Experiment 6 - Investigating the properties of buffers PDF

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Experiment 6 - Investigating the properties of buffers. High Scoring...

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REPORT COVER SHEET AND DECLARATION School of Chemistry Universityy of Melbour Melbourne The Universit ne Laboratory Report Cover Sheet Student Name:

Matthew Dang

Student Number:

1173696

Subject Name & Code:

Chemistry 1 – CHEM10003

Demonstrator:

Amani Alhifthi

Experiment Title:

Experiment 6: Investigating Buffers.

Due Date: 14/10/2020 By submit submitting ting work for asse assessment, ssment, I her hereby eby declare that I understa understand nd the Universi University’s ty’s policy on academic integrity and I declare that:

This laboratory report is my own original work and does not involve plagiarism or unauthorised collusion, except where due credit is given to the work of others. The report is based on results and spectra obtained by me during my laboratory session. • This laboratory report has not previously been submitted for assessment in this or any other subject. For the purposes of assessment, I give the assessor of this assignment the permission to: • Reproduce this laboratory report and provide a copy to another member of staff; and • Take steps to authenticate the assignment/laboratory report, including communicating a copy of this assignment to a checking service (which may retain a copy of the assignment on its database for future plagiarism checking). Feedback on Report: Feedback on your report and the mark you received will be available on the Online Practical Assignments page on Canvas. •

Plagiarism: Plagiarism is the act of representing as one's own original work the creative works of another, without appropriate acknowledgment of the author or source. Collusion: Collusion is the presentation by a student of an assignment as his or her own work, but which is in fact the result in whole or in part of unauthorised collaboration with another person or persons. Collusion involves the cooperation of two or more students in plagiarism or other forms of academic misconduct. Both collusion and plagiarism can even occur in group work. For examples of plagiarism, collusion and academic misconduct in group work please see the University’s policy on Academic Honesty and Plagiarism: https://academichonesty.unimelb.edu.au Plagiarism and collusion constitute cheating. Disciplinary action will be taken against students who engage in plagiarism and collusion as outlined in University policy. Proven involvement in plagiarism or collusion may be recorded on your academic file in accordance with Statute 13.1.18.

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Experiment 6: Investigating Buffers. Author(s): Matthew Dang, Sophie Black, Tamsyn Barnham, Nathan Munasinghe Arachchige Day/Time/Group number: Tuesday, 10:00am, Group B Abstract: (Summary of what you did and what you found out) In this three-part experiment, titrations of a weak base (sodium carbonate – Na2CO3) against a strong acid (hydrochloric acid – HCl) using phenolphthalein indicator and a glass electrode explored the changes in pH of the solution. This was graphically plotted in a graph of pH vs. the volume of HCl added to determine the key features of the equivalence point (pH of 8.90 with 10.20 mL added HCl) and the half-equivalence point (10.20 pH and 5.10 mL of HCl). While, the changes in pH in buffer system of equal concentrations of a weak base (CO32-) and its conjugate acid (HCO3-) with the addition of a strong acid (HCl) and strong base (NaOH) was compared against the changes in distilled water. Effectively, the mechanisms for buffer systems resisted a change in pH while distilled water was drastically affected. This was furthered investigated with the mathematical relationship between pH and the ratio of weak acid (HCO32-) and its conjugate base (CO32-) was related to the Henderson-Hasselbalch equation of the pH of the solution. Comparison was made using these findings with the pH of half-equivalence point in Part A (10.20) and the pKa (10.31). Introduction and Aim: (What is a buffer? Why are buffers important in chemical reactions in the lab? Why are they important in the environment or in the body?) A buffer solution contains large amounts of a weak acid and its conjugate base and functioned opposes the change in pH upon the addition of an acid or base. Chemical application across buffers moderates the pH of solutions involved in reaction. While natural buffers in living systems prevent sudden and adverse changes in the either the acidity or alkalinity of the environment habituated by organism. While in the body, buffering is vital to maintaining homeostasis and specifically a stable pH level within the internal environment allows for optimal enzymatic function such as the pH of carbonic acid-bicarbonate buffer in blood. The aim of the experiment was to investigate the pH changes and equilibrium within an acid-base reaction using an acid-base titration and graphical analysis of a titration curve. Additionally, the experiment also examined the buffer properties and behaviour of a weak base and its conjugate acid against distilled water. Experimental: (How did you perform your experiment?) Experiment conducted by Reziah Taylor, refer to First Year Chemistry Laboratory Online Assignment, Experiment E4.

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Results and Questions: PART A: Table 1 ACID – BASE TITRATION Initial burette reading: 20.70  0.05 mL BURETTE READING (mL)  0.05 mL

TOTAL VOLUME OF HCl ADDED (mL)

20.70 0.00 21.90 1.20 22.90 2.20 23.90 3.20 25.00 4.30 26.00 5.30 27.00 6.30 28.00 7.30 29.00 8.30 29.20 8.50 29.70 9.00 29.90 9.20 30.10 9.40 Insert Excel plot here:

pH

11.14 11.02 10.80 10.60 10.43 10.26 10.10 9.93 9.68 9.59 9.52 9.40 9.29

BURETTE READING (mL)  0.05 mL

TOTAL VOLUME OF HCl ADDED (mL)

pH

30.30 30.50 30.70 30.90 31.10 31.30 31.50 31.70 32.70 33.70 34.70 35.70

9.60 9.80 10.00 10.20 10.40 10.60 10.80 11.00 12.00 13.00 14.00 15.00

9.25 9.10 8.96 8.81 8.40 8.09 7.83 7.71 7.42 7.15 7.00 6.80

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Question 1 (from film) a.What is the point in the titration called, when the indicator (phenolphthalein) changes colour? The endpoint.

Indicators can be extracted from plant, e.g. the anthocyanins from red cabbage. Here are the colour changes observed for aqueous red cabbage extract added to various pH solutions:

https://mylespower.co.uk/2012/04/06/homemade-ph-indicator/

b. Would a few drops of red cabbage juice be a good indicator for the titration in Part A? Explain your answer. For the titration in Part A, it requires an indicator with a pH range that has an end-point or a colour change that occurs at the equivalence point to signify when the amount of base is completely neutralised by the acid titrant. Hence, drops of red cabbage juice would not be an appropriate indicator for this section of the experiment as a pH indicator because it changes colour depending on the acidity of the solution, not based on the end point.

Question 2 a. What is the pH of the solution and volume of HCl added at the equivalence point? Equivalence point:

total vol HCl = 10.20 mL

pH = 8.90

b. Determine the volume of HCl added at half the equivalence point (i.e. half the volume of HCl required to reach the equivalence point). What is the pH of the solution at this point (i.e. the half-equivalence point)? Half-equivalence point:

total vol HCl = 5.10 mL

pH = 10.20

Question 3 a. Before and after the half-equivalence point, strong acid has been added in small increments. In this region, how has the pH of the solution responded to addition of strong acid? After the addition of strong acid in small increments, the pH of the solution responded with a slight decrease before and after the half-equivalence point as the solution is still in the buffered region.

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b. Write a definition of a buffer solution. A buffer solution is an aqueous solution that is comprised of a mixture of a weak acid and its conjugate base and is characterised by its resistance to changes in pH when either OH- ions or H+ ions are added. If you added a small increment of strong base (e.g. NaOH) at the half-equivalence point of your titration, what do you think would happen to the pH of the solution? Similarly, as the half-equivalence point is within the buffered region of the titration curve, the slight addition of a strong base would result in only a slight increase in the pH of the solution.

c. Consider the ICE table below showing moles of each species when reacting the base with strong acid. Here we have added half the required amount of HCl: CO32- (aq)

Reaction:

+

H+ (aq)

HCO3- (aq)

→

Initial:

x

x/2

0

Change:

-x/2

-x/2

+x/2

Equilibrium:

x – x/2 = x/2

x/2 – x/2 = 0

x/2

This is the half-equivalence point. What is the relationship between the concentrations of CO32  and HCO3- at the half-equivalence point? At the half-equivalence point, the number of CO32- and HCO3- are equal since half of base in the buffer solution have been neutralised by the HCl, to which half of it was converted to its conjugate acid. PART B: INVESTIGATING THE BEHAVIOUR OF A SOLUTION CONTAINING EQUAL CONCENTRATIONS OF A WEAK BASE AND ITS CONJUGATE ACID Table 2 pH RESULTS FOR STRONG ACID OR BASE ADDED TO VARIOUS SOLUTIONS Initial pH

pH after adding 4 drops 2M HCl

pH after adding 4 drops 2M NaOH

Change in pH value (pH units)

5

Water

7.01

- 4.74

2.27

7.37

12.16

Buffer Solution: Sodium 10.05 9.86 carbonate/ Sodium hydrogen carbonate 10.06 10.22 solution Question 4 a. In the solution containing CO32  and HCO3- , an equilibrium exists:

4.79

-0.19 0.16

HCO3- (aq) + H2O (l) ⇄ CO32  (aq) + H3O+ (aq)

Write the Ka expression:

Ka =

[H3 O+ ][CO2− 3 ] − [HCO3 ]

b. Consider the ionic reaction that occurs when a few drops of strong acid, HCl(aq) are added OR a few drops of strong acid, NaOH(aq) are added. Think about how the concentrations of HCO3- (aq), CO32  (aq) will change and complete the Table 3. Remember that Ka is constant (if temp. stays constant). In Column 5, if [HCO3-] and [CO32-] change in the way you suggest, then how will [H3O+] change for Ka to remain constant? Table 3 Response of HCO3- (aq), CO32  (aq) solution to addition of Strong Acid or Base Add

Reacts with (HCO3- or CO32  )

Change in [HCO3-] (Increase/ decrease)

Change in [CO32-] (Increase/ decrease)

For Ka to remain constant, change in [H3O+] (Increase/decrease)

Change in pH (large/small Increase/decreas Increase/decrease e )

HCl

CO32-

Increase

Decrease

Increase

Small decrease

NaOH

HCO3-

Decrease

Increase

Decrease

Small increase

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Question 5 When significant amounts of the acid and conjugate base are both present, [H3O+] is determined by the equilibrium, maintaining the value of the acidity constant Ka (at room temperature). What do you think will happen if a large excess of strong acid or base (e.g. 5 mL) is added to   the solution containing CO32 /HCO 3 ? Explain your answer. (You can choose one case to explain, i.e. 5 mL strong acid added, or 5 mL strong base added). For the case of adding a large excess of strong acid, the amount of weak base (CO32-) will not be able to compensate the change and will be depleted through neutralisation. Thus, the buffer capacity of the buffer system is exceeded, and the solution fails to perform as a buffer leading to a drastic decrease in pH. PART C: DETERMINING A MATHEMATIC RELATIONSHIP BETWEEN pH AND   RATIO OF [CO32 /HCO 3] Using the solutions and pH measurements shown Buffer System

Sodium hydrogen carbonate (0.1 M NaHCO3) (Acid)

Sodium carbonate (0.1 M Na2CO3) (Conjugate Base)

pH

1

Volume added (mL) 22.5

Concentration (M) 0.09

Volume added (mL) 2.5

Concentration (M) 0.01

9.73

2

17.5

0.07

7.5

0.03

10.03

3 4

12.5 7.5

0.05 0.03

12.5 17.5

0.05 0.07

10.29 10.58

5

2.5

0.01

22.5

0.09

10.92

The following linear relationship was plotted

Fig 6.2 Plot of pH versus log10 ([CO32  ]/[HCO3 ]).

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From the equation of a straight line, y = mx + c, where m is the gradient and c is the yintercept, we can write an expression for the line as: pH = 10.31 + 0.6396log10 ([CO32  ]/[HCO3  ]) Given that the pKa(HCO3 ) = 10.2 ± 0.1, we can write a relationship between the pH of the solution and the concentration ratio of ([CO32  ]/[HCO3 ]) for our limited data set: pH = pKa + 0.6396log10 ([CO32  ]/[HCO3  ])

This general form of this relationship is known as the Henderson-Hasselbalch equation:

where Ka is the acid dissociation constant for a weak acid (like HCO3  ): HA (aq) + H2O (l) ⥫⥬ A- (aq) + H3O+ (aq) Ka measures the extent of the dissociation of the acid, like HCO3 , in water and pKa =  log10(Ka). Question 6 Compare the pKa value found from your graph in this section (Part C) with the pH at the halfequivalence point of your pH titration curve (Part A). They should be similar. Why? (Hint: From the titration in Part A, what was the ratio of [HCO3-] to [CO32-]?) For part A, at the half equivalence point, the concentration ratio of [HCO3-] to [CO32-] is 1:1 because the base (CO32-) is titrated and with half the amount of the acid (HCl) to produce the same concentration of its conjugate acid (HCO3-). As such, it is then calculated that: [CO2− 3 ] )=0 log10 ( [HCO− 3] Using the relationship of the Henderson-Hasselbalch equation, the equation of trendline from the graph from Part C results in a pH that is equal to the y-intercept/pKa. [CO32− ] 𝑝𝐻 = 𝑝𝐾𝑎 + 0.6396 ∙ log10 ( ) [HCO−3 ] 𝑝𝐻 = 𝑝𝐾𝑎 + 0 𝑝𝐻 = 𝑝𝐾𝑎 Thus, the pH of the solution at half equivalence point from Part A is estimated from the graph to be 10.20 and the pH of the solution derived from the equation of Part C is 10.31. Undoubtedly, the pKa value from the graph compared with the pH of the half-equivalence of the titration curve shared a similar pH, supported by the calculations and understanding of Henderson-Hasselbalch equation.

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PART D: BUFFERS IN ACTION Question 7 Rainwater normally has a pH that is slightly acidic due to dissolved CO2 forming HCO3- and H2CO3. High levels of air pollutants like oxides of nitrogen and sulphur, NOx and SO2, which react with H2O in the upper atmosphere, result in acid rain, HNO3 and H2SO4. Acid rain leads to dramatic changes in the pH of water bodies to unsustainable levels for aquatic life. Common pH levels of rivers and water bodies, between 6.5 and 8.5, are determined by the characteristics of the surrounding soil and riverbeds. Rivers running through limestone rock (CaCO3) show less pH change because of acid rain, as observed for the river in the film clip. Explain why the river water in the film required much more acid, compared to the distilled water sample, to be added before a noticeable change in pH. Include relevant reaction equation(s). The limestone in the rivers (CaCO3) dissolves into calcium and carbonate ions. CaCO3 (s) ⇌ Ca2+ (aq) + CO32- (aq) The carbonate ions will react with the H+ ions to form bicarbonate ions. This results in the H+ ions to be consumed as the carbonate ions will react with the hydronium ions, forming weaker carbonic acid and resulting in a slight decrease in pH. CO32- (aq) + H3O+ (aq) ⇌HCO3- (aq) + H2O (l) HCO3- (aq) + H3O+ (aq) → H2CO3- (aq) + H2O (l) As effect, the limestone acts a natural buffer for the river water to resist a dramatic change in pH introduced by acidic rain or the surface water of the river being in equilibrium with the atmosphere carbon dioxide that forms a concentration of carbonic acid (Shapley, 2011). While for the case of the distilled water sample, it contains no ions that would buffer the sample and would expectantly have drastic change in pH even with small additions of an acid or base. As such, the river water can maintain a stable pH and would require a larger amount of acid compared to distilled water to result in a noticeable change in pH. Conclusion: (What have you found out?) Ultimately, the titration curve from Part A was used in determining the equivalence and half-equivalence point of the neutralisation reaction. While the functionality of a buffer solution in Part B (containing weak base – Na2CO3 and its conjugate acid – NaHCO3) was examined. This revealed the slight pH changes accordingly with the addition of a strong acid (HCl) and a strong base (NaOH) to the buffer solution and contrastingly, the distilled water sample showed dramatic changes in pH due to a lack of a buffering system. Combined with the equation of the Henderson-Hasselbalch expression derived from Part C, it was demonstrated that the pH of the half-equivalence point from Part A (10.20) was alike to the pKa/y-intercept of the equation (10.31) due to concentration ratio of [HCO3-]/[CO32-] being the same at that point.

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References: Shapley, P. (2011). Limestone and Acid Rain. University of Illinois. Retrieved from http://butane.chem.uiuc.edu/pshapley/GenChem1/L26/3.html

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