Chapter 17, Problem 88GQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# What volume of 0.120 M NaOH must be added to 100. mL of 0.100 M NaHC2O4 to reach a pH of 4.70?

Interpretation Introduction

Interpretation:

The volume of 0.120 M,NaOH has to be calculated when it is mixed with 100 mL 0.100 M,NaHC2O4 to get the buffer solution of pH value equals to 4.70.

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.

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 NaHC2O4 with NaOH. The equilibrium can be represented as,

NaHC2O4(aq)+NaOH(aq)H2O(l)+Na2C2O4(aq)

Calculation of pH at various points is done as follows,

(1) The pH value before the titration can be calculated by using 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 NaHC2O4/Na2C2O4. The pH calculation for buffer solution is done by using Henderson-Hasselbalch 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.

Explanation

The volume of NaOH used is calculated below.

Given:

Refer to section 17.3 Â in the textbook for the value of Ka.

The value of Ka for NaHC2O4 is 6.4Ã—10âˆ’5.

The pKa value is calculated as follows;

pKa=âˆ’log(Ka)

Substitute, 6.4Ã—10âˆ’5 for Ka.

pKa=âˆ’log(6.4Ã—10âˆ’5)=4.19

Therefore, pKa value is 4.19.

The initial concentration of NaHC2O4 is 0.100Â molâ‹…Lâˆ’1.

The initial concentration of NaOH is 0.120Â molâ‹…Lâˆ’1.

The volume of NaHC2O4 is 100â€‰mL.

Conversion of 100Â mL into L.

(100â€‰mL)(1â€‰L1000â€‰mL)=0.100â€‰L

Let the volume of NaOH added xâ€‰mL.

Conversion of xâ€‰mL into L.

(xâ€‰mL)(1â€‰L1000â€‰mL)=0.00xâ€‰L

The total volume after the reaction is calculated as,

totalâ€‰volume=volumeâ€‰ofÂ NaHC2O4(L)Â +Â volumeÂ ofÂ NaOH(L)

totalâ€‰volumeÂ =Â 0.100(L)+0.00x(L)=(0.100+0.00x)L

Therefore, total volume after reaction is (0.100+0.00x)L.

The calculation of moles is done by using the expression,

Numberâ€‰ofÂ moles=concentration(molâ‹…Lâˆ’1)â‹…volume(L)

The ICE table (1) for the reaction between NaOH and NaHC2O4 is given below,

EquationNaHC2O4(aq)+NaOH(aq)â†’H2O(l)+Na2C2O4(aq)Initial(mol)0.01(0.00120x)0Change(mol)âˆ’0.00120xâˆ’0.00120x+0.00120xAfterÂ reaction(mol)(0

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