   Chapter 15, Problem 110CP

Chapter
Section
Textbook Problem

# A 0.400-M solution of ammonia was titrated with hydrochloric acid to the equivalence point, where the total volume was 1.50 times the original volume. At what pH does the equivalence point occur?

Interpretation Introduction

Interpretation: The titration of NH3 with HCl is given. The pH at which equivalence point occurs 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.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

Explanation

Explanation

Given

The concentration of NH3 is 0.400M .

Let the volume of NH3 is xL .

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 NH3 to obtain the number of moles as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.400M×xL=0.400xmoles

Total volume at equivalence point is 1.5xL .

Volume of HCl =TotalvolumeVolumeofNH3

Substitute the total volume and volume of NH3 in the above equation as,

Volume of HCl =TotalvolumeVolumeofNH3=1.5xLxL=0.5xL

The net ionic reaction is given as,

NH3+H+NH4+

As it is seen from the equation that 1mole of HCl is required for the neutralization of 1mole of NH3 . Therefore, 0.400xmoles of HCl are required for the neutralization of 0.400xmoles of NH3 . As HCl is a monoprotic and a strong acid , therefore the moles of H+ required for the neutralization of 0.400xmoles of NH3 is 0.400xmoles . This leads to the formation of 0.400xmoles of NH4+ .

Substitute the number of moles and volume of solution in equation (1) to obtain the concentration of NH4+ .

Concentration=NumberofmolesVolumeofsolutioninlitres=0.400xmoles1.5xL=0.27M

Make the ICE table for the hydrolysis of NH4+ .

NH4+(aq)+H2O(l)NH3+H3O+(aq)Initial:0.2700Change:xxxEquilibrium:0.27xxx

The hydrolysis constant Kh for the given reaction is,

Kh=[NH3][H3O+][NH4+] (2)

The relation between Kw and concentration of H+ and OH is given as,

Kw=[H+][OH]

Where,

• Kw is the ionic product of water (1

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