   # An unknown monoprotic acid reacts with NaOH according to the net ionic equation HA(aq) + OH – (aq) → A – (aq) + H 2 O(l) Calculate the molar mass of HA if 0.856 g of the acid requires 30.08 mL of 0.323 M NaOH. ### Chemistry & Chemical Reactivity

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

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 4.7, Problem 4CYU
Textbook Problem
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## An unknown monoprotic acid reacts with NaOH according to the net ionic equationHA(aq) + OH–(aq) → A–(aq) + H2O(l)Calculate the molar mass of HA if 0.856 g of the acid requires 30.08 mL of 0.323 M NaOH.

Interpretation Introduction

Interpretation:

Molar mass of unknown acid, HA used in the given reaction has to be determined.

Concept introduction:

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
• Concentrationofsubstance=Amountof substancevolumeofthesubstance
• Amountof substance=Concentrationofsubstance×volumeofthesubstance
• Titration is a method to determine the concentration of a substance in solution by adding a solution of known volume and concentration until the reaction is completed.

### Explanation of Solution

Balanced chemical equation for the given reaction is,

HA(aq)+OH(aq)A(aq)+H2O(l)

From the balanced chemical equation it is clear that acid (HA) and base (OH) react in a 1:1ratio.

And so the number of moles of acid and base used in the reaction is equal.

Therefore,

MolesAcid=Molesbase=(0.323mol/L)(0

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