   # A 0.210-g sample of an acid (molar mass = 192 g/mol) is titrated with 30.5 mL of 0.108 M NaOH to a phenolphthalein end point. Is the acid monoprotic, diprotic, or triprotic? ### 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 94AE
Textbook Problem
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## A 0.210-g sample of an acid (molar mass = 192 g/mol) is titrated with 30.5 mL of 0.108 M NaOH to a phenolphthalein end point. Is the acid monoprotic, diprotic, or triprotic?

Interpretation Introduction

Interpretation:

A 0.210g sample of an acid and its molar mass value is given. It is titrated with 30.5mL of 0.108M NaOH . The given acid is to be identified as a monoprotic, diprotic or triprotic.

Concept introduction:

An acid that donate a single proton per molecule to a given aqueous solution is known as a monoprotic acid. An acid that donate two protons per molecule to a given aqueous solution is known as a diprotic acid. An acid that donate three protons per molecule to a given aqueous solution is known as a triprotic acid.

To identify: If the given acid is monoprotic, diprotic or triprotic.

### Explanation of Solution

Explanation

Given

The mass of the given acid is 0.210g .

The molar mass of the given acid is 192g/mol .

The moles of the acid is calculated by the formula,

Numberofmoles=MassMolarmass

Substitute the value of mass and molar mass of the given acid in the above formula.

Numberofmoles=0.210g192g/mol=1.09×103mol

The number of moles can also be calculated by the formula,

Numberofmoles=Volume×Molarity

Substitute the value of volume and molarity of NaOH in the above formula.

Numberofmoles=0.0305L×0.108M=3.29×103mol

The number of moles of the given acid is 1.09×103mol .

The number of moles of NaOH is 3.29×103mol

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