# Determine the volume (in mL) of 1.00 M NaOH that must be added to 250 mL of 0.50 M CH 3 CO 2 H to produce a buffer with a pH of 4.50.

### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

Chapter
Section

### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 17, Problem 21PS
Textbook Problem
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## Determine the volume (in mL) of 1.00 M NaOH that must be added to 250 mL of 0.50 M CH3CO2H to produce a buffer with a pH of 4.50.

Interpretation Introduction

Interpretation:

The volume of 1.00M, NaOH has to be calculated when it is mixed with 0.50M,CH3COOH to get the buffer solution of pH value equals to 4.5.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base. There are four types of acid-base titrations.

(1) Strong acid-Strong base, in this type of titration a strong acid is titrated against a strong base for example, HCl is titrated against NaOH.

(2) Strong acid-Weak base, in this type of titration a strong acid is titrated against a weak base for example, HCl is titrated against NH4OH.

(3) Weak acid-Strong base, in this type of titration a weak acid is titrated against a strong base for example, CH3COOH is titrated against NaOH.

(4) Weak acid-Weak base, in this type of titration a weak acid is titrated against a weak base for example, CH3COOH is titrated against NH4OH.

For weak acid-strong base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of CH3COOH with NaOH. The equilibrium can be represented as,

CH3COOH(aq)+NaOH(aq)H2O(l)+CH3COONa(aq)

Calculation of pH at various points is done as follows,

(1) The pH value before the titration can be calculated by using the Ka and its relation with H3O+ ion concentration.

Ka=[H3O+](eq)[A](eq)[HA](eq) (1)

(2) The pH calculation just before the equivalence point,

As the addition of NaOH is done there will be formation of buffer solution CH3COOH/CH3COO. The pH calculation for buffer solution is done by using Henderson-Hesselbalch equation.

pH=pKa+log[conjugatebase][acid] (2)

At the midpoint of the titration, when concentration of acid and its conjugate base are equal pH value at midpoint will be given as;

pH=pKa+log[conjugatebase][acid]

Substitute, [conjugatebase]for[acid].

pH=pKa+log[conjugatebase][conjugatebase]=pKa+log(1)=pKa+0=pKa

Therefore, pH value at midpoint is equal to pKa.

(3) The pH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only OH ion and CH3COO. The OH will be produced due to the hydrolysis of acetate ion at equivalence point. The hydrolysis equilibrium is represented as,

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

By using the value of Kb for the acetate ion, concentration of  OH can be calculated.  Thus the value of pH is greater than 7 at equivalence point for the weak acid- strong base titrations.

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)

(4) The pH calculation after the equivalence point.

After the equivalence point there will be excess of OH ion in the solution and there will be very less amount of CH3COO ion. The amount of CH3COO produce can be neglected with respect to excess amount of OH.

Concentration of OH after equivalence point will be calculated by using the expression,

concentration=numberofmolestotalvolumeofsolution (4)

### Explanation of Solution

The volume of volume of NaOH used is calculated below.

Given:

Refer to table 16.2 in the textbook for the value of Ka.

The value of Ka for acetic acid is 1.8×105.

The pKa value is calculated as follows;

pKa=log(Ka)

Substitute, 1.8×105 for Ka.

pKa=log(1.8×105)=4.74

Therefore, pKa value is 4.74.

The initial concentration of CH3COOH is 0.50molL1.

The initial concentration of NaOH is 1.00molL1.

The volume of CH3COOH is 250mL.

Conversion of 250mL into L.

(250mL)(1L1000mL)=0.250L

Let the volume of NaOH added xmL.

Conversion of xmL into L.

(xmL)(1L1000mL)=0.00xL

The total volume after the reaction is calculated as,

totalvolume=volumeofCH3COOH(L) + volume of NaOH(L)

totalvolume = 0.250(L)+0.00x(L)=(0.250+0.00x)L

Therefore, total volume after reaction is (0.250+0.00x)L.

The calculation of moles is done by using the expression,

Numberof moles=concentration(molL1)volume(L)

The ICE table (1) for the reaction between NaOH and CH3COOH is given below,

EquationCH3COOH(aq)+NaOH(aq)H2O(l)+CH3COONa(aq)Initial(mol)0.125(0.00x)0Change(mol)0.00x0.00x0.00xAfterreaction(mol)(0.1250.00x)00.00x

From ICE table (1),

After the reaction there are only two species present, which are CH3COOH and its conjugate base CH3COONa. There is a formation of buffer takes place.

Number of moles of acetic acid left after reaction is (0.1250.00x).

Number of moles of acetate ion produced after the reaction is 0.00xmol.

Approximation, the value of 0.00x is very small on comparison to 0.125 . Therefore, 0.00x can be neglected with respect to 0.125

So, the Number of moles of acetic acid left after reaction are 0.125mol.

Concentration calculation is done by using the expression,

concentration = Numberof molestotal volume(molL1)

Substitute, 0

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