   # 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 &gt; 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. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 14, Problem 7RQ
Textbook Problem
46 views

## 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.

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,

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,

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_

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_

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

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