Prac 2 Potentiometric acid-base titrations and measuring the buffer capacity of a biological buffer PDF

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Potentiometric acid-base titrations and measuring the buffer capacity of a biological buffer
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EXPERIMENT 2 REPORT STUDENT NAME:

LAB SESSION AND DAY/DATE: Monday 11-2PM

STUDENT EMAIL ADDRESS:

DEMONSTRATOR’S NAME:

A. Data and Results – Titration: H3PO4 vs. NaOH Burette Total volume pH reading added 0.37 1.16 1.91 2.68 3.50 4.35 4.99 5.64 6.30 6.76 7.21 7.33 7.51 7.74 8.01 8.27 8.75 9.41 9.89 10.43 11.00 11.50 12.38 12.76 13.50 14.03 14.63 14.80 15.12 15.35 15.62 15.90 16.50 16.92

0.00 0.79 1.54 2.31 3.13 3.98 4.62 5.27 5.93 6.39 6.84 6.96 7.14 7.37 7.64 7.90 8.38 9.04 9.52 10.06 10.63 11.13 12.01 12.39 13.13 13.66 14.26 14.43 14.75 14.98 15.25 15.53 16.13 16.55

1.77 1.79 1.81 1.84 1.89 1.97 2.04 2.21 2.39 2.61 3.06 3.32 4.21 5.13 5.52 5.76 6.06 6.35 6.52 6.69 6.85 6.99 7.20 7.37 7.62 7.87 8.37 8.66 9.82 10.33 10.67 10.91 11.20 11.33

Burette reading

Total volume added

pH

17.52 17.97 18.49

17.15 17.60 18.12

11.47 11.56 11.62 1

18.98 19.45 20.00

18.61 19.08 19.63

11.68 11.74 11.78

2

Using Excel (or similar graph plotting program), plot pH (y axis) versus volume of titrant (base) added. Make sure you label the plots with appropriate titles and axis labels. Insert this graph into your report. Ensure the graph has a meaningful title and axes labels.

Titration of Phosphoric acid with Sodium Hydroxide 14 12

pH level

10 8 6 4 2 0 0

5

10

15

20

25

Volume of 0.10M NaOH (titrant) added (mL)

Figure 1: The titration curve of H3PO4 and NaOH A. Calculations and results Exact concentration of NaOH (on label of bottle): 0.10 mol/L Write balanced molecular equations for the reactions occurring at the three equivalence points on the plot (note the third equivalence point may be hard to see on the plot): A. H3PO4 (aq) + NaOH ⇌ NaH2PO4 (aq) + H2O (l) B. NaH2PO4 (aq) + NaOH ⇌ Na2HPO4 (aq) + H2O (l) C. Na2HPO4 (aq) + NaOH ⇌ Na3PO4 (aq) + H2O (l) Calculate the number of moles of the sodium phosphate (NaH2PO4) salt formed at the first equivalence point (based on the concentration and volume of base used). From this value, calculate the unknown starting concentration of phosphoric acid: 1st equivalence point: 7.14mL = 7.14 x 10-3L NaOH added n (NaOH) = C x V = 0.10M x 7.14 x 10-3L = 7.14 x 10-4 mol Mole ratio 1:1 between NaOH & NaH2PO4  7.14 x 10-4 mol of NaH2PO4 formed at the 1st equivalence point. H3PO4 also has this same number of mol because the ratio is 1:1. 20mL = 0.02L The unknown starting C of H3PO4: C (H3PO4) = n/V = 7.14 x 10-4 mol / 0.02L = 0.0357 = 3.57 x 10-2M 3

Using the Henderson-Hasselbalch equation, demonstrate that the pH at 50% neutralisation for a monoprotic acid/base titration is equal to the pKa for the acid. pH = pKa + log ([A-] /[HA]) Because at half-equivalence, exactly half of the acid has been neutralised, so [HA] = [A-] [A-] / [HA] = 1  pH = pKa + log (1) pH = pKa + 0 Therefore, pH = pKa when the concentration of acid & base is equivalent (i.e. equivalence point).

Using your titration curve estimate the pKa for NaH2PO4, compare this value to the literature value in the introduction. Explain your answer. Hint – think about the definition of pKa from the previous question. How can you determine the 50% neutralisation point for NaH2PO4? Literature values of the pKa are 2.15, 7.09 and 12.32, respectively. Because pH = pKa at 50% neutralisation, the pKa values shown in Figure 1 are: pKa,1 = 1.97 pKa,2 = 7.2 pKa,3 = 11.62  The literature and the estimated pKa values are similar. A. Titration Questions 1.

Why is the equilibrium between the acid NaH 2PO4, and its conjugate base Na2HPO4, a suitable buffer for maintaining intracellular pH (pH 6.9-7.3)? Justify your answer using the titration curve from part A.

When small amounts of acid or base are added, buffers have the ability to counteract pH changes that occurred. Buffers are most effective in resisting pH change when the acid: base ratio is 1:1. The acid: base ratio is 1:1 at 50% neutralisation; and as illustrated in Figure 1, the 50% neutralisation point for NaH2PO4 has a pKa of 7.2. The phosphoric acid was able to maintain a pH between 6.9 to 7.3 when NaOH was gradually added. Therefore, the equilibrium between the acid NaH 2PO4, and its conjugate base Na2HPO4 is a suitable buffer for maintaining intercellular pH around 6.9-7.3.

2.

What other pH ranges could phosphoric acid and its conjugates bases be used to generate buffers for. Give the chemical equation for the equilibrium for each buffer.

The literature value given are: pKa,1 = 2.15 pKa,2 = 7.09 pKa,3 = 12.32

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Therefore, phosphoric acid and its conjugates bases can be used to generate buffers from: 1.95 to 2.35 6.89 to 7.29 12.12 to 12.52 The chemical equation for the equilibrium for each buffer is: A. H3PO4 (aq) ⇌ H+ (aq) + H2PO4- (aq) B. H2PO4- (aq) ⇌ H+ (aq) + HPO42- (aq) C. HPO42-(aq) ⇌ H+ (aq) + PO43- (aq) C1. Buffer titration – Data and Results Buffer Solution Volume (mL) NaH2PO4 Volume (mL) Na2HPO4 Theoretical pH Solution added Starting pH pH after 0.050 mL pH after 0.100 mL pH after 0.150 mL pH after 0.200 mL pH after 0.250 mL

1

2

3

3

4

5

9.5

7.5

5.0

5.0

2.5

0.5

0.5

2.5

5.0

5.0

7.5

9.5

5.81

6.61

7.09

7.09

7.57

8.37

HCl

HCl

HCl

NaO H

NaO H

NaO H

5.89

6.38

6.90

6.91

7.29

7.88

4.52

5.84

6.44

7.08

8.05

9.56

2.68

3.36

5.84

7.59

9.58

10.46

2.41

2.68

3.34

9.14

11.23

10.85

2.26

2.45

2.68

10.36

11.55

11.13

2.18

2.26

2.44

10.88

11.73

11.29

Use the Henderson-Hasselbalch equation to predict the theoretical pH of each of your buffer solutions. Show your working for one buffer solution. The rest can be computed and added to the table above. Buffer 1 example Find concentration: Na2HPO4 NaH2PO4 Ci x Vi = Cf x Vf Ci x Vi = Cf x Vf 0.1M x 0.5mL = Cf x 100mL 0.1M x 9.5mL = Cf x 100mL -3 Cf = (0.1 x 0.5)/100 = 5 x 10-4 M Cf = (0.1 x 9.5)/100 = 9.5 x 10 M Find pH: 5

pH = pKa + log ([A-] /[HA]) pH = 7.09 + log (5 x 10-4)/ (9.5 x 10-3) = 5.81 In excel, create an X-Y scatter plot of your data using the table format below. Note for solution 4, which was titrated with both HCl and NaOH, put all values in the same column.

Volume NaOH added

Volume HCl added

Buffer solutions Volume (mL) 0.250 0.200 0.150 0.100 0.050 0.000 0.050 0.100 0.150 0.200 0.250

X-Axis

1

2

-0.250 -0.200 -0.150 -0.100 -0.050 0.000 0.050 0.100 0.150 0.200 0.250

2.18 2.26 2.41 2.68 4.52 5.89

2.26 2.45 2.68 3.36 5.84 6.38

3

4

5

2.44 2.68 3.34 5.84 6.44 6.90 7.08 8.05 7.59 9.58 9.14 11.23 10.36 11.55 10.88 11.73

9.56 10.46 10.85 11.13 11.29

Create an X-Y scatter plot for all of the buffer solutions on one plot. Use the column marked X-axis, for the X-axis values of the plot. This will arbitrarily assign HCl additions with negative values. This is not a real value, only so the HCl and NaOH plots appear as mirror images. Insert a copy of this plot to this report. Ensure the graph has a meaningful title and axes labels. You can use text boxes to label the two sides of the x-axis as “volume of 1.0M HCl added”, or “volume of 1.0 M NaOH added”.

Buffer Capacity of different phosphate buffer solutions Volume of 1.0M NaOH added

12

pH of solution

10 8 6 4 2 0

0

2

4

6

8

10

Volume of 1.0M HCL added Buffer 1

Buffer 2

Buffer 3

Buffer 4

Buffer 5

Figure 2: Buffer capacity of 5 different phosphate buffer solutions. C1. Buffer capacity 6

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Next create a second plot of the data so that only the first 3 data points are used (0 - 0.1 mL), so that you are only using the linear region of all the plots. Add a trendline (line of best fit), and label it with the equation. Record the slope of the trendline for each plot in the table below.

Volume HCL added Volume NaOH added

Volume (mL)

X-Axis

1

0.100 0.050 0.000 0.050 0.100

-0.100 -0.050 0.000 0.050 0.100

2.68 4.52 5.89

Buffer solutions 2 3 4 3.36 5.84 6.38

32.1

Slope

30.2

5

5.84 6.44 6.90 7.08 7.59

8.05 9.58

9.56 10.46

8.28

30.6

18

Buffer Capacity of different phosphate buffer solutions 12 10

pH of solution

8 6 4 f(x) = 0.04 0.28 x + 2.85 0.16 f(x) = 0.25 x + 0.12 f(x) = − 0.11 x + 2.1 f(x) = − 0.09 x + 1.73

2 0

0

2

4

6

8

10

Volume of 1.0M HCL added Buffer 1 Buffer 3 Buffer 5

Linear (Buffer 1) Linear (Buffer 3) Linear (Buffer 5)

Buffer 2 Buffer 4

Linear (Buffer 2) Linear (Buffer 4)

Figure 3: The Buffer capacity graph showing the trendline of 5 different phosphate buffer solution. The slope of the line of best fit is measuring how much the pH of the solution changes, for each millilitre of acid or base being added, where smaller value of the slope will indicate smaller change in pH per mL of acid or base added and larger value will indicate larger change in pH per mL of acid or base added. Therefore, by comparing the values of the slopes for each buffer solution, we can determine which one has the highest buffer capacity.

7

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C2. Dilute buffer titration Buffer Solution added Starting pH pH after 0.050 mL pH after 0.100 mL pH after 0.150 mL

3

0.1X 3

3 NaO H

0.1X 3

HCl

HCl

6.90

6.89

6.91

6.89

6.55

2.63

7.19

10.76

6.10

2.19

7.69

11.63

3.87

1.98

8.70

11.85

NaOH

C. Buffer Questions 1. Using the buffer capacity graph, which of the buffer solutions has the highest buffer capacity? How does the starting pH of your best buffer compare to the p Ka of NaH2PO4 ⇄ Na2HPO4? The role of buffer is to counteract pH changes that occurred when a small amount of acid or base is added. It is known that a smaller value of the slope will indicate smaller change in pH per mL of acid or base added and larger value will indicate larger change in pH per mL of acid or base added. As demonstrated in Figure 3, buffer 3 has the smallest slop of 8.28. Furthermore, in theory, buffers with the best capacity has an acid-base ratio of 1:1, and this buffer meets the condition. Therefore, buffer 3 has the highest buffer capacity. The starting pH level of buffer 3 is 6.90, which is close to the pKa of NaH 2PO4 ⇄ Na2HPO4, which is 7.09. You are running a biochemical assay with an enzyme that needs to maintain a constant pH of 6.6 in order to operate effectively. Using starting stock solutions of 0.2 M of NaH 2PO4 and Na2HPO4, how much of each stock would you need for 100 mL of buffer at a final concentration of 0.01 M and a pH of 6.6. Hint: First, calculate the dilution to determine total amount of phosphate stock required. Then use the Henderson-Hasselbalch equation to calculate the ratio of acid to base. 2.

1/ Dilution: Ci x Vi = Cf x Vf 0.2M x Vi = 0.01M x 100mL Vi = 1/0.2 = 5mL of phosphate stock needed 2/ Acid to Base ratio: pH = 6.6 pKa = 7.09 6.6 = 7.09 + log([base]/[acid]) log([base]/[acid]) = -0.49 [base]/[acid] = 0.324 Therefore, in regard to the 5 ml of stock solution, there is 1.62 ml of base (Na2HPO4), and 3.38 ml of acid (NaH2PO4). 8

3.

In section C2, you measured the change in pH of a dilute buffer solution, upon addition of acid or base. What effect did diluting the buffer have on its buffer capacity? Explain why this occurs.

It is shown in the Dilute buffer titration table that the pH level increased or decreased at a faster rate in the diluted solution (0.1 x 3) compared to the original solution. This indicates that the original solution has a higher buffer capacity than the diluted solution. The buffer capacity is measured by the amount of H 3O+ and OH- presented in the solution. By diluting with water, the added concentration of H 2O increases the amount of H 3O+ formed when adding acid (HCl), thus decreasing the pH level of the solution. Water also increases the amount of OH- formed when adding base (NaOH), thus increasing the pH of the solution. Therefore, diluting causes a buffer solution to decrease its buffer capacity.

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