Book cover for General Chemistry: Principles and Modern Applications

General Chemistry: Principles and Modern Applications

Ralph H. Petrucci, F. Geoffrey Herring, Jeffry D. Madura, Carey Bissonnette

ISBN #9780132931281

11th Edition

3,230 Questions

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

This chapter covers the principles of solubility and complex-ion equilibria, emphasizing the role of the solubility product constant (Ksp) in predicting and manipulating precipitation reactions. Key topics include the mathematical relationship between solubility and Ksp, the impact of the common-ion effect on equilibrium, and practical techniques such as fractional precipitation. Additionally, the chapter highlights the interplay between pH and solubility and connects equilibrium concepts with real-world applications in analytical and industrial chemistry.

Learning Objectives

1

Define and calculate the solubility product constant (Ksp) and understand its significance in predicting precipitation.

2

Explain the relationship between solubility and Ksp, including the impact of common ions and pH on solubility equilibria.

3

Analyze the limitations of the Ksp concept and apply criteria for precipitation and its completeness in various chemical systems.

4

Demonstrate the methodology of fractional precipitation and its use in separating ions in complex mixtures.

5

Apply concepts of complex-ion equilibria to real-world problems and interpret experimental data using equilibrium principles.

Key Concepts

CONCEPT

DEFINITION

Ksp (Solubility Product Constant)

A numerical constant that represents the product of the molar concentrations of the ions of a sparingly soluble compound, each raised to the power of its stoichiometric coefficient at equilibrium.

Solubility

The maximum amount of a solute that can dissolve in a given amount of solvent at a specified temperature, typically expressed in mol/L or other concentration units.

Common-Ion Effect

The decrease in solubility of an ionic compound when a common ion is added to the solution, shifting the equilibrium according to Le Chatelier's principle.

Fractional Precipitation

A technique used to separate ions in a mixture by selectively precipitating one ion at a time through careful control of solution conditions and reagent addition.

Precipitation

The process by which a solid forms and separates from a solution when the product of the ionic concentrations exceeds the solubility product (Ksp) of the compound.

Complex-Ion Equilibria

The equilibrium established when metal ions form coordination complexes with ligands, affecting the solubility and reactivity of the species involved.

Example Problems

Example 1

Write $K_{\text {sp }}$ expressions for the following equilibria. For example, for the reaction $\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}(\mathrm{aq})+$ $\mathrm{Cl}^{-}(\mathrm{aq}), K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]$ (a) $\mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})$ (b) $\operatorname{Ra}\left(\mathrm{IO}_{3}\right)_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Ra}^{2+}(\mathrm{aq})+2 \mathrm{IO}_{3}^{-}(\mathrm{aq})$ (c) $\mathrm{Ni}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s}) \rightleftharpoons 3 \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{PO}_{4}^{3-}(\mathrm{aq})$ (d) $\mathrm{PuO}_{2} \mathrm{CO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{PuO}_{2}^{2+}(\mathrm{aq})+\mathrm{CO}_{3}^{2-}(\mathrm{aq})$

Example 2

Write solubility equilibrium equations that are described by the following $K_{\text {sp expressions. For example, } K_{\mathrm{sp}}}=$ $\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right] \quad$ represents $\quad \mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}(\mathrm{aq})+$ $\mathrm{Cl}^{-}(\mathrm{aq})$ (a) $K_{\mathrm{sp}}=\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3}$ (b) $K_{\mathrm{sp}}=\left[\mathrm{BiO}^{+}\right]\left[\mathrm{OH}^{-}\right]$ (c) $K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2}$ (d) $K_{\mathrm{sp}}=\left[\mathrm{Pb}^{2+}\right]^{3}\left[\mathrm{AsO}_{4}^{3-}\right]^{2}$

Example 3

The following $K_{\mathrm{sp}}$ values are found in a handbook. Write the solubility product expression to which each one applies. For example, $K_{\mathrm{sp}}(\mathrm{AgCl})=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=$ $1.8 \times 10^{-10}$ (a) $K_{\mathrm{sp}}\left(\mathrm{Cr} \mathrm{F}_{3}\right)=6.6 \times 10^{-11}$ (b) $K_{\mathrm{sp}}\left[\mathrm{Au}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]=1 \times 10^{-10}$ (c) $K_{\mathrm{sp}}\left[\mathrm{Cd}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=2.1 \times 10^{-33}$ (d) $K_{\mathrm{sp}}\left(\mathrm{SrF}_{2}\right)=2.5 \times 10^{-9}$

Example 4

Calculate the aqueous solubility, in moles per liter, of each of the following. (a) $\mathrm{BaCrO}_{4}, K_{\mathrm{sp}}=1.1 \times 10^{-5}$ (b) $\mathrm{PbBr}_{2}, K_{\mathrm{sp}}= \times 10^{-5}$ (c) $\mathrm{CeF}_{3}, K_{\mathrm{sp}}=8 \times 10^{-16}$ (d) $\operatorname{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}, K_{\mathrm{sp}}=2.1 \times 10^{-20}$

Example 5

Arrange the following solutes in order of increasing molar solubility in water: $\mathrm{AgCN}, \quad \mathrm{AgIO}_{3}, \quad \mathrm{AgI}$ $\mathrm{AgNO}_{2,} \mathrm{Ag}_{2} \mathrm{SO}_{4}$. Explain your reasoning.

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Step-by-Step Explanations

QUESTION

Given a salt AB2 that dissociates as A^(2+) + 2B^(-) with a Ksp of 4.0 x 10^-5, how do you calculate its molar solubility?

STEP-BY-STEP ANSWER:

Step 1: Write the dissolution equation: AB2(s) ⇌ A^(2+) (aq) + 2B^(-) (aq).
Step 2: Let the molar solubility be s, so the concentration of A^(2+) is s and concentration of B^(-) is 2s.
Step 3: Write the expression for Ksp: Ksp = [A^(2+)] [B^(-)]^2 = s(2s)^2.
Step 4: Simplify the expression: Ksp = s(4s^2) = 4s^3.
Step 5: Solve for s: 4s^3 = 4.0 x 10^-5, so s^3 = 1.0 x 10^-5. Then, s = (1.0 x 10^-5)^(1/3).
Step 6: Calculate the cube root of 1.0 x 10^-5: s ≈ 2.15 x 10^-2 M.
Final Answer: The molar solubility of AB2 is approximately 2.15 x 10^-2 M.

Calculating Solubility from Ksp

QUESTION

How does the addition of a common ion affect the solubility of a sparingly soluble salt, and how can this be calculated?

STEP-BY-STEP ANSWER:

Step 1: Consider a salt MX dissolving as M^(+) + X^(-) with a known Ksp.
Step 2: If an external source adds X^(-) with concentration C, the solubility equilibrium shifts left, decreasing solubility.
Step 3: Write the equilibrium expression: Ksp = [M^(+)] [X^(-)].
Step 4: Let the solubility in presence of the common ion be s, then [M^(+)] = s and [X^(-)] = C + s.
Step 5: Since s is small compared to C, approximate [X^(-)] ≈ C, so Ksp ≈ s * C.
Step 6: Solve for s: s ≈ Ksp / C.
Final Answer: The solubility s is approximately Ksp divided by the concentration of the common ion, C, showing decreased solubility.

Effect of the Common Ion on Solubility

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Common Mistakes

  • Incorrectly setting up the equilibrium expression for Ksp by not raising ion concentrations to the correct stoichiometric powers.
  • Neglecting the common-ion effect or assuming it has no impact on the solubility of a salt.
  • Overlooking the influence of pH changes on the solubility equilibria, especially in systems with amphoteric species.
  • Failing to recognize the limitations of the Ksp concept in systems where additional equilibria (such as complex ion formation) influence solubility.
  • Assuming that the calculated solubility is unaffected by ionic strength or activity coefficients in non-ideal solutions.