   Chapter 4.8, Problem 4.10CYU

Chapter
Section
Textbook Problem

A 25.0-mL sample of vinegar (which contains the weak acid acetic acid, CH3CO2H) requires 28.33 mL of a 0.953 M solution of NaOH for titration to the equivalence point. What is the mass of acetic acid (molar mass = 60.05 g/mol), in grams, in the vinegar sample, and what is the concentration of acetic acid in the vinegar?CH3CO2H(aq) + NaOH(aq) → NaCH3CO2(aq) + H2O(l)

Interpretation Introduction

Interpretation:

The mass of acetic acid and its concentration in the given 25.0mL sample of vinegar should be determined.

Concept introduction:

Numberofmole=GivenmassofthesubstanceMolarmass

• 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
Explanation

Balanced chemical equation for the given reaction is,

CH3CO2H(aq)+NaOH(aq)NaCH3CO2(aq)+H2O(l)

The amount (moles) of NaOH available can be calculated by using the equation

Amountof substance=Concentrationofsubstance×volumeofthesubstance

AmountofNaOH =CNaOH×VNaOH =0.953molNaOHL×0.02833LNaOH =0.02700molNaOH

The balanced equation for the reaction shows that 1mol of acetic acid requires 1mol of sodium hydroxide. This is the required stoichiometric factor to obtain the amount of acetic acid present.

Thus the amount of acetic acid can be calculated as follows,

0.02700molNaOH×1molCH3CH2OH1molNaOH=0

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