   # A 6.50-g sample of a diprotic acid requires 137.5 mL of a 0.750 M NaOH solution for complete neutralization. Determine the molar mass of the acid. ### 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 6, Problem 138CP
Textbook Problem
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## A 6.50-g sample of a diprotic acid requires 137.5 mL of a 0.750 M NaOH solution for complete neutralization. Determine the molar mass of the acid.

Interpretation Introduction

Interpretation: The molar mass of unknown acid has to be calculated.

Concept introduction: The mass of a compound to the amount of the substance gives the molar mass of an element/compound.

Molarmass(g/mol)=massofagivencompound(ingrams)amountofsubstance(inmoles)

### Explanation of Solution

Explanation

Record the given data

Mass of the sample=6.50gram

Volume of NaOH required=137.5mL

Molarity of NaOH required=0.750M

The volume and molarity of NaOH are given along with the sample of mass as shown above.

To calculate the moles in the unknown acid

Let H2A be the unknown diprotic acid

MolesofH2A=0.13750.750molNaOHL×1molH2A2molNaOH

=0.0516mol

H2A(aq)+2NaOH(aq)2H2O(l)+Na2A(aq)

The mole of the diprotic acid is calculated by plugging the values of volume and molarity of sodium hydroxide

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