EBK EXPLORING CHEMICAL ANALYSIS
EBK EXPLORING CHEMICAL ANALYSIS
5th Edition
ISBN: 9781319416942
Author: Harris
Publisher: VST
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Chapter 10, Problem 10.1P
Interpretation Introduction

Interpretation:

Chemistry that goes into each region of titration between OH and H+ ions has to be explained.

Concept Introduction:

In an acid base-kind of titration, progress of reaction is quantified by neutralization of analyte when definite volume of titrant is added. The indicators used are sensitive to pH changes. Suitability of such indicator is governed by their ability to detect end point accurately when the equivalence point is reached.

Expert Solution & Answer
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Explanation of Solution

Before the equivalence point is reached any volume of H+ consumes the corresponding stoichiometric volume of OH in accordance with neutralization principle. However, concentration of OH ion remains in excess at any point before titration. Hence pH at this point is more than 7. Manner pH is calculated as follows:

  [OH]left=[OH]initial[OH]consumed

Net volume is calculated as follows:

  Net volume=Volume of analyte in flask+Volume added

New concentration of [OH] is calculated as follows:

  [OH]=[OH]leftNet volume

Expression to calculate [H+] from ionic product of water is as follows:

  [H+]=Kw[OH]

With [H+] thus obtained  pH is determined by expression as follows:

  pH=log[H+]

Once equivalence is reached entire OH ion has been neutralized by sufficient H+ ion. Thus there is no excess of OH and thus pH at this point is 7. At this point, there is equal concentration of OH and H+ ion.

Beyond equivalence point, all added H+ will lead to excess concentration of H+ ion and hence pH at this point is less than 7. Here excess H+ is determined as follows:

  Excess H+=(Volume)(Molarity)

Net volume is calculated as follows:

  Net volume=Volume of analyte in flask+Volume added

New concentration of [H+] is calculated as follows:

  [H+]=Excess H+(mmol)Net volume(mL)

With [H+] thus obtained, pH is again determined by expression as follows:

  pH=log[H+]

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