Book cover for Chemistry: The Molecular Nature of Matter

Chemistry: The Molecular Nature of Matter

Neil D. Jespersen, James E. Brady, Alison Hyslop

ISBN #9781118413920

7th Edition

3,064 Questions

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53,557 Students Helped

Homework Questions

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

This chapter explores the solubility equilibria of sparingly soluble salts, emphasizing the role of Ksp and molar solubility in predicting precipitation. It outlines how factors like common ion effects, pH changes, and complex ion formation alter solubility, and introduces problem-solving tools, including concentration tables and equilibrium calculations, for both laboratory analysis and industrial applications. The techniques of selective precipitation and ligand addition are highlighted as essential for both qualitative analysis and separation processes.

Learning Objectives

1

Describe the principles of solubility equilibria including the definitions and applications of Ksp and molar solubility.

2

Analyze the influence of common ions, pH changes, and complex ion formation on the solubility of sparingly soluble salts.

3

Apply concentration tables and equilibrium calculations to predict precipitate formation and to manipulate equilibria for selective precipitation.

4

Utilize problem-solving tools to design strategies for qualitative analysis and industrial applications involving solubility equilibria.

Key Concepts

CONCEPT

DEFINITION

Ksp (Solubility Product Constant)

A constant that describes the maximum concentration of ions in a saturated solution of a sparingly soluble salt.

Molar Solubility

The number of moles of a solute that can dissolve per liter of solution before the solution becomes saturated.

Common Ion Effect

The decrease in solubility of a sparingly soluble salt when a common ion is added to the solution.

Complex Ion

A species formed when a central metal ion binds to one or more ligands, which can enhance solubility through complexation.

Selective Precipitation

A technique used to separate ions in a solution by carefully controlling conditions to precipitate one ion while leaving others in solution.

pH Influence

The change in solubility that can occur due to the acidic or basic nature of the solution, affecting equilibria particularly in salts of weak acids or bases.

Example Problems

Example 1

What is the difference between an ion product and an ion product constant?

Example 2

Use the following equilibrium to demonstrate why the $K_{\mathrm{sp}}$ expression does not include the concentration of $\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}$ in the denominator: $$ \mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3 \mathrm{Ba}^{2+}(a q)+2 \mathrm{PO}_{4}^{3-}(a q) $$

Example 3

What is the common ion effect? How does Le Châtelier's principle explain it? Use the solubility equilibrium for $\mathrm{AgCl}$ and the addition of $\mathrm{NaCl}$ to the solution to illustrate the common ion effect.

Example 4

With respect to $K_{\mathrm{sp}}$, what conditions must be met if a precipitate is going to form in a solution?

Example 5

What limits the accuracy and reliability of solubility calculations based on $K_{\mathrm{sp}}$ values?
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Step-by-Step Explanations

QUESTION

Given the salt AB with a dissolution equilibrium: AB(s) ⇌ A⁺(aq) + B⁻(aq) and a solubility product Ksp = [A⁺][B⁻] = 1.0 × 10⁻¹⁰, calculate its molar solubility.

STEP-BY-STEP ANSWER:

Step 1: Represent molar solubility as 's'. Since AB dissolves into one A⁺ and one B⁻, [A⁺] = s and [B⁻] = s.
Step 2: Write the solubility product expression: Ksp = s × s = s².
Step 3: Substitute the given Ksp value: s² = 1.0 × 10⁻¹⁰.
Step 4: Solve for s by taking the square root: s = √(1.0 × 10⁻¹⁰) = 1.0 × 10⁻⁵ M.
Final Answer: The molar solubility is 1.0 × 10⁻⁵ M.

Calculating Molar Solubility for a Sparingly Soluble Salt

QUESTION

How does the addition of a common ion affect the solubility of a sparingly soluble salt?

STEP-BY-STEP ANSWER:

Step 1: Identify the common ion in the solution. For example, if salt AB is in solution and ion A⁺ is added from another source, it acts as a common ion.
Step 2: Recognize that an increase in the concentration of A⁺ shifts the dissolution equilibrium to the left according to Le Chatelier’s Principle.
Step 3: Understand that this shift reduces the concentration of B⁻ and results in lower solubility of the salt.
Final Answer: The addition of a common ion decreases the solubility of a sparingly soluble salt by shifting the equilibrium to favor the undissolved solid.

Effect of the Common Ion on Solubility

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Common Mistakes

  • Assuming that Ksp values directly provide molar solubility without considering stoichiometry.
  • Neglecting the impact of common ions or pH changes when calculating solubility.
  • Overlooking the role of complex ion formation, which can dramatically enhance solubility contrary to expectations.
  • Confusing the conditions under which selective precipitation can be applied, leading to improper separation of ions.