Skip to main content
Pearson+ LogoPearson+ Logo
Ch.16 - Aqueous Equilibria: Acids & Bases
All textbooksMcMurry 8th EditionCh.16 - Aqueous Equilibria: Acids & BasesProblem 157
Chapter 16, Problem 157

We’ve said that alkali metal cations do not react appreciably with water to produce H3O+ ions, but in fact, all cations are acidic to some extent. The most acidic alkali metal cation is the smallest one, Li+, which has Ka = 2.5 * 10^-14 for the reaction: Li(H2O)4+ (aq) + H2O (l) ⇌ H3O+ (aq) + Li(H2O)3(OH) (aq). This reaction and the dissociation of water must be considered simultaneously in calculating the pH of Li+ solutions, which nevertheless have pH ≈ 7. Check this by calculating the pH of a 0.10 M LiCl solution.

Verified step by step guidance
1
Step 1: Identify the relevant reactions. The primary reaction is the hydrolysis of the hydrated lithium ion: Li(H2O)4+ (aq) + H2O (l) ⇌ H3O+ (aq) + Li(H2O)3(OH) (aq). Additionally, consider the autoionization of water: 2H2O (l) ⇌ H3O+ (aq) + OH- (aq).
Step 2: Write the expression for the equilibrium constant (Ka) for the hydrolysis reaction: Ka = [H3O+][Li(H2O)3(OH)] / [Li(H2O)4+]. Given Ka = 2.5 * 10^-14, use this to find [H3O+].
Step 3: Assume that the initial concentration of Li(H2O)4+ is equal to the concentration of LiCl, which is 0.10 M. Set up an ICE (Initial, Change, Equilibrium) table to determine the changes in concentrations at equilibrium.
Step 4: Use the small x approximation, assuming that the change in concentration of Li(H2O)4+ is small compared to its initial concentration. This simplifies the equilibrium expression to solve for [H3O+].
Step 5: Calculate the pH using the formula pH = -log[H3O+]. Since the solution is expected to have a pH close to 7, verify that the calculated pH is reasonable given the small Ka value.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Acid-Base Chemistry

Acid-base chemistry involves the study of proton donors (acids) and proton acceptors (bases). In aqueous solutions, acids increase the concentration of H3O+ ions, while bases increase OH- ions. Understanding the behavior of cations, such as Li+, in water is crucial, as they can influence the pH by participating in acid-base reactions.
Recommended video:
Guided course
02:00
Arrhenius Acids and Bases

Dissociation of Water

Water can dissociate into H3O+ and OH- ions, establishing a dynamic equilibrium. This self-ionization is essential for understanding pH, as pure water has a neutral pH of 7, where the concentrations of H3O+ and OH- are equal. The presence of cations like Li+ can shift this equilibrium, affecting the overall pH of the solution.
Recommended video:
Guided course
04:52
Percent Dissociation Example

pH Calculation

pH is a measure of the acidity or basicity of a solution, calculated as the negative logarithm of the H3O+ ion concentration. In the case of LiCl, the pH must account for both the contribution of Li+ ions and the dissociation of water. Calculating the pH involves determining the equilibrium concentrations of H3O+ and OH- ions, which can be influenced by the cation's acidity.
Recommended video:
Guided course
01:30
pH Calculation Example
Related Practice
Textbook Question
In the case of very weak acids, 3H3O+ 4 from the dissociation of water is significant compared with 3H3O+ 4 from the dissociation of the weak acid. The sugar substitute saccharin 1C7H5NO3S2, for example, is a very weak acid having Ka = 2.1 * 10-12 and a solubility in water of 348 mg/100 mL. Calculate 3H3O+ 4 in a saturated solution of saccharin. (Hint: Equilibrium equations for the dissociation of saccharin and water must be solved simultaneously.)
Textbook Question
In aqueous solution, sodium acetate behaves as a strong electrolyte, yielding Na+ cations and CH3CO2 - anions. A particular solution of sodium acetate has a pH of 9.07 and a density of 1.0085 g/mL. What is the molality of this solution, and what is its freezing point?
Textbook Question

During a certain time period, 4.0 million tons of SO2 were released into the atmosphere and subsequently oxidized to SO3. As explained in the Inquiry, the acid rain produced when the SO3 dissolves in water can damage marble statues: CaCO3(s) + H2SO4(aq) → CaSO4(aq) + CO2(g) + H2O(l) (a) How many 500 pound marble statues could be damaged by the acid rain? (Assume that the statues are pure CaCO3 and that a statue is damaged when 3.0% of its mass is dissolved.)

Textbook Question

A 1.000 L sample of HF gas at 20.0 °C and 0.601 atm pressure was dissolved in enough water to make 50.0 mL of hydrofluoric acid. (a) What is the pH of the solution?

Textbook Question

A 1.000 L sample of HF gas at 20.0 °C and 0.601 atm pressure was dissolved in enough water to make 50.0 mL of hydrofluoric acid. (b) To what volume must you dilute the solution to triple the percent dissociation?