pka titration iar

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Investigating the pH of Weak Acid and Strong Base Titration

with Ethanoic Acid and Sodium Hydroxide

IB HL Chemistry Internal AssessmentKevin Omana - Mendoza

April 29, 2013


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Once I have graphed the points from the titration, I will find the first derivative line in

order to find the equivalence point. That is, the point at which the amount of acid is

equal to amount of base. At this point, a rapid increase in pH occurs because the acid in

the beaker has been neutralized and as such, the excess of base remains and any

additional NaOH will increase the pH once the reaction reaches equivalence point. I will

also calculate the second derivative in order to confirm the values returned by the first

derivative.

In order to find the equivalence point using the first derivative, I will find the peak of the

first derivative, which indicates the greatest slope of the titration curve, where the

equivalence point occurs. This value can be confirmed by checking if the maximum point

on the first derivative curve corresponds to the point at which the second derivative has

a negative value. Once I find the equivalence point, I will divide the volume by two in

order to find the half equivalence point.

For a visual aid in identifying when the equivalence point is reached, I will use the

indicator Phenolphthalein, which has a pH range of 8.3-10.0 and is therefore the most

suitable. The suitability of Phenolphthalein is because the experiment is a weak acid-strong base titration, and as such, due to salt hydrolysis, the equivalence point will be at

a pH greater than 7. The salt in the reaction, CH3COONa, is made up of the weak acid

CH3COO- and the strong base Na+. Thus, the salt resulting from the titration will be

basic, causing the equivalence point pH to be greater than 7.

The Ka and pKa values are found using the expression:

KaH

A HA

moldm-3


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Which can be adjusted, using the general equation for this reaction, to:

KaH CH3COO

CH3COOH

moldm-3

Here is where the importance of the half equivalence point comes in. At the half

equivalence point, pH is equal to pKa and exactly half of the weak acid will have been

converted into its conjugate base and thus the concentrations of the weak acid and

conjugate base will be the same. In this reaction that would mean

CH3COO= CH3COO

At which point they would cancel out leaving the expression to read:

Ka H At this point, I would take the log of both sides in order to get the equation:

log(Ka) log(H)

This step is important because at the half equivalence point, pKa is equal to pH and the

pH equation is:

pH log(H

)

And thus, if pH=pKa at the half equivalence point, we can substitute pKa in for pH. Doing

this, we would then need to adjust the equation: log(Ka) log(H)

To read:

Ka


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Which can also be read as :

The literature values for Ka and pKa are2:

Variables:

Independent Variable: The volume of 0.1M NaOH added to a 25 ml solution of 0.1M

CH3COOH

Dependent Variables: The Recorded p

Aspect 2: Controlling the vairables:

Control: Temperature, Pressure and Molarity of NaOH and CH3COOH (0.1M)

2IB HL Chemistry Data Booklet

Ka 4.76

Ka 1.74 10 5

Ka 10 pKa


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Materials:

Safety goggles NaOH (0.1M) CH3COOH (0.1M) CBL Unit pH probe (0.01pH) Graphing Calculator (Ti-84) 50 cm3 Burette (0.05 cm3) 25 cm3Pipette (0.06 cm3) Phenolphthalein Indicator 1 pH probe holder 1 pipette filler

Magnetic stirrer Stirring magnet Buffer solution

with pH of 4

Buffer solutionwith pH of 7

Buffer solutionwith pH of 10

1 Burette Stand 1 Burette Clamp 4 250mL

beakers


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Diagram of materials:


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Aspect 3: Developing a method for data collecting

PROCEDURE

1. Put on safety goggles2. Set up the CBL unit, making sure the pH probe is in the correct channel and the

channel is set up to look for a pH probe.

3. Calibrate the CBL unit using the 4 and 10 pH buffer solutions to createbenchmarks and check calibration using the pH 7 buffer.

4. Condition burette by rinsing with 0.1 M NaOH 3 times, making sure that thereare no air bubbles trapped anywhere in the burette.

5. Fill burette to 50.00 cm3 (0.05cm3) and place securely in the burette clamp onthe stand.

6. Using the pipette filler, fill the pipette to 25.00 cm3 (0.06 cm3) with 0.1 MCH3COOH.

7. Empty Pipette into a clean, 250ml beaker.8. Place beaker underneath the burette on top of the magnetic stirrer unit. (which

at this time should be plugged in)

9. Place Stirring magnet in CH3COOH solution.10.Place the pH probe in the CBL probe stand and Position the pH probe so that it is

inside the beaker of CH3COOH.

11.Add deionized water to the beaker of CH3COOH such that the pH probe is in thesolution but without touching the sides of the beaker or the stirring magnet.

12.Carefully add two drops of phenolphthalein indicator to solution.13.Switch on the magnetic stirrer such that the magnet stirs the solution gently

without splashing.

14.Conduct one quick titration as a benchmark by Opening the burette to allowNaOH to be added to the beaker. When the solution begins to turn pink, slow the

flow of NaOH. Once the solution remains pink for 30 seconds, record the volume


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of NaoH.

15. Fill burette to 50.00 cm3 (0.05cm3) and place securely in the burette clamp onthe stand.

16.Using the pipette filler, fill the pipette to 25.00 cm3 (0.06 cm3) with 0.1 MCH3COOH.

17.Empty Pipette into a clean, 250ml beaker.18.Place beaker underneath the burette on top of the magnetic stirrer unit.19.Place Stirring magnet in CH3COOH solution.20.Place the pH probe in the CBL probe stand and Position the pH probe so that it is

inside the beaker of CH3COOH.

21.Add deionized water to the beaker of CH3COOH such that the pH probe is in thesolution but without touching the sides of the beaker or the stirring magnet.

22.Carefully add two drops of phenolphthalein indicator to solution.23.Switch on the magnetic stirrer such that the magnet stirs the solution gently

without splashing.

24.Using the CBL unit, record the initial pH of the CH3COOH.25.Begin a slow titration, taking pH measurements every 1.00 cm3 (o.o5 cm3) until

20.00 cm3

(o.o5 cm3

) has been added.

26.At this point begin to take pH measurements ever 0.50 cm3(0.05 cm3) until thesolution turns purple for 30 seconds and for 2.00 (0.05 cm

3) cm

3following. Then

return to taking pH measurements every 1.00 cm3

(0.05 cm3) until 30.00 cm

3

(0.05 cm3) has been added.

27.Repeat steps 15-26 for three trials.Chemical Safety:

-Wear Safety Goggles at all times during the experiment

-NaOH is highly corrosive, and can cause severe burns to areas of skin contact.3

-Ethanoic Acid can cause severe eye damage and potentially cause burns to areas of

contact.4

3http://www.jtbaker.com/msds/englishhtml/s4034.htm

4http://cartwright.chem.ox.ac.uk/hsci/chemicals/ethanoic_acid.html


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Data Collection and Processing:Aspect 1: Recording Raw Data:

The software on the CBL unit and the graphing calculator allowed for theelectronic recording of data. These data were then imported directly

from the graphing calculator onto the computer using the program

Logger Pro. Hence, the Raw Data tables appear in computer type, with

the exception of qualitative observations, which are handwritten in.

We made the decision not to do an entire curve for the quick titration dueto the fact that the quick titration was done in order to get a benchmark

as to where the equivalence point would occur, not as a trial. Conducting

the quick titration as a trial would skew results because it was done less

carefully that the other three trials. End pH and end Volume are

measured from the point at which the indicator turned pink for 30

seconds.

First and Second Derivative values were added to the raw data table afterthe experiment had been completed. They are present for ease of

understanding and lack of redundancy of data.

CONDITIONS:

Room Temperature (20.4C 0.1C


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Slow Titration Trial One

Volume of 0.1

M NaOH

added (0.05cm3)

pH (0.01) First Derivative SecondDerivative

0.00 2.65 0.287 -0.037302469

1.00 2.96 0.253 -0.04045216

2.00 3.19 0.212 -0.043143519

3.00 3.41 0.152 -0.033907407

4.00 3.45 0.129 -0.011888889

5.00 3.63 0.144 -0.002666667

6.00 3.77 0.136 -0.004552469

7.00 3.90 0.128 -0.004421296

8.00 4.01 0.131 -0.008256173

9.00 4.18 0.119 -0.017214506

10.00 4.27 0.087 -0.018580247

11.00 4.32 0.079 -0.01654321

12.00 4.45 0.058 -0.014205247

13.00 4.45 0.032 0.004094136

14.00 4.45 0.061 0.025476852

15.00 4.54 0.115 0.022585859

16.00 4.77 0.104 0.017913925

17.00 4.72 0.114 0.036127739

18.00 4.90 0.236 0.006798321

19.00 5.08 0.137 -0.047075429

20.00 5.22 0.074 -0.028891906

20.50 5.17 0.035 0.035599434

CH3COOH(aq)+NaOH(aq)CH3COONa+H2O


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Volume of 0.1

M NaOH

added (0.05cm3)

pH (0.01) First Derivative SecondDerivative

21.00 5.20 0.133 0.113640285

21.50 5.34 0.194 0.106243901

22.00 5.39 0.251 0.072613536

22.50 5.62 0.257 0.073302469

23.00 5.66 0.255 0.23212963

23.50 5.80 0.410 0.586419753

24.00 6.02 0.769 1.066851852

24.50 6.43 1.44 1.48383157

25.00 7.11 2.63 1.096857363

25.50 9.45 3.06 -0.260353836

26.00 10.81 2.21 -1.334483277

26.50 11.69 1.21 -1.332266468

27.00 12.03 0.504 -0.647237948

28.00 12.30 0.265 -0.238925794

29.00 12.48 0.185 -0.058544974

30.00 12.57 0.219 0.017795194

Slow Titration Trial One (Cont) CH3COOH(aq)+NaOH(aq)CH3COONa+H2O


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Slow Titration: Trial Two

Volume of 0.1

M NaOH

added (0.05cm3)

pH (0.01) First Derivative Second Derivative

0.00 2.60 0.277 -0.0188595679012

1.00 2.87 0.271 -0.0311520061728

2.00 3.23 0.214 -0.0362893518519

3.00 3.27 0.179 -0.0256921296296

4.00 3.55 0.171 -0.0213341049383

5.00 3.64 0.145 -0.0226296296296

6.00 3.86 0.111 -0.0129398148148

7.00 3.82 0.110 0.00229166666667

8.00 4.04 0.144 -0.00621141975309

9.00 4.19 0.110 -0.0244444444444

10.00 4.27 0.066 -0.0185185185185

11.00 4.27 0.064 -0.00364197530864

12.00 4.40 0.067 0.00316358024691

13.00 4.41 0.075 0.00529475308642

14.00 4.54 0.088 0.00113888888889

15.00 4.63 0.066 0.00823709315376

16.00 4.63 0.087 0.0273279461279

17.00 4.77 0.131 0.0350587355032

18.00 4.9 0.206 -0.00786744428411

19.00 5.07 0.111 -0.0266020230651

20.00 5.08 0.0954 0.000266824327935

20.5 5.26 0.120 0.0445620189995



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Volume

of 0.1 M NaOH

added (0.05cm

3)


21.00 5.21 0.155 0.0936204438566

21.50 5.39 0.243 0.0794427542622

22.00 5.49 0.255 0.0327744708995

22.50 5.66 0.248 0.0500823045267

23.00 5.75 0.240 0.187283950617

23.50 5.8 0.408 0.39024691358

24.00 6.12 0.612 0.496697530864

24.50 6.57 0.865 0.645617283951

25.00 6.79 1.37 0.705237654321

25.50 8.06 1.69 0.488727366255

26.00 8.51 1.89 0.14447197027

26.50 9.90 2.04 -0.503833994709

27.00 11.12 1.23 -0.67616872428

28.00 11.89 0.681 -0.456538359788

29.00 12.25 0.384 -0.313806363904

30.00 12.39 0.223 -0.213043062904

Slow Titration Trial two (Cont) CH3COOH(aq)+NaOH(aq)CH3COONa+H2O


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Slow Titration: Trial Three

Volume of 0.1

M NaOH

added (0.05cm3)


0.00 2.56 0.288 -0.0345185185185

1.00 2.87 0.258 -0.0360138888889

2.00 3.14 0.205 -0.0308148148148

3.00 3.23 0.189 -0.0197731481481

4.00 3.50 0.188 -0.0258888888889

5.00 3.68 0.132 -0.0280586419753

6.00 3.72 0.111 -0.0141280864198

7.00 3.86 0.124 -0.0153317901235

8.00 4.04 0.083 -0.0183333333333

9.00 4.00 0.066 -0.00422067901235

10.00 4.14 0.080 0.00611882716049

11.00 4.18 0.082 0.0128549382716

12.00 4.27 0.115 0.00907407407407

13.00 4.45 0.116 -0.00533950617284

14.00 4.49 0.103 -0.0195910493827

15.00 4.67 0.072 -0.0291217732884

16.00 4.67 0.025 -0.0145821188071

17.00 4.72 -0.015 0.0473912538079

18.00 4.46 0.139 0.133896103896

19.00 4.90 0.218 0.0422805148996

20.00 5.03 0.197 0.0192196916252

20.50 5.31 0.216 0.0192015563801



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Volume of 0.1

M NaOH

added (0.05cm3)

pH (0.01)First Derivative Second Derivative

21.00 5.26 0.203 0.0683933715461

21.50 5.44 0.322 0.0491801146384

22.00 5.67 0.290 -0.019608686067

22.50 5.76 0.218 0.0742181069959

23.00 5.80 0.298 0.330956790123

23.50 6.03 0.514 0.682222222222

24.00 6.26 0.889 1.14456790123

24.50 6.61 1.82 1.22917989418

25.00 8.24 2.44 0.634010361552

25.50 9.23 2.62 -0.320664388595

26.00 11.17 2.11 -1.21445537289

26.50 11.67 0.999 -1.23195965608

27.00 11.98 0.492 -0.582060405644

28.00 12.39 0.221 -0.283509920635

29.00 12.39 0.079 -0.148482069371

30.00 12.39 0.022 -0.0878203997648

Slow Titration Trial three (Cont) CH3COOH(aq)+NaOH(aq)CH3COONa+H2O


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Aspect 2: Processing Raw Data

Overview: Graphing the results from each trial, I took the first and second derivative in

order to find the equivalence point, which occurs at the maximum point of the first

derivative and the point where the second derivative crosses the X-axis and becomes

negative. I then half the X value of the equivalence point to find the half equivalence

point which is equivalent to the pKa value. Furthermore, the Ka value can be calculated

via the pKa value according to the following formula

I will also do analysis of uncertainties in the experiment according to the following data

and equations

Uncertainty values:

Pipette- 0.06 cm3

Burette- 0.05 cm3

pH Probe- 0.01pH

Temperature -0.1C

%Uncertainty =uncertainty

observed valuex100

Ka 10pKa

Ka


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re%Temperatu+Probe%pH+%Burette+%Pipette=tyUncertain%Total

I will go through sample calculations for trial one, including pKa, Ka, and uncertainty,

although all data will be analyzed and presented.

Trial one, sample calculations

Equivalence point: 25.50 cm30.06 cm

3NaOH

Half equivalence point: 12.75 cm30.05 cm

3NaOH

pH at half equivalence point = 4.48 0.01

Because at the half equivalence point, pH = pKa,

pKa = 4.48 0.01

Ka 104.48

Ka3.31105

Calcuations of error:

Pipette uncertainty: 0.06 cm3/ 25 cm

3(100%) = 0.24%

Burette uncertainty: 0.05 cm3 / 25.50 cm

3(100%) = 0.19%

a 10

pKa


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pH probe uncertainty: 0.01 / 4.48 (100%) = 0.22%

Temperature uncertainty: 0.1 / 20.4 (100%) = 0.49%

Total % Uncertainty = 0.24% + 0.19% + 0.22% + 0.49%

Total % Uncertainty = 1.14 %

Uncertainty of pKa = 1.14% x 4.48 = 0.05

Uncertainty of Ka = 1.14% x 3.31105= 3.77 x 10

-7

Experimental values of pKa and Ka with uncertainties

Equivalence

Point

0.05 cm3

Half

Equivalence

Point (cm3)

pH at

Equivalence

point

0.01

pH at Half-

Equivalence

Point 0.05

pKa

value

0.05

Ka value

3.77 x 10-7

Trial 1 25.50 cm3

NaOH

12.75 cm3

NaOH

9.45 4.48 4.48 3.31105

Trial 2 26.48 cm3

NaOH

13.24 cm3

NaOH

9.90 4.43 4.43 3.72 x 10-5

Trial 3 25.51 cm3

NaOH

12.75 cm3

NaOH

9.23 4.40 4.40 3.91 x 10-5


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Aspect 3: Presenting Processed Data


Half Equivilance PointpH= 4.73

Equivilance PointpH = 9.45

Base Hydrolysis


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CH3COOH(aq)+NaOH(aq)CH3COONa+H2OPH

NaOH


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CH3COOH(aq)+NaOH(aq)CH3COONa+H2OPH


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AVERAGE Ka and pKa Values:

Average Ka=Trial1 + Trial 2 + Trial 3

3

Average Ka=3.3110

-5+ 3.7210-5+ 3.9810

-5

3

Average Ka= 3.67 x 10-53.77 x 10

-7

Average pKa=Trial1 + Trial 2 + Trial 3

3

Average pKa=4.48 + 4.43 + 4.40

3

Average pKa= 4.44 0.05

Given Literature Values:

pKa= 4.76

Ka= 1.74 x 10-5

I will calculate the percent error and then compare it to the total percent uncertainty to

establish whether or not any discrepancy was due to random or systematic error.

Percent Error:

% error=|experimental value- literature value|

literature value100

% error Ka=| 3.6710

-5 - 1.7410-5 |

|1.74

10

-5

|

100

% error Ka= 110.00%

% error pKa=| 4.44 - 4.76 |

|4.76|100


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% error pKa= 6.72

Conclusion and Evaluation

Aspect 1: Concluding

I was able to successfully titrate 0.1 M ethanoic acid and 0.1 M sodium hydroxide. Also,

both the readings from the phenolphthalein indicator and the pH probe show

characteristics of a weak acid-strong base titration. As I explained in my hypothesis, the

salt in the reaction, CH3COONa, which has a stron conjugate base component of CH-

3COO-according to the equation:

CH3COO-(aq)+ H2O(l)CH3COOH(aq)+ OH

-

Thus, the salt resulting from the titration will be basic, causing the equivalence point pH

to be greater than 7. This characteristic was consistently demonstrated. I also was able

to gather data for three titration curves with a starting pH of ethanoic acid around 3.00

0.01 and an ending pH of approximately 130.01. As per my hypothesis, we were able

to use this lab to calculate Ka from pH =pKa, Ka= 10-pka

. We took our hypothesis and

through calculations, we were able to not only find the pKa, but apply it and find the Ka

of all the trials. We had an Average Kaof 3.67 x 10-5

which is 110% off the literature

value of 1.74 x 10-5

and an average pKa value of 4.44 which is 6.72% off of the literature

value of 4.76. The % error can be offset slightly by 1.14% total percent uncertainty

within the experiment. In the titration, we hypothesized that even the smallest value

could have offset the Ka, and as it is proven with our data, the 6.72% resulted in a 110%

difference from the literature value. Where this is especially significant is with the Ka

value, which was the only value derived experimentally. Although the 1.14% uncertainty

does not completely offset the 6.72% error in regards to Ka, it does offer a partial

explanation as to the error is both Ka and pKa. It must be noted that because of the use

of a logarithmic function to calculate Ka from pKa, even a tiny error in pKa can have a

significant effect on Ka. Thus, the more important percent error is in relation to pKa


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because any error in pKa will have such a significant impact on Ka. Even a discrepancy

of 0.05 can have a significant effect on final results when involved in a logarithmic

function. The rest of the errors were either random or systematic.

Aspect 2: Evaluating Procedure

1. The temperature during our experiment was 20.4C, whereas the temperature atwhich the literature values were calculated was 24.85C, and temperature affects

the Ka and pKa meaning that there would be some discrepancy between our

values and the literature values.

2. We used deionized water during our experiment as opposed to distilled water.Deonized water contains more impurities than distilled water and thus could

have affected our results.

3. To be more ready, we should have taken more measurements, that way therewould be a smaller window for room error. This room for error could be a factor

in to our 110% window of error.

4. Another factor in this was the burette and pipette which could have easilydripped too much or too little of the NaOH into the BaOH, causing a room for

error. More preparation could have been done to avoid this. This effect wouldinfact cause the 110% difference as even one +/- 0.05 of NaOH and BaOH would

affect not only the equivilance point but also affect the base hydrolysis and the

overall pKa and Ka.

5. Our equivalence point volumes did not divide to find the half-equivalence pointsin a manner that would reflect accurate measurements taken. Because data was

only taken ever 1cm3during the times when the half-equivalence points would

occur, the pH values of the half equivalence points are only approximationsbased on the trends in the graph and hence there is a large room for error.

Aspect 3: Improving the Investigation

1. While there is no certain way to account for the temperature discrepancy, itis a factor that should have taken into account when looking at percent


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uncertainty, which could explain the discrepancies between experimental

values and literature values.

2. Using distilled instead of deionized water would eliminate or greatly reduceany error associated with the impurities in deionized water.

3.References:

Green, John, and Sadru Damji. "Acids and Bases." Chemistry. Melton: IBID,1998. 218.

IB HL Chemistry Data Booklet http://www.jtbaker.com/msds/englishhtml/s4034.htm http://cartwright.chem.ox.ac.uk/hsci/chemicals/ethanoic_acid.html
http://www.jtbaker.com/msds/englishhtml/s4034.htmhttp://www.jtbaker.com/msds/englishhtml/s4034.htmhttp://www.jtbaker.com/msds/englishhtml/s4034.htm

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