   # Malonic acid (HO 2 CCH 2 CO 2 H) is a diprotic acid. In the titration of malonic acid w ith NaOH, stoichiometric points occur at pH = 3.9 and 8.8. A 25.00-mL sample of malonic acid of unknown concentration is titrated with 0.0984 M NaOH, requiring 31.50 mL of the NaOH solution to reach the phenolphthalein end point. Calculate the concentration of the initial malonic acid solution. (Sec Exercise 113.) ### 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 118CP
Textbook Problem
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## Malonic acid (HO2CCH2CO2H) is a diprotic acid. In the titration of malonic acid w ith NaOH, stoichiometric points occur at pH = 3.9 and 8.8. A 25.00-mL sample of malonic acid of unknown concentration is titrated with 0.0984 M NaOH, requiring 31.50 mL of the NaOH solution to reach the phenolphthalein end point. Calculate the concentration of the initial malonic acid solution. (Sec Exercise 113.)

Interpretation Introduction

Interpretation: The diprotic nature of Malonic acid and value of pH for the titration of Malonic acid and NaOH is given. The initial concentration of Malonic acid is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

To determine: The initial concentration of Malonic acid.

### Explanation of Solution

Explanation

Given

The concentration of NaOH is 0.0984M .

The volume of NaOH is 31.50mL .

The volume of Malonic acid is 25.0mL .

The pH at stoichiometric points is 8.8 and 3.9 .

The conversion of mL into L is done as,

1mL=0.001L

The conversion of all the given species into l is done as,

31.50mL=31.50×0.001L=0.0315L25.0mL=25.0×0.001L=0.025L

The concentration of any species is given as,

Concentration=NumberofmolesVolumeofsolutioninlitres (1)

Rearrange the above equation to obtain the value of number of moles.

Numberofmoles=Concentration×Volumeofsolutioninlitres

Substitute the value of concentration and volume of NaOH to obtain the number of moles as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0

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