Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393912340
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 15, Problem 15.112QA
Interpretation Introduction

To find:

The color of red cabbage juice at equivalence point when 25 mL of a 0.10 M solution of acetic acid is titrated with 0.10M NaOH.

Expert Solution & Answer
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Answer to Problem 15.112QA

Solution:

Red cabbage juice would have yellow color since pH of the solution at equivalence point is in the alkaline region.

Explanation of Solution

1) Concept:

 We are asked to find the color of red cabbage juice at equivalence point when a weak acid, acetic acid and a strong base, sodium hydroxide are titrated. Since red cabbage juice is a sensitive acid-base indicator, we need to find pH of the solution at equivalence point. Red cabbage juice changes its color as the pH of the solution changes. It shows red color at acidic pH and yellow at alkaline pH.

The equivalence point of a titration is the point at which chemically equivalent quantities of base and acids have been mixed. So, to reach equivalence point 25 mL of 0.10M NaOH is required. We can calculate the moles of the weak acid present and the moles of OH- added using their respective volumes and molarities. As the NaOH is added to the acetic acid solution, the following reaction occurs:

CH3COOH aq+ OH -(aq) CH3COO -(aq)+H2Oaq

At equivalence point,  neither acetic acid nor NaOH remains. The only species present at the equivalence point will be sodium acetate (CH3COONa). From the dissociation reaction of sodium acetate and creating an ICE table, we can find the pH of the solution.

We can calculate [CH3COO-] and then use equilibrium reaction of acetate with water to determine the pH of the solution. The Ka for acetic acid, which can be found in Appendix A5.1, equals 1.74 × 10-5. The pH at equivalence point is greater than 7.00 because the salt sodium acetate is a basic salt made up of a weak acid and a strong base.

2) Formula:

i) Kb= KwKa

ii) pH+pOH=14

iii) pH= -log[H+]

3) Given:

i) Volume of acetic acid is 25 mL=0.025L

ii) Molarity of acetic acid is 0.10M

iii) Molarity of NaOH is 0.10M

4) Calculations:

The initial quantity of CH3COOH in the sample is

25 mL ×1 L1000 mL × 0.10 mol CH3COOH L=0.0025 mol CH3COOH

At equivalence point, mol CH3COOH=mol NaOH=0.0025 mol

The quantity of NaOH required to reach the equivalence point is:

0.0025 mol NaOH 0.1 mol/L=0.025L=25.0 mL

We can use a modified RICE table to determine how many moles of CH3COOH will remain and how many moles of CH3COO- have been produced.

Reaction CH3COOH (aq)+ CH3COOH ( Mol) OH- OH-(Mol) CH3COO -(aq) CH3COO- ( Mol)
Initial 0.0025 0.0025 0
Change -0.0025 -0.0025 +0.0025
Final 0 0 0.0025

The total sample volume is  25mL+25mL=50mL or 0.0500L

The concentration of acetic acid and acetate ion is

CH3COO-= 0.0025 mol0.050 L=0.050M

Because the product of the neutralization reaction is a weak base, we must consider the reaction of the weak base with water to calculate H3O+ at equilibrium, and thus, the pH of the solution.

The equilibrium reaction of acetate with water is as follows:

CH3COO -(aq)+H2O (l)  CH3COOH (aq)+ OH -(aq)

The equilibrium constant for this reaction is Kb= KwKa  where Ka is the acid ionization constant of acetic acid.  We define x as [OH-] produced by the reaction of acetate with water. The complete RICE table of concentrations is as follows:

Reaction CH3COO -aq+H2O l  [CH3COO-] CH3COOH (aq)+ [CH3COOH] OH- (aq) [OH-]
Initial 0.050 0 0
Change -x +x +x
Equilibrium 0.050-x x x

Kb= [CH3COOH] [OH-] [CH3COO-]= xx  0.050-x~ x20.050

Kb= KwKa= 1.00 × 10-141.74 × 10-5 =5.747 × 10-10= x20.050

On solving this equation we get x=0.00005385. Thus [OH-]=0.000053605M

pOH= -log[OH-]

pOH= -log(0.00005385)

pOH=5.27

pH+pOH=14

pH=14-5.27

pH=8.73

pH=8.73

Since pH of the solution at equivalence point is in the alkaline region red cabbage juice would have yellow color at this pH.

Conclusion:

As expected for the titration of a weak acid and a strong base, the pH at the equivalence point is greater than 7.00  because the salt formed in the reaction is a basic salt which then reacts with water to produce OH-.

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Chapter 15 Solutions

Chemistry: An Atoms-Focused Approach

Ch. 15 - Prob. 15.11QACh. 15 - Prob. 15.12QACh. 15 - Prob. 15.13QACh. 15 - Prob. 15.14QACh. 15 - Prob. 15.15QACh. 15 - Prob. 15.16QACh. 15 - Prob. 15.17QACh. 15 - Prob. 15.18QACh. 15 - Prob. 15.19QACh. 15 - Prob. 15.20QACh. 15 - Prob. 15.21QACh. 15 - Prob. 15.22QACh. 15 - Prob. 15.23QACh. 15 - Prob. 15.24QACh. 15 - Prob. 15.25QACh. 15 - Prob. 15.26QACh. 15 - Prob. 15.27QACh. 15 - Prob. 15.28QACh. 15 - Prob. 15.29QACh. 15 - Prob. 15.30QACh. 15 - Prob. 15.31QACh. 15 - Prob. 15.32QACh. 15 - Prob. 15.34QACh. 15 - Prob. 15.35QACh. 15 - Prob. 15.36QACh. 15 - Prob. 15.37QACh. 15 - Prob. 15.38QACh. 15 - Prob. 15.39QACh. 15 - Prob. 15.40QACh. 15 - Prob. 15.41QACh. 15 - Prob. 15.42QACh. 15 - Prob. 15.43QACh. 15 - Prob. 15.44QACh. 15 - Prob. 15.45QACh. 15 - Prob. 15.46QACh. 15 - Prob. 15.47QACh. 15 - Prob. 15.48QACh. 15 - Prob. 15.49QACh. 15 - Prob. 15.50QACh. 15 - Prob. 15.51QACh. 15 - Prob. 15.52QACh. 15 - Prob. 15.53QACh. 15 - Prob. 15.54QACh. 15 - Prob. 15.55QACh. 15 - Prob. 15.56QACh. 15 - Prob. 15.57QACh. 15 - Prob. 15.58QACh. 15 - Prob. 15.59QACh. 15 - Prob. 15.61QACh. 15 - Prob. 15.62QACh. 15 - Prob. 15.63QACh. 15 - Prob. 15.64QACh. 15 - Prob. 15.65QACh. 15 - Prob. 15.66QACh. 15 - Prob. 15.67QACh. 15 - Prob. 15.68QACh. 15 - Prob. 15.69QACh. 15 - Prob. 15.70QACh. 15 - Prob. 15.71QACh. 15 - Prob. 15.72QACh. 15 - Prob. 15.73QACh. 15 - Prob. 15.74QACh. 15 - Prob. 15.75QACh. 15 - Prob. 15.76QACh. 15 - Prob. 15.77QACh. 15 - Prob. 15.78QACh. 15 - Prob. 15.79QACh. 15 - Prob. 15.80QACh. 15 - Prob. 15.81QACh. 15 - Prob. 15.82QACh. 15 - Prob. 15.83QACh. 15 - Prob. 15.84QACh. 15 - Prob. 15.85QACh. 15 - Prob. 15.86QACh. 15 - Prob. 15.87QACh. 15 - Prob. 15.88QACh. 15 - Prob. 15.89QACh. 15 - Prob. 15.90QACh. 15 - Prob. 15.91QACh. 15 - Prob. 15.92QACh. 15 - Prob. 15.93QACh. 15 - Prob. 15.94QACh. 15 - Prob. 15.95QACh. 15 - Prob. 15.96QACh. 15 - Prob. 15.97QACh. 15 - Prob. 15.98QACh. 15 - Prob. 15.99QACh. 15 - Prob. 15.100QACh. 15 - Prob. 15.101QACh. 15 - Prob. 15.102QACh. 15 - Prob. 15.103QACh. 15 - Prob. 15.104QACh. 15 - Prob. 15.105QACh. 15 - Prob. 15.106QACh. 15 - Prob. 15.107QACh. 15 - Prob. 15.108QACh. 15 - Prob. 15.109QACh. 15 - Prob. 15.110QACh. 15 - Prob. 15.111QACh. 15 - Prob. 15.112QACh. 15 - Prob. 15.113QACh. 15 - Prob. 15.114QACh. 15 - Prob. 15.115QACh. 15 - Prob. 15.116QACh. 15 - Prob. 15.117QACh. 15 - Prob. 15.118QACh. 15 - Prob. 15.119QACh. 15 - Prob. 15.120QACh. 15 - Prob. 15.121QACh. 15 - Prob. 15.122QACh. 15 - Prob. 15.123QACh. 15 - Prob. 15.124QACh. 15 - Prob. 15.125QACh. 15 - Prob. 15.126QACh. 15 - Prob. 15.127QACh. 15 - Prob. 15.128QACh. 15 - Prob. 15.129QACh. 15 - Prob. 15.130QACh. 15 - Prob. 15.131QACh. 15 - Prob. 15.132QACh. 15 - Prob. 15.133QACh. 15 - Prob. 15.134QACh. 15 - Prob. 15.135QACh. 15 - Prob. 15.136QACh. 15 - Prob. 15.137QACh. 15 - Prob. 15.138QACh. 15 - Prob. 15.139QACh. 15 - Prob. 15.140QACh. 15 - Prob. 15.141QACh. 15 - Prob. 15.142QACh. 15 - Prob. 15.143QACh. 15 - Prob. 15.144QACh. 15 - Prob. 15.145QACh. 15 - Prob. 15.146QACh. 15 - Prob. 15.147QACh. 15 - Prob. 15.148QACh. 15 - Prob. 15.149QACh. 15 - Prob. 15.150QACh. 15 - Prob. 15.151QACh. 15 - Prob. 15.152QA
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