Introductory Chemistry: An Active Learning Approach
Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN: 9781305079250
Author: Mark S. Cracolice, Ed Peters
Publisher: Cengage Learning
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Chapter 18, Problem 69E
Interpretation Introduction

Interpretation:

The Ka and the percent ionization of 0.22MHC4H5O3 is to be stated.

Concept introduction:

A strong acid easily releases protons to a base or when dissolved in water. It readily participates in an acid-base reaction. A weak acid undergoes only slight ionization in solution. It does not release protons easily. The major species present in a solution of weak acid is the unionized molecular species. However, some ions are present as minor species. The extent of the ionization is determined by the ionization constant.

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Answer to Problem 69E

The value of Ka of 0.22MHC4H5O3 is 2.6×104 and the percent ionization of 0.22MHC4H5O3 is 3.5%.

Explanation of Solution

The formula to calculate the concentration of hydrogen ions, [H+] is given below.

pH=log[H+]…(1)

The pH of 0.22MHC4H5O3 is 2.12.

Substitute the value of pH in equation (1).

2.12=log[H+][H+]=antilog(2.12)=7.6×103M

The equation for the dissociation HC4H5O3 is shown below.

HC4H5O3(aq)C4H5O3(aq)+H+(aq)

The dissociation constant, Ka for the above reaction is given below.

Ka=[C4H5O3][H+][HC4H5O3]…(2)

For the reaction given above, the concentration of the cation is equal to that of the anion.

[C4H5O3]=[H+]

The value of [H+] is 7.6×103M.

Therefore, the value of [C4H5O3] is 7.6×103M.

The concentration of HC4H5O3, [HC4H5O3] is 0.22M.

Substitute the values of [H+], [C4H5O3], and [HC4H5O3] in equation (2).

Ka=7.6×103×7.6×1030.22=2.6×104

Therefore, the value of Ka is 2.6×104.

The formula to calculate percent ionization is given below.

Percentionization=ConcentrationofionizedacidInitialconcentrationofacid×100%…(3)

The initial concentration is 0.22M.

The concentration of ionized acid, [C4H5O3] is 7.6×103M.

Substitute the initial concentration and the concentration of the ionized acid in equation (3).

Percentionization=7.6×103M0.22M×100%=3.5%

Therefore, the percent ionization of 0.22MHC4H5O3 is 3.5%.

Conclusion

The value of Ka of 0.22MHC4H5O3 is 2.6×104 and the percent ionization of 0.22MHC4H5O3 is 3.5%.

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Chapter 18 Solutions

Introductory Chemistry: An Active Learning Approach

Ch. 18 - Prob. 11ECh. 18 - Prob. 12ECh. 18 - Prob. 13ECh. 18 - Prob. 14ECh. 18 - Prob. 15ECh. 18 - Prob. 16ECh. 18 - Prob. 17ECh. 18 - Prob. 18ECh. 18 - Prob. 19ECh. 18 - Prob. 20ECh. 18 - Prob. 21ECh. 18 - Prob. 22ECh. 18 - Prob. 23ECh. 18 - Prob. 24ECh. 18 - Prob. 25ECh. 18 - Consider the following system at equilibrium at...Ch. 18 - Prob. 27ECh. 18 - Prob. 28ECh. 18 - Prob. 29ECh. 18 - Prob. 30ECh. 18 - Prob. 31ECh. 18 - Prob. 32ECh. 18 - Prob. 33ECh. 18 - Prob. 34ECh. 18 - Which direction of the equilibrium...Ch. 18 - Prob. 36ECh. 18 - Prob. 37ECh. 18 - Prob. 38ECh. 18 - Prob. 39ECh. 18 - Consider the following system at equilibrium at...Ch. 18 - Prob. 41ECh. 18 - Prob. 42ECh. 18 - Prob. 43ECh. 18 - Prob. 44ECh. 18 - Prob. 45ECh. 18 - Prob. 46ECh. 18 - Prob. 47ECh. 18 - Prob. 48ECh. 18 - Prob. 49ECh. 18 - Prob. 50ECh. 18 - The equilibrium between nitrogen monoxide, oxygen,...Ch. 18 - The equilibrium constant expression for a given...Ch. 18 - Prob. 53ECh. 18 - For the following system, K=4.86105 at 298K:...Ch. 18 - Prob. 55ECh. 18 - Prob. 56ECh. 18 - Prob. 57ECh. 18 - Question 57 and 58: In Chapter 9, we discussed how...Ch. 18 - Prob. 59ECh. 18 - A student measures the molar solubility of...Ch. 18 - Prob. 61ECh. 18 - Prob. 62ECh. 18 - Find the moles per liter and grams per 100mL...Ch. 18 - Prob. 64ECh. 18 - Prob. 65ECh. 18 - Prob. 66ECh. 18 - Prob. 67ECh. 18 - Ksp for silver hydroxide is 2.0108. Calculate the...Ch. 18 - Prob. 69ECh. 18 - Prob. 70ECh. 18 - Prob. 71ECh. 18 - Prob. 72ECh. 18 - Prob. 73ECh. 18 - Prob. 74ECh. 18 - Prob. 75ECh. 18 - Prob. 76ECh. 18 - Prob. 77ECh. 18 - Prob. 78ECh. 18 - Prob. 79ECh. 18 - Classify each of the following statements as true...Ch. 18 - Prob. 81ECh. 18 - Prob. 82ECh. 18 - Prob. 83ECh. 18 - Prob. 84ECh. 18 - Prob. 85ECh. 18 - Prob. 86ECh. 18 - Prob. 87ECh. 18 - Prob. 88ECh. 18 - Prob. 89ECh. 18 - Prob. 90ECh. 18 - Hard water has a high concentration of calcium and...Ch. 18 - Prob. 18.1TCCh. 18 - Prob. 18.3TCCh. 18 - a What happens to a reaction rate as temperature...Ch. 18 - Prob. 18.5TCCh. 18 - Write a brief description of the relationships...Ch. 18 - Prob. 2CLECh. 18 - Prob. 3CLECh. 18 - Prob. 4CLECh. 18 - Prob. 5CLECh. 18 - Prob. 1PECh. 18 - Prob. 2PECh. 18 - Prob. 3PECh. 18 - Prob. 4PECh. 18 - Prob. 5PECh. 18 - Prob. 6PECh. 18 - Prob. 7PECh. 18 - Prob. 8PECh. 18 - Prob. 9PECh. 18 - Prob. 10PECh. 18 - Prob. 11PECh. 18 - Prob. 12PECh. 18 - What is the molar solubility of calcium fluoride...Ch. 18 - Prob. 14PECh. 18 - Prob. 15PECh. 18 - Prob. 16PECh. 18 - Prob. 17PECh. 18 - Prob. 18PECh. 18 - Prob. 19PE
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