FOUND.OF COLLEGE CHEMISTRY
FOUND.OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781119234555
Author: Hein
Publisher: WILEY
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 16, Problem 20PE

(a)

Interpretation Introduction

Interpretation:

Concentration H+ for solution of 0.025 M lactic acid (HC3H5O2) has to be calculated.

Concept Introduction:

Consider ionization of a weak acid HA that is in equilibrium with its ions and is as follows:

  HA(aq)H+(aq)+A(aq)

The equilibrium constant for ionization of weak acid is called ionization constant Ka and is expressed as follows:

  Ka=[H+][A][HA]

Ka is defined as the ratio of concentration of products to concentration of reactants.

(a)

Expert Solution
Check Mark

Explanation of Solution

The ionization of lactic acid is as follows:

  HC3H5O2(aq)H+(aq)+C3H5O2(aq)

The expression for Ka for given reaction is as follows:

  Ka=[H+][C3H5O2][HC3H5O2]        (1)

Through chemical equation it is evident that one C3H5O2 is formed for every H+. Thus concentration of H+ is equal to concentration of C3H5O2. Thus consider x for H+ and C3H5O2. Value of x is small so concentration of HC3H5O2 can be considered as 0.025M.

Rearrange equation (1) for [H+].

  [H+]=Ka[HC3H5O2][C3H5O2]        (2)

Substitute x for [C3H5O2], x for [H+], 8.4×104 for Ka and 0.025 for [HC3H5O2] in equation (2).

  x=(8.4×104)(0.025)xx2=0.000021

Solve this equation for x.

  x=4.6×103

Hence, [H+] in 0.025 M lactic acid, HC3H5O2 is 4.6×103 M.

(b)

Interpretation Introduction

Interpretation:

pH for solution of 0.025 M lactic acid (HC3H5O2) has to be calculated.

Concept Introduction:

pH is defined as the concentration of hydrogen ion. It also explains about the acidity of solution.

(b)

Expert Solution
Check Mark

Explanation of Solution

The ionization of lactic acid is as follows:

  HC3H5O2(aq)H+(aq)+C3H5O2(aq)

The expression for Ka for given reaction is as follows:

  Ka=[H+][C3H5O2][HC3H5O2]        (1)

Through chemical equation it is evident that one C3H5O2 is formed for every H+. Thus concentration of H+ is equal to concentration of C3H5O2. Thus consider x for H+ and C3H5O2. Value of x is small so concentration of HC3H5O2 can be considered as 0.025M.

Rearrange equation (1) for [H+].

  [H+]=Ka[HC3H5O2][C3H5O2]        (2)

Substitute x for [C3H5O2], x for [H+], 8.4×104 for Ka and 0.025 for [HC3H5O2] in equation (2).

  x=(8.4×104)(0.025)xx2=0.000021

Solve this equation for x.

  x=4.6×103

Hence, [H+] in 0.025 M lactic acid, HC3H5O2 is 4.6×103 M.

pH is defined as negative logarithm of concentration of H+. It is calculated as follows:

  pH=log[H+]        (3)

Substitute 4.6×103 M for H+ in equation (3).

  pH=log[4.6×103]=log4.6+3log10=0.66+3=2.34 

Hence, pH in 0.025 M lactic acid, HC3H5O2 is 2.34 .

(c)

Interpretation Introduction

Interpretation:

Percent ionization of solution of 0.025 M lactic acid (HC3H5O2) has to be calculated.

Concept Introduction:

Consider a weak acid HA that is in equilibrium with its ions and is as follows:

  HA(aq)H+(aq)+A(aq)

Percent ionization for a weak acid is defined as the ratio of concentration of H+ or A to initial concentration of HA that is multiplied with 100 and is expressed as follows:

  Percent ionization=([H+]or[A][HA])100

(c)

Expert Solution
Check Mark

Explanation of Solution

The ionization of lactic acid is as follows:

  HC3H5O2(aq)H+(aq)+C3H5O2(aq)

The expression for Ka for given reaction is as follows:

  Ka=[H+][C3H5O2][HC3H5O2]        (1)

Through chemical equation it is evident that one C3H5O2 is formed for every H+. Thus concentration of H+ is equal to concentration of C3H5O2. Thus consider x for H+ and C3H5O2. Value of x is small so concentration of HC3H5O2 can be considered as 0.025M.

Rearrange equation (1) for [H+].

  [H+]=Ka[HC3H5O2][C3H5O2]        (2)

Substitute x for [C3H5O2], x for [H+], 8.4×104 for Ka and 0.025 for [HC3H5O2] in equation (2).

  x=(8.4×104)(0.025)xx2=0.000021

Solve this equation for x.

  x=4.6×103

Hence, [H+] in 0.025 M lactic acid, HC3H5O2 is 4.6×103 M.

The percent ionization of HC3H5O2 is as follows:

  Percent ionization=([H+][HC3H5O2])100=(4.6×103 M0.025 M)100=18.4 %

Hence, percent ionization of HC3H5O2 is 18.4 %.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A. B. b. Now consider the two bicyclic molecules A. and B. Note that A. is a dianion and B. is a neutral molecule. One of these molecules is a highly reactive compound first characterized in frozen noble gas matrices, that self-reacts rapidly at temperatures above liquid nitrogen temperature. The other compound was isolated at room temperature in the early 1960s, and is a stable ligand used in organometallic chemistry. Which molecule is the more stable molecule, and why?
Where are the chiral centers in this molecule? Also is this compound meso yes or no?
PLEASE HELP! URGENT!

Chapter 16 Solutions

FOUND.OF COLLEGE CHEMISTRY

Ch. 16.6 - Prob. 16.11PCh. 16.6 - Prob. 16.12PCh. 16.7 - Prob. 16.13PCh. 16.7 - Prob. 16.14PCh. 16.7 - Prob. 16.15PCh. 16.8 - Prob. 16.16PCh. 16 - Prob. 1RQCh. 16 - Prob. 2RQCh. 16 - Prob. 3RQCh. 16 - Prob. 4RQCh. 16 - Prob. 5RQCh. 16 - Prob. 6RQCh. 16 - Prob. 7RQCh. 16 - Prob. 8RQCh. 16 - Prob. 9RQCh. 16 - Prob. 10RQCh. 16 - Prob. 11RQCh. 16 - Prob. 12RQCh. 16 - Prob. 13RQCh. 16 - Prob. 14RQCh. 16 - Prob. 15RQCh. 16 - Prob. 16RQCh. 16 - Prob. 17RQCh. 16 - Prob. 18RQCh. 16 - Prob. 19RQCh. 16 - Prob. 20RQCh. 16 - Prob. 21RQCh. 16 - Prob. 22RQCh. 16 - Prob. 23RQCh. 16 - Prob. 24RQCh. 16 - Prob. 25RQCh. 16 - Prob. 26RQCh. 16 - Prob. 27RQCh. 16 - Prob. 1PECh. 16 - Prob. 2PECh. 16 - Prob. 3PECh. 16 - Prob. 4PECh. 16 - Prob. 5PECh. 16 - Prob. 6PECh. 16 - Prob. 7PECh. 16 - Prob. 8PECh. 16 - Prob. 9PECh. 16 - Prob. 10PECh. 16 - Prob. 11PECh. 16 - Prob. 12PECh. 16 - Prob. 13PECh. 16 - Prob. 14PECh. 16 - Prob. 15PECh. 16 - Prob. 16PECh. 16 - Prob. 17PECh. 16 - Prob. 18PECh. 16 - Prob. 19PECh. 16 - Prob. 20PECh. 16 - Prob. 21PECh. 16 - Prob. 22PECh. 16 - Prob. 23PECh. 16 - Prob. 24PECh. 16 - Prob. 25PECh. 16 - Prob. 26PECh. 16 - Prob. 27PECh. 16 - Prob. 28PECh. 16 - Prob. 29PECh. 16 - Prob. 30PECh. 16 - Prob. 31PECh. 16 - Prob. 32PECh. 16 - Prob. 33PECh. 16 - Prob. 34PECh. 16 - Prob. 35PECh. 16 - Prob. 36PECh. 16 - Prob. 37PECh. 16 - Prob. 38PECh. 16 - Prob. 39PECh. 16 - Prob. 40PECh. 16 - Prob. 41PECh. 16 - Prob. 42PECh. 16 - Prob. 43PECh. 16 - Prob. 44PECh. 16 - Prob. 45PECh. 16 - Prob. 46PECh. 16 - Prob. 47PECh. 16 - Prob. 48PECh. 16 - Prob. 49AECh. 16 - Prob. 50AECh. 16 - Prob. 51AECh. 16 - Prob. 52AECh. 16 - Prob. 53AECh. 16 - Prob. 54AECh. 16 - Prob. 55AECh. 16 - Prob. 56AECh. 16 - Prob. 57AECh. 16 - Prob. 58AECh. 16 - Prob. 59AECh. 16 - Prob. 60AECh. 16 - Prob. 61AECh. 16 - Prob. 62AECh. 16 - Prob. 63AECh. 16 - Prob. 64AECh. 16 - Prob. 65AECh. 16 - Prob. 66AECh. 16 - Prob. 67AECh. 16 - Prob. 68AECh. 16 - Prob. 69AECh. 16 - Prob. 70AECh. 16 - Prob. 71AECh. 16 - Prob. 72AECh. 16 - Prob. 73AECh. 16 - Prob. 74AECh. 16 - Prob. 75AECh. 16 - Prob. 76AECh. 16 - Prob. 77AECh. 16 - Prob. 78AECh. 16 - Prob. 79AECh. 16 - Prob. 80AECh. 16 - Prob. 81AECh. 16 - Prob. 83AECh. 16 - Prob. 84AECh. 16 - Prob. 85AECh. 16 - Prob. 86CECh. 16 - Prob. 87CECh. 16 - Prob. 88CECh. 16 - Prob. 89CE
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning
Text book image
General, Organic, and Biological Chemistry
Chemistry
ISBN:9781285853918
Author:H. Stephen Stoker
Publisher:Cengage Learning
General Chemistry | Acids & Bases; Author: Ninja Nerd;https://www.youtube.com/watch?v=AOr_5tbgfQ0;License: Standard YouTube License, CC-BY