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Strength of Vinegar by Strength of Vinegar by Acid-Base TitrationAcid-Base Titration

Final Exercise – 105 pointsFinal Exercise – 105 points

CH 3COOHNaOH

Vinegar

Lye

2

? QUESTIONS ?? QUESTIONS ?

How, and with what accuracy and precision can you determine the

concentration of a solution of NaOH?

Using that NaOH solution, with what accuracy and precision can you

determine the concentration of a solution of acetic acid - by titration?

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Concepts: Strong/Weak Acids Acid Dissociation/Ka Concentration End point / Equivalence pointTitration curves Logarithms Indicator Mole Relationships pH & pKa TitrationTechniques: Titration / Back-titration pH Measurement

Apparatus: Burets pH Meter

Primary StandardPrimary Standard

Weighing by Difference Weighing by Difference **Preparing solutions of precise Preparing solutions of precise concentrationconcentration

StandardizationStandardization

Volumetric Volumetric FlaskFlaskAnalytical BalanceAnalytical Balance

Pipet

From last From last weekweek

This week

4* Remember the first exercise!

potassium hydrogen phthalate (KHPKHP)

Last exercise involved acid-base titration in which concentration of NaOH was given.

In this exercise, you STANDARDIZE the NaOH.

StandardizationAnalytical determination of purity or concentration of

substance through reaction with substance of verifiedverified composition and puritycomposition and purity (Primary

Standard )This STANDARDIZATION involves reaction of the base, NaOH, with a PRIMARY STANDARD, the weak acid:

O

O H

O

O -

K +

H

phthalic acid5

High purityHigh purityStableStable

Not hygroscopicNot hygroscopicNot efflorescentNot efflorescentHigh solubilityHigh solubility

High molar High molar massmass

KHPKHP is a monoprotic weak acid. One available proton.

[K+] + HC8H4O4- + OH- C8H4O4

= + H2O + [K+]

The stoichiometry of the reaction is:

Acetic acid, CH3COOH, is also a monoprotic weak acid.

1 mol KHP1 mol KHP 1 mol NaOH1 mol NaOH

1 mol CH1 mol CH33COOH COOH 1 mol NaOH 1 mol NaOH

STOICHIOMETRYSTOICHIOMETRY

CH3COOH + OH- CH3COO- + H2O

It’s pKIt’s pKaa is 4.7 is 4.7

pKpKaa is 5.4 is 5.4

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An aqueous solution, can have [ H+ ] = 0.1 and

[ OH- ] = 0.1 at the same time

A. B.

0%0%

A. TrueB. False

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An aqueous solution, can have [ H+ ] = 0.1 and

[ OH- ] = 0.1 at the same time

B False

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[H+] and [OH-] are dependent on one another in aqueous solutions.

[ H+ ] [ OH- ] = 1.0 X 10-14

pH at equivalence point is ~8.5

Phenolphthalein is a reasonable

indicator

Equivalence Point

mL of NaOH added

Half equivalence

point

pKa = 4.7

pH ~ 8.5

[ H+ ] [ A- ] Ka = ---------------

[ HA ]

[ HA ] [ H+ ] = Ka ----------

[ A- ]

[ HA ] -log [ H+ ] = - log Ka - log -------

[ A- ]

[ HA ] pH = pKa - log -----------

[ A- ] pH = pKa

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Titration curve for Acetic Acid with NaOH looks like this

The pH at the equivalence point in the titration of acetic acid with sodium

hydroxide is

10A. B. C. D.

0% 0%0%0%

A. the pKa of acetic acid

B. ½ the pKa of acetic acid

C. 7D. Larger than 7

Equivalence Point

mL of NaOH added

pH ~ 8.5

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D. Larger than 7

It does not matter in which order youA.) Standardize NaOH, B.) Titrate UnknownC.) Determine pH of unknown

PROCEDUREPROCEDURE

A. STANDARDIZATION Prepare solution of known concentration of primary standard, KHP

• Weigh sample BY DIFFERENCE*• Dissolve fully in ERLENMEYERERLENMEYER• Bring to total volume in VOLUMETRIC VOLUMETRIC FLASK**FLASK**

TA’s will assign order to minimize

waiting

Must transfer ALL the solution

*Web supplement****Follow manual procedureFollow manual procedure 12

CALCULATIONS – KHP Solution

Weight of vial + KHP 15.4371 g Weight of vial + remaining KHP 12.3495 gWeight of KHP transferred 3.0876 g Volume of KHP Solution 0.2500 L

Weight of KHP 250

mL

Weigh by difference!!!

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CALCULATIONS – KHP Solution

Weight of vial + KHP 15.4371 g Weight of vial + remaining KHP 12.3495 gWeight of KHP transferred 3.0876 g Volume of KHP Solution 0.2500 LMolarity of KHP Solution: (3.0876 g / 204.22 g/mol) / 0.2500 L = 0.06047 M

Molar Mass of KHP

Weight of KHP

Volume of Volumetri

c Flask

4 Sig Figs because

of0.015119 Mols of

KHP

If calculated concentration of KHP If calculated concentration of KHP differs more than 10% from 0.06 M, differs more than 10% from 0.06 M,

you have either made an error in you have either made an error in weight of KHP or made a calculation weight of KHP or made a calculation

error.error.

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0.06 0.06 0.006 or between 0.054 and 0.006 or between 0.054 and 0.0660.066

PROCEDURE PROCEDURE (cont’d)(cont’d)

Determine concentration of stock NaOH Solution ( NominallyNominally 0.1 M )

with

(Delivered from

buret 1buret 1 )

measured volumes of NaOH Solution

(unknownunknown concentration)

Titrate

measured volumes of standard KHP solution (knownknown concentration)

(Delivered from

buret 2buret 2 )15

In STANDARDIZATION, BOTH REAGENTS are delivered by a

BURET

Can "BACKTITRATEBACKTITRATE“I.e., if end point is overshot, can

recover by adding more KHP and continuing titration, e.g.,

KHP

NaOH

1. Add measured volume of KHP(~35mL)2. Titrate to Phenolphthalein Endpoint 3. If overshoot endpoint, add more KHP4. Titrate to Phenolphthalein Endpoint again

MMaakkee s suurree y yoouu llaabel bel tthhee bbuurreettss..

Do at most 3 titrations for the Do at most 3 titrations for the standardizationstandardization

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Having determined concentration of KHP solution, calculate volume of ~ 0.1 M NaOH needed to react with 35 mL of KHP.

35 mL of our KHP solution (0.06047 M) contains:

What volume of 0.1 M NaOH will contain 2.1 mmol of NaOH?

m moles = Molarity (mmol/mL) X Volume (m L)

2.1 mmol = 0.1 (mmol/mL) X V (mL)

V = 2.1 / 0.1 = 21 mL 20 mL

35 mL X 0.06047 mmol / mL = 2.1 mmol of KHPAmt = Concentration X Volume

Amt = Concentration X Volume

Why do we make this calculation?

So we will know when to begin to add the NaOH slowly!

That will require 2.1 mmol of NaOH

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How many mL of 0.10 M NaOH would 35 mL of 0.10 M KHP require?

A. B. C. D.

0% 0%0%0%

A. 25 mLB. 35 mLC. 0.006407 MD. Whatever

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B 35 mL

Stoichiometry is 1 ↔ 1

35 mL X 0.10 M = 3.5 mmol KHP

3.5 mmol KHP ↔ 3.5 mmol NaOH

0.10 M = 0.10 mmol/mL

3.5 mmol / 0.10 mmol/mL = 35 mL

So, the answer is,

How many mL of ~0.10 M NaOH would 35 mL of 0.10 M KHP require?

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PROCEDURE - UNKNOWNPROCEDURE - UNKNOWN

Unknowns are solutions of acetic acid in water.

We titrate aliquots* of (undiluted) unknown. In this exercise, aliquot is a 5 mL pipet-ful (5.00 0.01) of solution’

* aliquots are fixed, repetitive fractions of a solution

• Dilute unknown with distilled water (~ 40 mL)• Add Phenolphthalein as indicator• Titrate unknown to phenolphthalein end point.

Since aliquots are identical, can track precision of titration by calculating percent error involumes of NaOH used.

Remember: you cannot backtitrate in this caseRemember: you cannot backtitrate in this case

Adding water doesn’t change the amountamount of acetic

acid.

Mandrake Root Extract

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E.g., suppose volumes of NaOH are

Avg Vol = (25.37 + 25.84 + 24.93) / 3 = 25.38

Avg Dev= ( + 0.01 + 0.46 + 0.45) / 3 = 0.31

% Dev = 100 X 0.31 / 25.38 = 1.2 %

25.37 ,25.84 , 24.93

Do at most 5 Do at most 5 titrations for titrations for

unknown (incl unknown (incl prelim)prelim)

Since all other numbers in subsequent calculations are the same for any run, that will be percent deviation in the final result.

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The darkestdarkest pink color is the best signal for the equivalence point in the titration of acetic acid with NaOH using phenolphthalein

22A. B.

0%0%

A. TrueB. False

The lightestlightest pink color is the best signal for the equivalence point in the titration

of acetic acid with NaOH using phenolphthalein

BB = False

A few drops after that point is reached, the pH of the solution increases rapidly and the

solution gets as deeply colored as it is able. On further addition of base, solution lightens due to

dilution.Titration Curve

2

3

4

5

6

7

8

9

10

11

12

13

0 5 10 15 20 25 30 35 40

Volume of added NaOH

pH

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Before end point

At end point

Just pastJust past end point

KHP & Unknown solutions are colorless. No

interference from beverage color.

Way pastWay past end point 24

CALCULATIONS – NaOH STANDARDIZATIONMolarity of KHP Solution: 0.06047 M

KHP buret reading, final 37.44 mL 38.77 KHP buret reading, initial 3.68 mL 4.73Volume of KHP titrated 33.76 mL 34.04

NaOH buret reading, final 28.73 mL 31.83NaOH buret reading, initial 4.52 mL 7.88Volume of NaOH used 24.21 mL 23.95

mmol of KHP titrated 33.76 mL X 0.06047 M = 2.042 mmol 2.202mmol of NaOH used 2.042 mmol 2.202

Molarity of NaOH 2.042 mmol / 24.21mL = 0.08435 M 0.08526 M

Stoichiometry is 1 to 1

From Part 1

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CALCULATIONS – NaOH STANDARDIZATIONMolarity of KHP Solution: 0.06047 M

KHP buret reading, final 37.44 mL 38.77 KHP buret reading, initial 3.68 mL 4.73Volume of KHP titrated 33.76 mL 34.04

NaOH buret reading, final 28.73 mL 31.83NaOH buret reading, initial 4.52 mL 7.88Volume of NaOH used 24.21 mL 23.95

mmol of KHP titrated 33.76 mL X 0.06047 M = 2.042 mmol 2.058mmol of NaOH used 2.042 mmol 2.058

Molarity of NaOH 2.042 mmol / 24.21mL = 0.08435 M 0.08592 M

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We have three values for our NaOH concentration. Are they in reasonable agreement?Average = (0.08435 + 0.08592 + 0.08539) / 3 = 0.08522 M

Avg Dev = (0.00087 + 0.00070 + 0.00017) / 3 = 0.00058 M

Pct Dev = 100 X 0.00058 / 0.08522 = 0.68%

Suppose the 3rd titration produces 0.08539 M

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CALCULATIONS - UNKNOWN

Volume of Unknown 5.00 mL 5.00 5.00

Concentration of NaOH solution 0.08522 M

NaOH buret, final 22.47 mL 21.16 23.72 NaOH buret, initial 3.15 mL 2.37 4.22 Volume NaOH used 19.32 mL 18.79 19.50

mmol of NaOH used19.32 * 0.08522 = 1.646 mmol 1.585 1.645

mmol of Acetic Acid titrated = 1.646 mmol 1.585 1.645

Acetic Acid Concentration1.646 / 5.00 = = 0.329 M 0.317 0.329

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From Standardizati

on

3 Sig Figs

Stoichiometry is 1 to 1

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CALCULATIONS - UNKNOWN

Volume of Unknown 5.00 mL 5.00 5.00

Concentration of NaOH solution 0.08522 M

NaOH buret, final 22.47 mL 21.16 23.72 NaOH buret, initial 3.15 mL 2.37 4.22 Volume NaOH used 19.32 mL 18.79 19.50

mmol of NaOH used18.79 * 0.08522 = 1.646 mmol 1.601 1.645

mmol of Acetic Acid titrated = 1.646 mmol 1.601 1.645

Acetic Acid Concentration1.601 / 5.00 = =0.329 M 0.320 0.329

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CALCULATIONS - UNKNOWN

Volume of Unknown 5.00 mL 5.00 5.00

Concentration of NaOH solution 0.08522 M

NaOH buret, final 22.47 mL 21.16 23.72 NaOH buret, initial 3.15 mL 2.37 4.22 Volume NaOH used 19.32 mL 18.79 19.50

mmol of NaOH used19.50 * 0.08522 = 1.646 mmol 1.601 1.662

mmol of Acetic Acid titrated = 1.646 mmol 1.601 1.662

Acetic Acid Concentration1.662 / 5.00 = =0.329 M 0.320 0.332

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30

The concentration of the sodium hydroxide to be used in computations in this exercise is the value given on the stock solution

container.

A. B.

0%0%

A. TrueB. False

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The concentration of the sodium hydroxide to be used in computations in this exercise is the value given on the stock solution

container. .

B = False

The purpose of the standardization is to determine the precise concentration

of the NaOH solution.

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Avg Conc of Acetic Acid =

(0.329 M + 0.320 M + 0.332 M) / 3 = 0.327 M

Avg Deviation =

(0.002 + 0.007 + 0.005) / 3 = 0.005 M

Pct Deviation =

100 X 0.005 / 0.327 = 1.5%

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Read & Record BuretsRead & Record Burets to to 2 DECIMAL PLACES

Read & Record Weights Read & Record Weights to to 4 DECIMAL 4 DECIMAL PLACES

24.64

mL

4.6427 g

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Begin each titration with Begin each titration with buret reading between buret reading between

0.00 and 5.00 mL0.00 and 5.00 mL

STANDARDIZATION – Precision

1. WEIGH 3g of solid KHP using the analytical balance ( 0.0004 g )

Precision: 100 X 0.0004 / 3.0000 0.01 %

2. PREPARE A SOLUTION of KHP using 250 mL volumetric flask ( 0.05 mL )

Precision: 100 X 0.05 / 250.00 0.02 %

3. TITRATE measured volumes of KHP and NaOH using burets ( 0.05 mL )

Precision: KHP 100 X 0.05 / 30.00 0.2 % NaOH 100 X 0.05 / 30.00 0.2 %

Concentration of NaOH determined with PRECISION of approximately 0.01 + 0.02 + 0.2 + 0.2 = 0.43 %

ANALYSIS OF ERRORSANALYSIS OF ERRORS

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TITRATION OF UNKNOWN – Precision

1. TRANSFER aliquot of unknown using a 5 mL transfer pipet ( 0.01 mL )

Precision: 100 X 0.01 / 5.00 = 0.2 %

Note that 1 drop (0.05 mL) ↔ 1%

2. TITRATE aliquot using standardized NaOH, again using a buret ( 0.05 mL )

Precision: (Buret)100 x 0.05 / 30.00 = 0.2 %

OVERALL PRECISION of determination of concentration of unknown = 0.2 + 0.2 =

0.4 %

ANALYSIS OF ERRORSANALYSIS OF ERRORS

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Suppose you over-titrate by 10 drops I.e., you go past the end point by 10 drops (0.5 mL).

• Unknowns range in concentration from 0.5 M to 1.0 M.

• 5 mL of the unknown contains between 2.5 and 5.0mmol of acetic acid.

• That will require between 25 and 50 mL of ~0.1 M NaOH

The error represented by 0.5 mL will range from: 100 X 0.5 / 25 = 2 % to 100 X 0.5 / 50 = 1 %

Errors in Errors in indentifying the end point indentifying the end point are not a are not a major contribution to errors in the accuracy major contribution to errors in the accuracy of the concentration of the vinegar of the concentration of the vinegar unknown.unknown.

How large an error in accuracy results?

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What about the pipet?

The volume of the sample you are using is 5.00 mL, delivered by pipet.

The intrinsic precision of the 5 mL pipet is 0.01 mL

The volume of a drop is ~ 0.05 mL. The volume of a drop is ~ 0.05 mL. Each drop represents: Each drop represents:

100 X 0.05/5.00 = 1% 100 X 0.05/5.00 = 1% error in the volume delivered!error in the volume delivered!

BUTBUT

100 X 0.01 / 5.00 = 0.2%

1. Incomplete transfer of solid KHP into flask (caused by weighing using a watch glass or paper)Weigh by difference!Weigh by difference!

2. Failure to dissolve KHP completelyMake sure no KHP particles remain undissolvedMake sure no KHP particles remain undissolved

3. Incomplete transfer of KHP solution from Erlenmeyer to volumetric flaskTransfer completely and rinse with distilled Transfer completely and rinse with distilled

water from wash bottlewater from wash bottle

4. Failing to bring KHP solution to correct volumeUse dropper for last 0.5 mLUse dropper for last 0.5 mL ! !

PITFALLS TO AVOIDPITFALLS TO AVOID

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5. Incomplete mixing of KHP solution in volumetric flask

Mix thoroughly both before and after bringing Mix thoroughly both before and after bringing to final volumeto final volume

6. Improper use of the pipet Make sure you use the pipet as instructed

7. Contamination of unknown by allowing contact with syringe while pipeting

Discard any sample for which this happensDiscard any sample for which this happens

8. Over-titrating unknownStop at earliest detectable pink colorStop at earliest detectable pink colorGRADE ON THIS EXERCISE WILL DEPEND

ON both AccuracyAccuracy and PrecisionPrecision40

NEXT WEEKNEXT WEEK

SUSB-054 – Part 1

Gasometric Determination of Sodium Bicarbonate in a Mixture

DO PRELAB to SUSB-054 - 1

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