \(25^{\circ} \mathrm{C}\) with cobalt carbonate \(\left(\mathrm{CoCO}_{3}\right)\) contains \(2.71 \mathrm{mg}\) of \(\mathrm{CoCO}_{3} .\) Calculate the solubility-product constant for this salt at \(25^{\circ} \mathrm{C}\).

Compare the titration of a strong, monoprotic acid with a strong base to the titration of a weak, monoprotic acid with a strong base. Assume the strong and weak acid solutions initially have the same concentrations. Indicate whether the following statements are true or false. (a) More base is required to reach the equivalence point for the strong acid than the weak acid. (b) The \(\mathrm{pH}\) at the beginning of the titration is lower for the weak acid than the strong acid. \((\mathbf{c})\) The pH at the equivalence point is 7 no matter which acid is titrated.

Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid HA and a base B are mixed. The pH of the resulting solution is 9.2 . (a) Write the chemical equation and equilibrium-constant expression for the reaction between HA and B. (b) If \(K_{a}\) for HA is \(8.0 \times 10^{-5}\), what is the value of the equilibrium constant for the reaction between HA and B? (c) What is the value of \(K_{h}\) for B?

A weak monoprotic acid is titrated with \(0.100 \mathrm{M} \mathrm{NaOH}\). It requires \(25.0 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution to reach the equivalence point. After \(12.5 \mathrm{~mL}\). of base is added, the pH of the solution is 4.16 . Estimate the \(\mathrm{p} K_{a}\) of the weak acid.

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