Problem 87
Question
What is the difference between molar solubility and solubility product?
Step-by-Step Solution
Verified Answer
Answer: The key differences between molar solubility and solubility product are:
1. Molar solubility is a quantitative measure of the maximum concentration of a solute that can be dissolved in a solvent, whereas solubility product is an equilibrium constant for the dissolution of a sparingly soluble ionic compound.
2. Molar solubility is expressed in moles per liter (mol/L), while solubility product is a dimensionless number.
3. The relationship between molar solubility and solubility product depends on the stoichiometry of the dissolution reaction.
1Step 1: Definition of Molar Solubility
Molar solubility is a measure of the maximum amount of a substance (solute) that can dissolve in a given volume of a solvent to form a saturated solution. It is expressed in units of moles per liter (mol/L). Molar solubility informs us about the maximum concentration of a solute that can be dissolved in the solvent to maintain equilibrium.
2Step 2: Definition of Solubility Product
Solubility Product (K_sp) is the equilibrium constant for the dissolution of a sparingly soluble ionic compound in a solvent, most commonly water. It describes the relationship between the concentrations of the ions and the overall solubility of a compound. A higher K_sp value for a compound indicates higher solubility. The solubility product is mathematically expressed as the product of the molar concentrations of the ions, each raised to the power of their coefficients in the balanced chemical equation.
3Step 3: Difference Between Molar Solubility and Solubility Product
Molar solubility and solubility product are related, but they are not the same thing. The key differences between the two concepts are:
1. Molar solubility is a quantitative measure of the maximum concentration of a solute that can be dissolved in a solvent, whereas solubility product is an equilibrium constant for the dissolution of a sparingly soluble ionic compound.
2. Molar solubility is expressed in moles per liter (mol/L), while solubility product is a dimensionless number.
3. The relationship between molar solubility and solubility product depends on the stoichiometry of the dissolution reaction. For example, for the dissolution of a compound AB into A^+ and B^-, the relationship is K_sp = [A^+][B^-] = (molar solubility)^2.
In conclusion, molar solubility and solubility product are two different concepts that describe the solubility of a compound. While molar solubility gives us an idea about the maximum amount of solute that can dissolve in a solvent, solubility product helps us understand the equilibrium between the dissolved and undissolved forms of a sparingly soluble ionic compound.
Key Concepts
Solubility ProductEquilibrium ConstantIonic CompoundsStoichiometry
Solubility Product
The solubility product constant, denoted as \(K_{sp}\), is a key concept in the study of ionic compounds and their solubility in water. It provides a valuable insight into how a sparingly soluble compound behaves in a solution. The solubility product is a specific form of the equilibrium constant. It applies to the dissolution of ionic compounds in water, where an equilibrium is established between the solid and its ions in solution.
- For a compound \(AB\) that dissolves to form \(A^+\) and \(B^-\), the equation for the solubility product is \(K_{sp} = [A^+][B^-]\).
- This product is considered at equilibrium when the solution is saturated, meaning no more solute can dissolve.
- Each ion's concentration is taken to the power of its coefficient in the balanced dissolution equation.
Equilibrium Constant
The equilibrium constant is a vital concept in chemistry, encapsulating the idea that chemical reactions can reach a point where the rates of the forward and reverse reactions are equal. This balance, known as equilibrium, can be described mathematically using an equilibrium constant.
- This constant provides a ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their coefficients in the balanced equation.
- For dissolutions, the equilibrium constant specific to solubility is the solubility product \(K_{sp}\).
- A larger value of an equilibrium constant indicates that the products of the reaction are favored.
Ionic Compounds
Ionic compounds play a crucial role in solubility, as they are often the substances in question when discussing solubility products. These compounds consist of positive and negative ions bound together by ionic bonds, forming a crystal lattice structure. When they dissolve in water, they dissociate into their constituent ions.
- For example, sodium chloride \(NaCl\) dissolves to form \(Na^+\) and \(Cl^-\) ions.
- The solubility of these compounds depends on the strength of the ionic bonds and the polarity of water.
- Not all ionic compounds dissolve easily; some are said to be sparingly soluble, contributing minimally to the solution's concentration of ions.
Stoichiometry
Stoichiometry is the bridge that connects the chemical equation with the quantities of reactants and products in a reaction. It quantifies the relationships between the amounts of substances participating in a chemical process. This concept is essential when working with solubility products and dissolution reactions.
- It allows us to calculate how much product forms from a given quantity of reactants using ratios derived from the balanced chemical equation.
- In solubility calculations, stoichiometry helps us determine the relationship between molar solubility and \(K_{sp}\).
- The coefficients in a chemical equation indicate the proportions of ions, guiding how we calculate concentrations in a saturated solution.
Other exercises in this chapter
Problem 82
What is the pH of \(0.50 M \mathrm{CrCl}_{3} ?\)
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What is the pH of \(1.00 M \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} ?\)
View solution Problem 88
Give an example of how the common-ion effect limits the dissolution of a sparingly soluble ionic compound.
View solution Problem 89
Which cation will precipitate first as a carbonate mineral from an equimolar solution of \(\mathrm{Mg}^{2+}, \mathrm{Ca}^{2+},\) and \(\mathrm{Sr}^{2+} ?\)
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