CH1801 C1 Titration Techniques integrated with Matlab PDF

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School of Chemical and Biomedical Engineering

CH 1801 BG 1801 Titration Techniques Location: N1.2-01-01

Name:

_______________________________________

Matric Number:

_______________________________________

Group:

_______________________________________

Date of experiment:

_______________________________________ GRADE: _____________

TITRATION TECHNIQUES Objectives: In the first part of this laboratory experiment, you will titrate a polyprotic acid (eg. Phosphoric Acid) with a strong base titrant (eg. NaOH solution). The titration curve will be plotted and the relationship between the shape of titration curves and the pKa values will also be demonstrated. You are required to use Matlab to process all the data, plot graphs and carry out calculations. Please refer to the Matlab manual in Appendix. In the second part of this laboratory experiment, you will gain the understanding the principles behind potentiometric redox titration. You will get familiarized with the use of a combination electrode. Introduction: Part1 One method a chemist can use to investigate acid-base reactions is by titration. A pH titration is performed by adding small, precise amounts of standard base to an acid solution of known concentration. The pH is recorded methodically and is plotted vs. the volume of base added to the acid solution. The result of this plot is an “S” shaped curve. A single “S” curve is observed for a monoprotic acid and multiple “S” curves are observed for polyprotic acids. The inflection point of each “S” curve (the middle of the “S”) is indicative of the endpoint. These points (also known as equivalence points) occur when the dissociated proton from the acid and the OHfrom the base are stoichiometrically equivalent in solution. The equivalence point is needed to determine the concentration of an acid or base. Acids that contain more than one acidic (ionizable) hydrogen (proton) are called polyprotic or polybasic acids. The dissociation of polyprotic acids occurs in a stepwise fashion, with one proton being lost at a time. For example, the generic triprotic acid will dissociate as shown in Reactions (1) through (3). The equilibrium constants for these reactions are symbolized by Kan. The trailing subscript "a" indicates that the equilibrium constant describes an acid dissociation reaction. The trailing subscript "n" is written as a number and indicates which proton is being dissociated. Phosphoric Acid (H3PO4) is a triprotic acid with three ionizable protons. The reaction for the dissociation of each of these protons has its own value of Ka, as indicated in the following reactions: [H+][H2PO4-] H3PO4

≡H

+

+ H2PO4

-

= 7.11x10-3

Ka1 =

pKa1=2.15

(1)

pKa2=7.20

(2)

[H3PO4]

H2PO4

-

≡H

+

+ HPO4

2-

Ka2

[H+][HPO42-] = ---------------- =6.34x10-8 [H2PO4-]

2

HPO4

2-

≡H

+

+ PO4

3-

Ka3

[H+][PO43-] = --------------[HPO42-]

=4.20x10-13

pKa3=12.38

(3)

If the values of Ka1, Ka2, and Ka3 differ by several orders of magnitude, the dissociation may be regarded as taking place in a stepwise fashion. This is true for two of the ionizable protons of phosphoric acid. The third hydrogen of phosphoric acid is so slightly dissociated (Ka3 = 4.2 x 10-13) that it does not yield an end point of any practical value. The typical titration curve for phosphoric acid (Figure 1), therefore, consists of two parts, each of which resembles the titration curve of a monoprotic weak acid with a strong base.

Figure 1 Titration Curve for Phosphoric Acid with NaOH

nstants

3

At equal acid and conjugate base concentrations, pH=pKa1. There are three such points for phosphoric acid. They are labeled on Fig 2. These points are important in the prediction of the titration curves. They correspond to points where half of an equivalent of proton has been consumed by addition of strong base. Thus, the point where pH=pKa1 is halfway to the first equivalence point. Where pH=pKa2 is halfway between the first and second equivalence points, etc. The solution has maximum buffer capacity at these points. In other words, there is maximum resistance to changes in pH. The equivalence points can also be identified in the fraction plot. At the first equivalence point , [H3PO4] approaches zero. This occurs when [H2PO4-] is a maximum. One can see this point in the relative concentration plot. It occurs at a pH that is halfway between the two points with maximum buffer capacity. In fact, we can expect that the first equivalence point will occur at a pH of

Similarly, the second equivalence point, laying halfway between the points where pH=pKa2 and pH=pKa3 is

To summarize, without even performing the titration, or solving the fifth power polynomial equation that governs the proton concentration, we would predict the following pH at the halfway and equivalence points

Part 2 One of the most common methods in detecting the equivalence point of an acid-base or redox titration is by means of a visual indicator which undergoes a color change at the end point. Another important method is by potentiometric measurements using a pH meter or combination electrode. This experiment will involve the use of potentiometric analysis to determine the Fe2+ concentration of a solution by titration against a Ce4+ solution. This is a redox reaction and the half reactions can be written as: Fe2+(aq)  Fe3+(aq) + eCe4+(aq) + e-  Ce3+(aq)

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The equivalence point is detected by monitoring the potential of the cell formed by a combination electrode in contact with the solution. At the equivalence point, there will be a sharp change of cell potential. One advantage of using potentiometric titration over titrations involving a visual indicator is that there is no need for a color change in the solution to indicate the equivalence point. The principle behind potentiometric redox titration is equivalent to that of a galvanic cell. In the galvanic cell setup, the standard hydrogen electrode (SHE) can be used as a reference electrode to measure the reduction potential of many processes, such as: Fe3+(aq)  Fe2+(aq) + eA diagram showing the apparatus is shown in Figure 1b, and the cell notation is given as: Pt(s) | Fe2+(1 M), Fe3+(1 M) || H+(1 M), | H2 (1 atm) | Pt(s) Such a setup, however, is not very convenient to use in titrations. To make it more portable, two major adjustments are needed. First, the standard hydrogen reference electrode is replaced by another reference electrode. A common reference is the silver-silver chloride electrode, which is used in this experiment. The standard reduction potential for the AgCl|Ag couple in 3 M KCl solution is +0.207 V at 25°C: AgCl(s) + e-  Ag(s) + Cl-(aq) Since the standard reduction potential of the AgCl|Ag system with reference to the SHE is known, the voltage measurements obtained from the AgCl|Ag reference can be converted to the voltage with respect to the standard hydrogen reference by the formula: ESHE = Emeasured + 0.207 V Second, the AgCl|Ag electrode, wire, and salt bridge are intelligently assembled together to produce what is know as a combination electrode (Figure 1a), which can then be linked to a readout to measure the voltage of the Fe2+/Fe3+ system.

Pt Combination electrode

Fe2+/Fe3+ electrode (a)

Fe2+/Fe3+ electrode (b)

Figure 1. Setup of (a) a combination electrode (b) a regular galvanic cell In this experiment, the concentration of Fe2+ and Fe3+ ions will be changing as Ce4+ is gradually added. The potential of the solution will be measured by means of a combination electrode. In 5

essence, the potentiometric titration setup is simply a modified galvanic cell, with the reference electrode making up half the cell and the solution to be titrated making up the other half. The Nerst equation for the Fe2+/Fe3+ half cell as follows: E = EFeo 3 + ,Fe 2 + −

RT [ Fe 2 + ] log + nF [ Fe 3 ]

(1)

o E Fe 3 + Fe 2 + = 0.771 V ,

As the Fe2+(aq) and Fe3+(aq) concentrations change during the titration, there will also be a change in cell voltage. As the test solution also contains Ce4+(aq) and Ce3+(aq) ions, making up the Ce4+/Ce3+ half cell, another way to express the electrode potential is: o E = ECe − 4+ , Ce3 +

RT [ Ce3 + ] log nF [ Ce 4+ ]

(2)

o E Ce 4 + , Ce 3 + = 1.61 V

The electrode potentials of Equations (1) and (2) must be the same since they share the same electrode. These two equations can be used to derive the titration curve, an example of which is shown in Figure 2. The equivalence point occurs at the inflexion point (where the gradient is steepest). Theoretically, Equation (1) can be used to plot the first part of the curve since the concentrations of both Fe2+ and Fe3+ are known. Beyond the equivalence point, Equation (2) can then used to plot the titration curve as [Ce3+(aq)] and [Ce4+(aq)] are known. Near the equivalence point, Equations (1) and (2) can be combined to determine the cell voltage. 1.800

1.600

E/volts

1.400 1.200

1.000 0.800

Equivalence point

0.600

0.400 0.00

50.00

100.00

150.00

200.00

V/mL

Figure 2. Potentiometric titration curve for Fe2+/Ce4+ system. 6

An alternative way of determining the equivalence point is to plot the first and second derivative plots of the titration curve. In the first derivative plot, the equivalence point will be located at the peak maximum. Figure 3 shows how the values for the first and second derivative plots are obtained. Figure 4 shows the first derivative plot. For the second derivative graph, the equivalence point is located at the point where the curve intersects the x-axis. Figure 5 shows an example of the second derivative plot. Note that only the points close to the equivalence point are needed to plot the derivative curves. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

19 17

B

C A

D B

E C

F D

G E

H F

I G

J H

K I

L J

Derivative Titration: Redox First derivative Vol. of Ce 4+ added V/mL 35.45 35.50 35.55 35.60 35.65 35.70 35.75

E/volts 0.630 0.650 0.680 0.800 0.860 0.890 0.910

Vavg 1

∆E

35.475 35.525 35.575 35.625 35.675 35.725

0.020 0.030 0.120 0.060 0.030 0.020

(A5+A6)/2 B6-B5 A6-A5

Cell F6 = ∆E/∆V 1 = Cell G7 = V ave2 =

D6/E6 (C6+C7)/2

20 18 21 19

Cell H7 = ∆(∆E/∆V 1 ) = F7-F6

22 20 23 21 24 22

Cell J7 = ∆ E/∆V = H7/I7 1st Derivative: Plot V avg 1 vs. ∆E/∆V1

2

∆V1 Vavg 2 ∆V 1 ∆E/∆

∆(∆ ∆E/∆V 1)

∆V 2

2 ∆2 E/∆ ∆V

(∆(∆E/∆V 1)/∆V2)

Cell C6 = V ave1= Cell D6 = ∆E = Cell E6 = ∆V 1 =

Cell I7 = ∆V2 =

Second derivative

0.05 0.05 0.05 0.05 0.05 0.05

0.400 0.600 2.400 1.200 0.600 0.400

35.500 35.550 35.600 35.650 35.700

0.20 1.80 -1.20 -0.60 -0.20

0.050 0.050 0.050 0.050 0.050

4.00 36.00 -24.00 -12.00 -4.00

Copy all formulas down to end

C7-C6 2

2

2

2nd Derivative: Plot V avg 2 vs. ∆ E/∆V

Figure 3. First and second derivative calculations.

7

2.60 2.40 2.20 2.00 1.80

∆V ∆E/∆

1.60 1.40

Equivalence point

1.20 1.00 0.80 0.60 0.40 0.20 0.00 35.50

35.55

35.60

35.65

35.70

35.75

V/mL

Figure 4. First derivative curve. 40.00

30.00

Equivalence point

20.00

∆ 2E/∆ ∆ V2

10.00

0.00 35.50

35.55

35.60

35.65

35.70

-10.00

-20.00

-30.00

V/mL

Figure 5. Second derivative curve.

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References: 1. G. C. Christian, “Analytical Chemistry”, 6th edition, John Wiley & Sons, Inc., New Jersey, 2004, pages 354-366, 380-382, 414-421, 433-437. 2. D. C. Harris, “Quantitative Chemical Analysis”, 3rd edition, W. H. Freeman and Company, New York, 1991.

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Equipment/Materials: Phosphoric acid (0.1M H3PO4) Sodium Hydroxide (0.1M NaOH) solution pH/mV meter and probe 100 mL beakers Burette Retort stand and clamps Pipette and pipette bulb Magnetic stirrer and stirring bar De-ionized water Combination electrode Measuring cylinder Pipette 0.01 M Fe(NH4)2(SO4)2, acidified 0.05 M Ce(NH4)4(SO4)4, acidified

Procedure (Part 1): 1. Calibrate the pH meter with the standard buffers. (Technician will do this before the lab class.) 2. Pipette accurately 10mL of 0.1M phosphoric acid into a 100mL beaker. 3. Place the beaker on the stirring plate and put the stir bar into the beaker. 4. Insert the pH probe into the beaker. Add sufficient De-ionized water to the beaker to cover the bulb of the pH electrode completely (approx 10mL). Ensure that the stir bar does not hit the tip of the electrode. The electrode is very fragile so be careful not to damage it when the stir bar is rotating. 5. Turn on the stirrer. 6. Rinse the burette with some NaOH Solution. (Technical staff will demonstrate how this is done.) 7. Fill the burette above the 0.00 mL mark with NaOH solution, open the stopcock and drain to remove any air bubbles. Adjust the level of the NaOH solution so that it is at the 0.00 mL mark and that the tip of the burette is filled. 8. Begin titrating, by adding NaOH in 2mL increments. Take down the pH reading after each addition. Take numerous readings (ie add smaller volume of NaOH for each step) near the equivalences points, where the pH changes more rapidly. 9. Continue the titration until the titrant in the burette reaches 35.0 mL. (OR) Continue the titration until the pH reaches above 11.5. 10. Repeat the experiment. The first run can be used to estimate the region where the equivalence points lie. In subsequent runs, larger volumes of titrant can be added at regions far from the equivalence point, while smaller volumes are added near the equivalence point. 11. Each student should have a chance to do the experiment. 12. Repeat step 2 to 10 with 20mL of unknown sample but with smaller increments (eg. 0.2mL increments) of NaOH. 13. Plot the data collected for both the phosphoric acid and unknown sample titration [pH (yaxis) vs. volume of NaOH (x-axis)] on the graph paper. Answer the questions below. 10

Procedure (Part 2): 1. Pipette 25 mL of a 0.01 M Fe(NH4)2(SO4)2 solution into a 100ml plastic beaker. This solution contains some acid to prevent oxidation of the Fe2+ ions by the atmosphere. 2. Place the beaker on the stirring plate and put the stir bar into the beaker. 3. Insert the combination electrode into the beaker. Ensure that the stir bar does not touch the tip of the electrode. The electrode is very fragile so be careful not to damage it when the stir bar is rotating. 4. Turn on the stirrer. 5. Fill the burette with 0.05M Ce(NH4)4(SO4)4 solution (orange solution) and note down the starting volume. 6. Titrate the Fe(NH4)2(SO4)2 solution with the Ce(NH4)4(SO4)4 solution from the burette in a stepwise manner. One student could read the amount of Ce(NH4)4(SO4)4 solution added at each step while another student could take note of the reading on the pH/mV meter. Take numerous readings (ie add very small volumes of Ce(NH4)4(SO4)4 solution for each step) near the equivalence point, where the potential changes rapidly.* Continue to titrate till well beyond the equivalence point (about 10 ml of titrant). 7. Repeat the titration a few more times. Every student should have a chance to do the experiment. Use the best of 2 results for the report.

*Note: The first run can be used to estimate the region where the equivalence point lies. In subsequent runs, larger volumes of titrant can be added at regions far from the equivalence point, while smaller volumes are added near the equivalence point.

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TITRATION TECHNIQUES LOG SHEET Name____________________ Date_____________________

Group ID____________________

Name of Acid Used ________________ Concentration of NaOH _________________ Vol of Acid sample ________________ Vol of unknown Sample_________________ Titration Data (Part1): 0.1M H3PO4 Titration 1 Vol. of NaOH pH

Titration 2 Vol. of NaOH pH

Unknown sample Titration 1 Vol. of NaOH pH

Titration 2 Vol. of NaOH pH

12

Use Matlab to process all the data, plot graphs and carry out calculations. Attach Matlab code in the lab report. Please refer to the Matlab manual in Appendix.

Question (Part1): 1. From the titration graph of 0.1M H3PO4, find the volume of base needed to reach the a) first equivalence point and b) second equivalence point 2. Subtract the volume of NaOH needed to reach the first equivalence from that needed to reach the second equivalence point. Using this volume, molarity and volume of H3PO4 used, calculate the molarity of NaOH. 3. Find the pKa1, pKa2, pKa3 and compare with literature values. Explain the difference if any. 4. Using information from the titration curve of the unknown sample (eg, volume of NaOH used at first equivalence point), molarity of NaOH to calculate the concentration of phosphoric acid (H3PO4) in the unknown sample.

Useful equations:

pKan = -log(Kan)

pH = -log[H+]

Part 2 Answer the following questions in your Log Sheet. 1. What is the concentration of Fe2+(aq) in 0.01 M of Fe(NH4)2(SO4)2 solution? 2. For the two sets of data using the 0.01 M Fe(NH4)2(SO4)2 solution, 13

a) Plot the graph of potential, E, versus the volume of Ce(NH4)4(SO4)4 solution added, V. b) Plot the graph of ∆E/∆V versus V. This is the first derivation plot. c) Plot the graph of ∆2E/∆V2 versus V. This is the second derivative plot. d) Obtain the volume of Ce(NH4)4(SO4)4 solution added at the equivalence point from graphs (a), (b) and (c). Which graph should give the most accurate answer? Why? 3. Is the volume of Ce(NH4)4(SO4)4 added at equivalence point similar to the the theoretical value? Suggest reasons for any discrepancies. 4. From Equation (1), what is the theoretical voltage when [Fe2+] = [Fe3+]? What is value obtained from the titration experiment? Suggest reasons for any discrepancies. 5. What are the sources of error in the experiment? 6. Suggest ways to improve the experiment. (Submit all the tables and the graphs for your report)

Physical constants Gas constant: R = 8.314 J/mol-K Faraday constant: F = 96485 C/mol

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LOG SHEET (Part 2) Name: Vol. of 4+ Ce added V/mL

Group: First derivative

E/volts

Vavg 1

∆E

Date:_______ Second derivative

∆ V1

∆E/∆ ∆ V1

Vavg 2

∆(∆ ∆E/∆ ∆V1)

∆ V2

2 2 ∆V ∆ E/∆

(∆(∆E/∆V1)/∆V2

15

Appendix Introduction to Matlab •

This lab is meant for self-learning. There are no assigned lab hours.

•

In case of questions, you can contact the teaching assistant through e-mail.
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