TitrationsTitrations pHpH Titrant volume, mL Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of

TitrationsTitrationsTitrationsTitrations

pHpHpHpH

Titrant volume, mLTitrant volume, mLTitrant volume, mLTitrant volume, mLCopyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved

Acid-Base TitrationsAcid-Base TitrationsAcid-Base TitrationsAcid-Base Titrations

Adding NaOH from the buret to acetic acid in the flask, a weak Adding NaOH from the buret to acetic acid in the flask, a weak acid. In the beginning the pH increases very slowly.acid. In the beginning the pH increases very slowly.

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Additional NaOH is added. pH rises as Additional NaOH is added. pH rises as equivalence point is approached.equivalence point is approached.

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Additional NaOH is added. pH increases and then levels Additional NaOH is added. pH increases and then levels off as NaOH is added beyond the off as NaOH is added beyond the equivalence pointequivalence point..

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QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final equivalence point. What is the pH of the final solution?solution?

QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final equivalence point. What is the pH of the final solution?solution?

Equivalence Equivalence pointpointEquivalence Equivalence pointpoint

pH of solution of pH of solution of benzoic acid, a benzoic acid, a weak acidweak acid


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QUESTION:QUESTION: How do we calculate pH at How do we calculate pH at the equivalence point?the equivalence point?

QUESTION:QUESTION: How do we calculate pH at How do we calculate pH at the equivalence point?the equivalence point?

Equivalence Equivalence pointpointEquivalence Equivalence pointpoint



QUESTION:QUESTION: How do we How do we calculate calculate pH before pH before titration?titration?

QUESTION:QUESTION: How do we How do we calculate calculate pH before pH before titration?titration?

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Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions

HBz reacts completely to produce Bz- HBz reacts completely to produce Bz- Part 1 calculation:Part 1 calculation: stoichiometry gives [Bz stoichiometry gives [Bz--] ]

produced is 0.020 M.produced is 0.020 M.

BzBz- - + H+ H22O O HBz + OH HBz + OH- - KKbb = 1.6 x 10 = 1.6 x 10-10-10

[Bz[Bz--]] [HBz][HBz] [OH[OH--]] equilibequilib 0.020 - x0.020 - x xx xx

x = [OHx = [OH--] = 1.8 x 10] = 1.8 x 10-6-6, pOH = 5.75, and , pOH = 5.75, and pH = 8.25pH = 8.25

HBz reacts completely to produce Bz- HBz reacts completely to produce Bz- Part 1 calculation:Part 1 calculation: stoichiometry gives [Bz stoichiometry gives [Bz--] ]

produced is 0.020 M.produced is 0.020 M.

BzBz- - + H+ H22O O HBz + OH HBz + OH- - KKbb = 1.6 x 10 = 1.6 x 10-10-10

[Bz[Bz--]] [HBz][HBz] [OH[OH--]] equilibequilib 0.020 - x0.020 - x xx xx

x = [OHx = [OH--] = 1.8 x 10] = 1.8 x 10-6-6, pOH = 5.75, and , pOH = 5.75, and pH = 8.25pH = 8.25

QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?What is the pH of the final solution?

Kb 1.6 x 10-10 = x2

0.020 - xKb 1.6 x 10-10 =

x2

0.020 - x

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Half-way Half-way pointpointHalf-way Half-way pointpoint

[HBz] = [Bz[HBz] = [Bz--]]

At midpoint of titration…At midpoint of titration…

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Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions

HBzHBz + H+ H22O O H H33OO++ + Bz + Bz- - KKaa = 6.3 x 10 = 6.3 x 10-5-5

pH = pKa + log ( [BzpH = pKa + log ( [Bz- - ]/[HBz] )]/[HBz] )

pH = 4.20 + log (1)pH = 4.20 + log (1)

pH = 4.20pH = 4.20

HBzHBz + H+ H22O O H H33OO++ + Bz + Bz- - KKaa = 6.3 x 10 = 6.3 x 10-5-5

pH = pKa + log ( [BzpH = pKa + log ( [Bz- - ]/[HBz] )]/[HBz] )

pH = 4.20 + log (1)pH = 4.20 + log (1)

pH = 4.20pH = 4.20

QUESTION:QUESTION: You titrate 100. mL of a 0.025 M You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH solution of benzoic acid with 0.100 M NaOH What is the pH at the half-way point?What is the pH at the half-way point?

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Half-way Half-way pointpointHalf-way Half-way pointpoint

……can calculate pH if know [Bzcan calculate pH if know [Bz- - ]/[HBz] ]/[HBz]

At any point of the titration…At any point of the titration…

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pH = pKapH = pKapH = pKapH = pKa

……can calculate pH if know [Bzcan calculate pH if know [Bz- - ]/[HBz] ]/[HBz]

At any point of the titration…At any point of the titration…

If pH > pKa, If pH > pKa, [Bz[Bz- - ]>[HBz]]>[HBz]If pH < pKa, If pH < pKa,

[HBz]>[Bz[HBz]>[Bz--]]

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TitrationsTitrations pHpH Titrant volume, mL Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of