Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Textbook Question
Chapter 14, Problem 7RQ

Sketch the titration curve for a weak acid titrated by a strong base. When performing calculations concerning weak acid–strong base titrations, the general two-slep procedure is to solve a stoichiometry problem first, then to solve an equilibrium problem to determine the pH. What reaction takes place in the stoichiometry part of the problem? What is assumed about this reaction?

At the various points in your titration curve, list the major species present after the strong base (NaOH, for example) reacts to completion with the weak acid, HA. What equilibrium problem would you solve at the various points in your titration curve to calculate the pH? Why is pH > 7.0 at the equivalence point of a weak acid-strong base titration? Does the pH at the halfway point to equivalence have to be less than 7.0? What does the pH at the halfway point equal? Compare and contrast the titration curves for a strong acid–strong base titration and a weak acid–strong base titration.

Expert Solution & Answer
Check Mark
Interpretation Introduction

Interpretation: The titration curve for a weak acid titrated by a weak base is to be sketched. The reaction takes place in the stoichiometric part of the problem is to be stated. The equilibrium problem, at the various points in the titration curve to calculate the pH is to be stated. The reason for the value of pH is greater than 7 at the equilibrium point for the titration of weak acid-strong base is to be stated. Whether the pH at the halfway point to equivalence have to be less than 7 is to be stated. The titration curves for a strong acid-strong base titration are to be compared and to be stated.

Concept introduction: The reaction between an acid and a base takes place with the formation of a salt and a water molecule.

The hydrogen ion concentration of the solution is known as pH of the solution.

It is the negative logarithm of Hydrogen ion concentration.

At the equivalence point the moles of the acid and base in the solution are same.

At the half equivalence point the concentration of titrant is just half of the initial amount.

For the given value of dissociation constant the pH=pKa.

To determine: The sketch of titration curve for a weak acid-strong base titration; The reaction takes place in stoichiometric part of the problem; The assumption about this reaction; The list of the major species present after the completion reaction between strong base with weak acid; The equilibrium problem which involve at the various points in the calculation of pH; The reason for the value of pH>7 at the equivalence point in the weak acid-strong base titration. If the pH at the halfway point to equivalence point is less than 7; The pH at the halfway point; The comparison between the titration curves of a strong acid-strong base titration and a weak acid-strong base titration.

Explanation of Solution

Explanation

A titration curve is plot of pH Vs volume of titrant. In such plots the pH is taken as independent variable while volume of titrant is taken as independent variable. Such plots are also known as pH curves.

In the calculation of weak acid-strong base titration, the two step procedure is involved. First one is stoichiometric problem and second is equilibrium problem. The step wise procedure is stated as follows,

  • Stoichiometric problem: The completion reaction is to be assumed in the weak acid-strong base titration.
  • Equilibrium problem: The pH at the equivalence point and the position of weak acid equilibrium is determined.

The titration of 50.0mlof0.10M acetic acid (Ka=1.8×105) with 0.10MNaOH is to be assumed. The pH curve for this titration is given as,

Chemistry: An Atoms First Approach, Chapter 14, Problem 7RQ

Figure 1

The equivalence point occurs on the addition of 50mlofNaOH. Here the amount of hydroxide ion is exactly equal to the original amount of acid. At the equivalence point the pH is greater than 7 because the acetate ion present at this point is a base and it reacts with water to produce hydroxide ion.

The reaction involves in the part of stoichiometric problem is given as,

OH-+CH3COOHCH3COO-+H2OBeforereaction10×0.10=1.0mmol50.0×0.10=5.0mmol0Afterrecation05.01.0=4.0mmol1.0mmol

Here the amount of Hydroxide ion is lesser than the acid. Therefore, it consumed completely and called as limiting reactant.

The pH at various points is calculated by representing the volumes of added NaOH as,

When 0.0ml0.10MNaOH has been added.

In this situation only weak acid that is acetic acid id present in the solution.

Therefore the equilibrium of CH3COOH is represented by the equation,

CH3COOHCH3COO+H+

Molarity of CH3COOH is 0.10M. The number of moles is calculated by the formula,

n=C×V

Where,

  • n is the number of moles.
  • C is the concentration of the solution.
  • V is the volume of the solution.

Substitute the values of concentrations and volume in the above equation.

n=C×V=0.100M×501000L=0.05mol

Now x is supposed to be the change in moles. The equilibrium reaction with the calculated moles is expressed in ICE (initial, change, equilibrium) table as,

CH3COOHH++CH3COOInitialmol0.0500Changeinmolx+x+xEquilibriummol0.05x+x+x

The acid dissociation constant (Ka) for this reaction is written as,

Ka=(x)(x)(0.05x)

Substitute the value of Kb in the above equation,

Kb=(x)(x)(0.05x)1.8×105=x2(0.05x)

Simplify the above equation.

x=2.9×104

Therefore, the concentration of H+ is 2.9×104.

The pH of the above given hydride ion is calculated as,

pH=log(H+)=log(2.9×104)=log(2.9)+4=3.53_

When 10.0ml0.10MNaOH has been added.

The species present in the mixture before the reaction are CH3COOH,OH,Na+ and H2O.

The stoichiometric problem part is represented as,

OH+CH3COOHCH3COO-+H2OBeforereaction10ml×0.10M=1.0mmol50.0ml×0.10M=5.0mmol0Afterrecation05.01.0=4.0mmol1.0mmol

The equilibrium problem part is given below.

Acetic acid is much stronger acid than water and CH3COO- is the conjugate base of CH3COOH. The pH of the acetic acid is determined as,

CH3COOH(aq)CH3COO(aq)+H+(aq)Ka=[H+][CH3COO][CH3COOH] (1)

Where,

  • Ka is the acid dissociation constant.
  • [H+] is the concentration of Hydrogen ion.
  • [CH3COO] is the concentration of CH3COO.
  • [CH3COOH] is the concentration of CH3COOH.

The concentration of [CH3COOH] is calculated as,

[CH3COOH]0=4.0mmol(50.0+10.0)ml[CH3COOH]0=4.060.0[CH3COOH]0=0.067

The concentration of [CH3COO] is calculated as,

[CH3COO]0=1.0mmol(50.0+10.0)ml[CH3COO]0=1.060.0[CH3COO]0=0.0167

The concentration of [H+] is approximately zero.

The ICE-table for the equilibrium reaction is given as,

CH3COOH(aq)H++CH3COOInitialconcentration0.06700.0167Changeconcentrationx+x+xEquilibriumconcentration0.067xx0.0167+x

Substitute the values of concentrations in the equation (1).

Ka=[H+][CH3COO][CH3COOH]1.8×105=x(0.0167+x)(0.067x)

Neglect the values of x, the above equation becomes,

1.8×105=x(0.0167)(0.067)x=0.12×1050.0167x=7.2×105

Therefore, the concentration of H+ is 7.2×105.

The pH of the above given hydride ion is calculated as,

pH=log(H+)=log(7.2×105)=log(7.2)+5=4.14_

When 25.0ml0.10MNaOH has been added.

The process is exactly similar as in part (b). Only volume is changed here that is 50.0+25.0=75.0ml.

The stoichiometric problem part is represented as,

OH-+CH3COOHCH3COO-+H2OBeforereaction25ml×0.10M=2.5mmol50.0ml×0.10M=2.5mmol0Afterrecation05.02.5=2.5mmol2.5mmol

The equilibrium problem part is given below.

Acetic acid is much stronger acid than water and CH3COO- is the conjugate base of CH3COOH. The pH of the acetic acid is determined as,

The concentration of [CH3COOH] is calculated as,

[CH3COOH]0=2.5mmol(50.0+25.0)ml[CH3COOH]0=2.575.0[CH3COOH]0=0.033

The concentration of [CH3COO] is calculated as,

[CH3COO]0=2.5mmol(50.0+25.0)ml[CH3COO]0=2.575.0[CH3COO]0=0.33

The concentration of [H+] is approximately zero.

The ICE-table for the equilibrium reaction is given as,

CH3COOH(aq)H++CH3COOInitialconcentration0.03300.033Changeconcentrationx+x+xEquilibriumconcentration0.033xx0.033+x

Substitute the values of concentrations in the equation (1).

Ka=[H+][CH3COO][CH3COOH]1.8×105=x(0.033+x)(0.033x)

Neglect the values of x, the above equation becomes,

1.8×105=x(0.033)(0.033)x=1.8×105

Therefore, the concentration of H+ is 1.8×105.

The pH of the above given hydride ion is calculated as,

pH=log(H+)=log(1.8×105)=log(1.8)+5=4.74_

This is the special point in the titration because it is the halfway to the equivalence point.

Here the  [H+]=Ka

Therefore, pH=pKa.

When 50.0ml 0.10MNaOH has been added.

This is the equivalence point of the titration. 5.0mmolOH has been added, which will just react with the 5.0mmolCH3COOH originally present. At this point the major species present are, Na+,CH3COO and H2O.

This solution contains CH3COO which is a base. Therefore, this base combines with proton that is Hydrogen ion. The only source of protons us water. So, the reaction is as follows,

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

The Kb for this reaction is written as,

Kb=[CH3COOH][OH][CH3COO]Kb=KwKaKb=1.0×10141.8×105Kb=5.6×1010

The concentration of [CH3COO] is calculated as,

[CH3COO]0=5.0mmol(50.0+50.0)ml[CH3COO]0=5.0100.0[CH3COO]0=0.050

The concentration of [H+] is approximately zero.

The concentration of [CH3COOH] is also approximately is zero.

The corresponding ICE-table is given as,

CH3COO(aq)+H2O(l)CH3COOH(aq)+OH(aq)Initial0.05000Changex+x+xEquilibrium0.050xxx

Therefore, the equilibrium constant is written as,

Kb=[CH3COOH][OH][CH3COO]5.6×1010=(x)(x)(0.050x)

The approximation is valid by the 5% rule. So,

5.6×1010=(x)(x)(0.050)x2=28.0×1012x=28.0×1012x=5.3×106

Therefore, the [OH] concentration in terms of Kw is written as,

[OH][H+]=Kw(5.3×106)[H+]=1.0×1014[H+]=1.0×10145.3×106[H+]=1.9×109

The pH of the above given hydride ion is calculated as,

pH=log(H+)=log(1.9×109)=log(1.9)+9=8.72_

The pH at the equivalence point of a titration of a weak acid with a strong base is always greater than 7 the reason is that the anion part of the acid is remains in the solution, as all of the acid has been converted to its conjugated base by the addition of strong acid.

Yes, from the titration curve it is clear that the pH at the halfway point to the equivalence is less than 7.

The value of pH at the halfway point of titration is equal to pKa. Mathematically it is given as,

pH=pKapH=log(Ka)

The comparison of the titration curve for strong acid-strong base and weak acid-strong base titration is given as,

  • The shape of both the curves is same after the equivalence point but before the equivalence point it is different.
  • In the beginning of the titration of the weak acid-strong base, the pH increases sharply than the strong acid-base titration.
  • The equivalence point for the strong acid-base titration occur at the pH 7.while for the weak acid-strong base titration the equivalence point occur at the pH>7 because of the high basicity of conjugated base of weak acid.
Conclusion

Conclusion

All the questions based on titration and titration curve have been rightfully stated

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Chapter 14 Solutions

Chemistry: An Atoms First Approach

Ch. 14 - What are the major species in solution after...Ch. 14 - Prob. 2ALQCh. 14 - Prob. 3ALQCh. 14 - Prob. 4ALQCh. 14 - Sketch two pH curves, one for the titration of a...Ch. 14 - Prob. 6ALQCh. 14 - Prob. 7ALQCh. 14 - You have a solution of the weak acid HA and add...Ch. 14 - The common ion effect for weak acids is to...Ch. 14 - Prob. 10QCh. 14 - Prob. 11QCh. 14 - Consider the following pH curves for 100.0 mL of...Ch. 14 - An acid is titrated with NaOH. The following...Ch. 14 - Consider the following four titrations. i. 100.0...Ch. 14 - Prob. 15QCh. 14 - Prob. 16QCh. 14 - How many of the following are buffered solutions?...Ch. 14 - Which of the following can be classified as buffer...Ch. 14 - A certain buffer is made by dissolving NaHCO3 and...Ch. 14 - Prob. 20ECh. 14 - Calculate the pH of each of the following...Ch. 14 - Calculate the pH of each of the following...Ch. 14 - Prob. 23ECh. 14 - Compare the percent ionization of the base in...Ch. 14 - Prob. 25ECh. 14 - Calculate the pH after 0.020 mole of HCl is added...Ch. 14 - Calculate the pH after 0.020 mole of NaOH is added...Ch. 14 - Calculate the pH after 0.020 mole of NaOH is added...Ch. 14 - Which of the solutions in Exercise 21 shows the...Ch. 14 - Prob. 30ECh. 14 - Calculate the pH of a solution that is 1.00 M HNO2...Ch. 14 - Calculate the pH of a solution that is 0.60 M HF...Ch. 14 - Calculate the pH after 0.10 mole of NaOH is added...Ch. 14 - Calculate the pH after 0.10 mole of NaOH is added...Ch. 14 - Calculate the pH of each of the following buffered...Ch. 14 - Prob. 36ECh. 14 - Calculate the pH of a buffered solution prepared...Ch. 14 - A buffered solution is made by adding 50.0 g NH4Cl...Ch. 14 - Prob. 39ECh. 14 - An aqueous solution contains dissolved C6H5NH3Cl...Ch. 14 - Prob. 41ECh. 14 - Prob. 42ECh. 14 - Consider a solution that contains both C5H5N and...Ch. 14 - Calculate the ratio [NH3]/[NH4+] in...Ch. 14 - Prob. 45ECh. 14 - Prob. 46ECh. 14 - Prob. 47ECh. 14 - Prob. 48ECh. 14 - Calculate the pH of a solution that is 0.40 M...Ch. 14 - Calculate the pH of a solution that is 0.20 M HOCl...Ch. 14 - Which of the following mixtures would result in...Ch. 14 - Prob. 52ECh. 14 - Prob. 53ECh. 14 - Calculate the number of moles of HCl(g) that must...Ch. 14 - Consider the titration of a generic weak acid HA...Ch. 14 - Sketch the titration curve for the titration of a...Ch. 14 - Consider the titration of 40.0 mL of 0.200 M HClO4...Ch. 14 - Consider the titration of 80.0 mL of 0.100 M...Ch. 14 - Consider the titration of 100.0 mL of 0.200 M...Ch. 14 - Prob. 60ECh. 14 - Lactic acid is a common by-product of cellular...Ch. 14 - Repeat the procedure in Exercise 61, but for the...Ch. 14 - Repeat the procedure in Exercise 61, but for the...Ch. 14 - Repeat the procedure in Exercise 61, but for the...Ch. 14 - Prob. 65ECh. 14 - In the titration of 50.0 mL of 1.0 M methylamine,...Ch. 14 - You have 75.0 mL of 0.10 M HA. After adding 30.0...Ch. 14 - A student dissolves 0.0100 mole of an unknown weak...Ch. 14 - Prob. 69ECh. 14 - Prob. 70ECh. 14 - Potassium hydrogen phthalate, known as KHP (molar...Ch. 14 - A certain indicator HIn has a pKa of 3.00 and a...Ch. 14 - Prob. 73ECh. 14 - Prob. 74ECh. 14 - Prob. 75ECh. 14 - Prob. 76ECh. 14 - Prob. 77ECh. 14 - Estimate the pH of a solution in which crystal...Ch. 14 - Prob. 79ECh. 14 - Prob. 80ECh. 14 - Prob. 81AECh. 14 - Prob. 82AECh. 14 - Tris(hydroxymethyl)aminomethane, commonly called...Ch. 14 - Prob. 84AECh. 14 - You have the following reagents on hand: Solids...Ch. 14 - Prob. 86AECh. 14 - Prob. 87AECh. 14 - What quantity (moles) of HCl(g) must be added to...Ch. 14 - Calculate the value of the equilibrium constant...Ch. 14 - The following plot shows the pH curves for the...Ch. 14 - Calculate the volume of 1.50 102 M NaOH that must...Ch. 14 - Prob. 92AECh. 14 - A certain acetic acid solution has pH = 2.68....Ch. 14 - A 0.210-g sample of an acid (molar mass = 192...Ch. 14 - The active ingredient in aspirin is...Ch. 14 - One method for determining the purity of aspirin...Ch. 14 - A student intends to titrate a solution of a weak...Ch. 14 - Prob. 98AECh. 14 - Prob. 99AECh. 14 - Consider 1.0 L of a solution that is 0.85 M HOC6H5...Ch. 14 - Prob. 101CWPCh. 14 - Consider the following acids and bases: HCO2H Ka =...Ch. 14 - Prob. 103CWPCh. 14 - Prob. 104CWPCh. 14 - Consider the titration of 100.0 mL of 0.100 M HCN...Ch. 14 - Consider the titration of 100.0 mL of 0.200 M...Ch. 14 - Prob. 107CWPCh. 14 - Prob. 108CPCh. 14 - A buffer is made using 45.0 mL of 0.750 M HC3H5O2...Ch. 14 - A 0.400-M solution of ammonia was titrated with...Ch. 14 - Prob. 111CPCh. 14 - Consider a solution formed by mixing 50.0 mL of...Ch. 14 - When a diprotic acid, H2A, is titrated with NaOH,...Ch. 14 - Consider the following two acids: In two separate...Ch. 14 - The titration of Na2CO3 with HCl bas the following...Ch. 14 - Prob. 116CPCh. 14 - A few drops of each of the indicators shown in the...Ch. 14 - Malonic acid (HO2CCH2CO2H) is a diprotic acid. In...Ch. 14 - A buffer solution is prepared by mixing 75.0 mL of...Ch. 14 - A 10.00-g sample of the ionic compound NaA, where...Ch. 14 - Prob. 121IPCh. 14 - Prob. 122MP
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  • Sketch the titration curve for a weak acid titrated by a strong base. When performing calculations concerning weak acidstrong base titrations, the general two-slep procedure is to solve a stoichiometry problem first, then to solve an equilibrium problem to determine the pH. What reaction takes place in the stoichiometry part of the problem? What is assumed about this reaction? At the various points in your titration curve, list the major species present after the strong base (NaOH, for example) reacts to completion with the weak acid, HA. What equilibrium problem would you solve at the various points in your titration curve to calculate the pH? Why is pH 7.0 at the equivalence point of a weak acid-strong base titration? Does the pH at the halfway point to equivalence have to be less than 7.0? What does the pH at the halfway point equal? Compare and contrast the titration curves for a strong acidstrong base titration and a weak acidstrong base titration.
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