## Buffers Laboratory Report

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## Buffers Laboratory Report

Aims
The major purpose of experiments is to compare the pH of solutions based on theoretical & practical calculations. Weak acids are known for their buffer properties. The experiment aims are to assess the change of pH after the addition of base and acid to the buffer. Moreover, practical laboratory experiments aim to examine destroy buffer after addition excessive volume of acid. Techniques adopted to proceed with experiments are stoichiometry, back titration, and usage of ICE tables. All buffer reactions on addition alkali or acid-based on Henderson-Hasselbalch equations according to M.Senozan (2001).

Buffers
Buffers are aqueous solutions made from a mixture of a weak acid and its conjugate base. The major properties of buffers are to decrease the impact of adding acid or alkali on the final pH solution. Therefore, buffers act a role neutralizing agent and significantly stabilized the pH of the solution. The addition of an excessive amount of base or acid destabilizes the buffer solution. PH of the solution is strictly associated with H+ concentration in solution. Conjugate acid & base ratio in solution depends on pKa & pKb values.

Equations
Molarity(M) =
ACN % error + x 100% Ka =
HA(aq) H+(aq) + A- (aq) Ka = 10 -pka pH = pKa +log10 []
pH = pKa +log10 []

Table 0.0.1 Reagents list Concentration [M] pKa values Initial value measured by pH probe
Acetic acid / CH3COOH 0,10 4,75 3,31
Sodium acetate / CH3COONa 0,10 7,45
Hydrochloric acid /HCl 0,10
Sodium hydroxide / NaOH 0,10
Deionised water / H2O
Tap water / H2O

Experiment no. 1

Table no. 1
Results experiment no. 1

Buffer Volume of CH3COOH (mL) 0,10 M Volume of CH3COONa (mL) 0,10 M First measured pH Second Measured pH Concordant
pH result Theoretical pH
A 10,0 10,0 4,90 4,91 4,91 4,75
B 15,0 5,0 4,41 4,42 4,42 4,27
C 18,0 2,0 3,97 3,98 3,98 3,79
D 2,0 18,0 5,84 5,85 5,85 5,70
E 5,0 15,0 5,08 5,09 5,09 5,22

Table no. 1.2
Theoretical calculations

Theoretical pH calculation for Buffer Buffer A.
Moles(CH3COOH) = 0,10M * 0,01dm3 = 0,001 moles
Moles(CH3COONa) = 0,10M * 0,01dm3 = 0,001 moles

pKa (CH3COOH) = according to Reusch (2021)

`` =``

=

pH= 4,75
Theoretical pH calculation for Buffer B.
Moles(CH3COOH) = 0,10M * 0,015dm3 = 0,0015 moles
Moles(CH3COONa) = 0,10M * 0,005dm3 = 0,0005 moles

``Theoretical pH calculation for Buffer C.``

Moles(CH3COOH) = 0,10M * 0,018dm3 = 0,0018moles
Moles(CH3COONa) = 0,10M * 0,002dm3 = 0,0002 moles

pH(Buffer C) = 3,79 Theoretical pH calculation for Buffer D.
Moles(CH3COOH) = 0,10M * 0,002dm3 = 0,0002moles
Moles(CH3COONa) = 0,10M * 0,018dm3 = 0,0018moles

pH(Buffer D) = 5,70
Theoretical pH calculation for Buffer E.
Moles(CH3COOH) = 0,10M * 0,005dm3 = 0,0005moles
Moles(CH3COONa) = 0,10M * 0,015dm3 = 0,0015moles

pH(Buffer E) = 5,22

Experiment no. 2

Table no. 2
Results theory vs practice.
Solution Initial measured PH Ph after addition of 2 ml HCl pH after addition of 2ml NaOH
Measured Predicted Measured Predicted
Buffer A 4,88 4,18 4,38 5,28 5,12
Tap water 7,70 2,45 2,67 11,73 11,96

Theoretical callculations
Moles(CH3COOH) = 0,10M * 0,01dm3 = 0,001 moles
Solution was split into two equal tubes.
0,001 moles / 2 = 0,0005 moles
Moles(CH3COONa) = 0,10M * 0,01dm3 = 0,001 moles
Solution was split into two equal tubes.
0,001 moles / 2 = 0,0005 moles
Moles(HCl) = 0,10M * 0,002 dm3 = 0,0002 moles

Reactions
CH3COO- + H+ → CH3COOH
ICE table & calculation after addition 2ml of 0,10 M HCl. Theoretical calculation.
Moles(HCl) = 0,10M * 0,002 dm3 = 0,0002 moles
Solution was split into two equal tubes.
0,001 moles / 2 = 0,0005 moles

Ice table after addition 2ml of 0,10M HCl. Theoretical calculation. CH3COOH
moles CH3COO-
moles Moles of H+
moles
Initial 0,0005 0,0005 0,0002
Change +0,0002 -0,0002 -0,0002
Equilibrium +0,0007 +0,0003 0

pKa(CH3COOH) = 4,75
Theoretical pH calculation for Buffer A (from experiment no. 1) pH= 4,38

Theoretical calculations. ICE table & calculation
After addition 2ml of 0,10 M NaOH to solution of 10ml tap water and 10ml of mixture in ratio 1:1 CH3COOH & CH3COONa both 0.1 M.

Moles(NaOH) = 0,10M * 0,002 dm3 = 0,0002 moles

Calculation for addition 10 ml tap water and 2 ml of NaOH.
[OH-] = [
[OH-] = 0,0090 M
pH=-log10[OH-]
ph=-log10[0,0090]
pH=2,04
PH=14 – 2,04
pH=11,96

Ice table & calculation
after addition 2ml of 0,10 M HCl to solution of 10ml tap water and 10ml of mixture in ratio 1:1 CH3COOH & CH3COONa both 0.1 M.

ICE table after addition 2ml of 0,10M NaOH. Theoretical calculations. Moles of
CH3COOH Moles of
CH3COO- Moles of
H+
Initial 0,0005 0,0005 0,0002
Change +0,0002 -0,0002 -0,0002
Equilibrium +0,0007 +0,0003 0

Calculation for addition 10 ml tap water and 2 ml of HCl to 10ml of water solution. Total volume equal 12 ml
[H+] = [
[H+] = 0,01666666666666 M
pH=-log10[H-]
ph=-log10[1,778151]
pH=1,78
PH=14 – 1,78
pH=11,96

Ice table after addition 2ml of 0,10M HCl to 10 ml of tap water. Theoretical calculation.
n(H2O)= 0,01 * 55,5 = 0,555 mol

ICE table after addition 2ml of 0,10M HCl to 10 ml of tap water. Theoretical calculation. HO- Moles of H+
moles
Initial 0,555 0,0002
Change -0,0002 +0,0002
Equilibrium +0,553 +0,002

pH=-log[H+]
pH=-log[0,002]
pH=2,67
[H+] = [
[H+] = 0,01666666666666 M
pH=-log10[H-]
ph=-log10[1,778151]
pH=1,78
PH=14 – 1,78
pH=11,96

Conclusion experiment no 2.
Buffer of weak acid and it conjugant base is solution effectively resist the changes of pH after addition of acidic or basic components. Buffer solution after addition of acid or alkali react with add component and minimalize the overall change in pH. Total change in pH is noticeable but not as significant as in solution without buffer. The experiment reveals that water after addition of base or acid capture the pH of addition. In comparison to buffer solution, it must to be taken to account that addition of base or acid do not change the magnitude of pH dramatically. Water in this case reveal the solvent properties.

Experiment no. 3

Table 3: Effect of dilution of weak acid and weak base

Dilution
Factor pH of 0,10M ethanoic acid pH of 0,10M sodium acetate
Measured Predicted Measured Predicted
1 2,70 2,87 7,82 8,87
10 3,16 3,37 6,98 8,38
100 3,49 3,87 6,60 7,88

Dilution Factor 1
ethanoic acid

pKa(CH3COOH)= 4,75
Ka = 10-pka
Ka = 10-4,75
Ka = 0,0000177828
[H+]=√Ka[HA]
[H+]=√0,0000177828*0,1
[H+]=0,00133335216533

pH = -log10(0,00133335216533)
PH=2,87 Dilution Factor 10
ethanoic acid

pKa(CH3COOH)= 4,75
Ka = 10-pka
Ka = 10-4,75
Ka = 0,0000177828
[H+]=√Ka[HA]
[H+]=√0,0000177828*
[H+]=√0,0000177828*0,01

[H+]=0,0004216965
pH = -log10(0,0004216965)
pH=3,37 Dilution Factor 100
ethanoic acid

pKa(CH3COOH)= 4,75
Ka = 10-pka
Ka = 10-4,75
Ka = 0,0000177828
[H+]=√Ka[HA]
[H+]=√0,0000177828*
[H+]=√0,0000177828*0,001

[H+]=0,0001333521

pH = -log10(0,0001333521)
pH=3,87

For dilution 1
sodium acetate

pKa (CH3COOH) = 4,75
pKa + pKb = 14
pKb = 14 – pKa
pKb = 14- 4,75
pKb = 9,25
Kb= 10-pKa
Kb= 0,000000000562341325
[OH-] =√Kb[B]
[OH-] =√0,000000000562341325*0,10
[OH-] =√0,0000000000562341
[OH-] = 0,000007498939925083
pOH= -log10(0,000007498939925083)
pOH= 5.1250001256
PH= 14 – 5.1250001256
pH= 8,87
For dilution 10
sodium acetate

[OH-] =√Kb[B]
[OH-] =√Kb []

[OH-] =√0,000000000562341325*0,01
[OH-] =0,000002371373705260310000000000

pOH = -log10(0,000002371373705260310000000000)
pOH = 5,62
pH= pOH + pH
14= 5,62 + pH
pH = 14 – 5,62
pH = 8,38
For dilution 100
sodium acetate

[OH-] =√Kb[B]
[OH-] =√Kb []

[OH-] =√0,000000000562341325*0,001
[OH-] =0,000002371373705260310000000000

pOH =-log10(0,000000749894209205539000000000)
pOH = 6,12

pH= pOH + pH
14= 6,12 + pH
pH = 14 – 6,12
pH = 7,88

Conclusion Experiment no. 3
There is significant difference between theoretical and practical results of pH. The trend with pH deviation rise with dilution factor. According to Senozan (2001) to obtain correct calculation of concentration [H+] we should conduct additional calculation for hydrolysis of A- and ionization of water. Calculation in this report do not take in account cases associated with ionization. However increasing pH differences after water addition indicate that those calculation are vital important for further research. Other causes of differentiation in acquired results are provided in section errors.
Cite Senozan 2001 “As we will show, the discrepancy between the exact and approximate calculation, even with moderate concentrations and PH values not far from the pKa, can be as much as 50%. (When Ka=10^-3 and the acid and base are 0,01M), and many buffer problems solved through Henderson-Hasselbach equation with the usual interpretation of [HA] and [A-] as the initial molarities – do not warrant an answer with more than the single significant figure.” This quote perfectly illustrate the deviation between theoretical and practical experiment.

Experiment no. 4

Table no. 4 Results
Based on data obtain from Mohammed Jenis DMU student.
Volume of NaOH added (ml) Measured pH Volume of NaOH
Added (ml) Measured pH Volume of NaOH
Added (ml) Measured pH Volume of NaOH
0 2,78 12 4,71 24 12,30 36 12,81
2 3,48 14 4,96 26 12,46 38 12,83
4 3,85 16 5,22 28 12,54 40 12,85
6 4,13 18 5,75 30 12,62 42 12,88
8 4,32 20 11,22 32 12,69 44 12,90
10 4,52 22 12,02 34 12,79 46 12,93

Titration curve is represented on Chart no 1.
Ph measured before addition of base equal 2,78. After addition of 10ml NaOH pH increased to 4,52. Addition additional 2 ml of NaOH reach to half equivalence point.It is situation when pH of solution equal to Pka of weak acetic acid. Proportion of conjugate base and conjugate acid are equal. It mean that after addition of total 12ml NaOH the buffer solution was established. The equivalence point should be reach after addition additional 24ml of NaOH based on practical experiment. After equivalence point after addition 46ml of NaOH. Base determine the pH condition.
Conclusion of Experiment no. 4
Compare to theoretical calculations half equivalence point should be reach after addition of 10ml of NaOH and equivalence point after addition of 20ml of NaOH. Theoretical and practical results differ by 20%. Other causes of differentiation in acquired results are provided in section errors.

Errors
Calibration of pH probe.
Error with reagents concentrations.
Errors associated with cross contamination with other reagents.
CO2 interaction with reagents.

Each experiment is prone to errors. Some of them are unavoidable. The most obvious are human errors associated with the addition of an inaccurate quantity of reagents. Reading the burette scale & record results with wrong coefficients are two majors of laboratory errors. Moreover, there are issues with an incorrect concentration of substances on reagents. Significant impact on provided results has pH device. Not calibrated pH devices provide fraudulent results. After and before each calculation probe device should be flushed with deionized water. Unclean laboratory equipment influences practical pH results.
All experiments was conducted in open space. Consideration of CO2 reacting with reagents from air must to be taken to account. Even small quantity of CO2 can significantly influence pH according to Senozan 2001. Therefore all conducted experiments are prone to CO2 from air.

pH device errors & concentration
Not calibrated probe device may be the main cause of observed differences. Based on both practical and theoretical calculations from Experiment no. 1 it can be deduced that the average pH difference between theory and practice equal reach 0,104 of pH. Thus the device was not calibrated before conducting experiments it is not possible to judge that the pH probe is the cause of the difference. Concentration can be accused of prone to results. At this stage of the experiment without additional equipment, it is not possible to indicate on the ingredient of difference. The use of litmus papers to assess the pH coefficient in the solution of individual reagents and comparing them with the calibration results of the pH meter would clearly indicate the source of the problem. Adopting to procedure calibration of the probe & check pH with an additional method will increase the precision and will reveal issues associated with the wrong concentration of reagents. However influence of CO2 in air can have more significant influence on results. This lab report do not examine this critical factor.

Recommendations
The main recommendation to improve the experiment is to ensure the concentration of reagents before any experiment will be conducted. Deviation will influence results. Usage of litmus papers to test reagents is wise advice before we start to use pH devices acordnig to Jennings et al. (December 2010). Calibration of pH probe with standardized solutions and comparison of results will offer significantly higher correctness and may help to calculate % error with higher precision. Usage of that two methods will provide a self-check of reagents concentration and measuring devices. Experiments conducted in close systems without access CO2 from air should provide more accurate results.

• References:
• William Reusch (2021) Virtual Textbook of Organic Chemistry, Available at: www2.chemistry.msu.eud/faculty/reusch/virttxjml/acidity2.htm (Accessed: 01.05.2021).
• Mohammed Jenis, DMU student, Acquired practical data to Experiment no. 4
• Henry N. & M.Senozan (November 2001) ‘The Henderson-Hasselbach Equation: Its History and Limitation’, Journal of Chemical Education, 78(11), pp.
• A Jennings et al. (December 2010) Titration and pH Measurement, Available at: https://www.researchgate.net/publication/229829145_Titration_and_pH_Measurement (Accessed: 06.05.2021).

## Introduction:

The main domain of this lab report is to introduce experimental procedures in chemistry while conceptualizing the extraction techniques, volumetric analysis and wet chemistry and introducing different equipment being used frequently in chemistry. This report therefore is based on results of experiment to identify the acid neutralizing capacity of antacid tablet and identify the number of moles of HCL neutralized by the antacid tablet and NaOH. The report also identified the number of tablets required to bring the HCL in the stomach to a more normal level and identify possible sources of error in this experiment.

Aim:

This experiment is aimed to identify the acid neutralizing capacity of antacid tablet and identify the real neutralizing power of tablet. The experiment was also aimed to identify the number of tablets required to neutralize the HCL in the stomach to more normal level. The acid neutralizing capacity of antacid tablet is examined by converting it into fine powder, adding HCL and add methyl orange indicator to note the change is color. NaOH to an accurately known volume was added and noted the color change at the end-point. The neutralizing capacity of antacid tablet is measured through back titration technique.
Results Presentation:

• Items Tablet 1 (Bisodol) Tablet 2 (Renee)
• Mass of tablet 3.6 3.84
• Mass of Crushed Tablet 1.28 1.19
• Concentration of HC (mol dm-3): 0.5 0.5
• Initial Volume 0.2 0.5
• Volume in the End 39.6 39.4
• Concentration of NaOH solution (mol dm-3) : 0.5 0.5

Calculations:

The amount (number of moles) of acid neutralised by the antacid can be determined by the total number of moles of HCl added minus the number of moles that were neutralized by the NaOH. Calculations based on findings of experiments are summarized as under.
Moles of HCL Added (nHCL) = CHCL * VHCL in ml
1,000
Bisodol (Tablet 1) = 0.5 * (39.4 / 1,000) in ml
= 0.0197 ml
Renee (Tablet 2) = 0.5 * (38.9 / 1,000) in ml
= 0.01945 ml
Moles of NaOH required for Back-Titration (nNaOH) = CNaOH * VNaOH in ml
1,000
Bisodol (Tablet 1) = 0.5 * (16.2 / 1,000) in ml
= 0.0081 in ml
Renee (Tablet 2) = 0.5 * (12.8 / 1,000) in ml
= 0.0064 in ml
Moles of Acid Neutralized = nHCL – nNaOH
Bisodol (Tablet 1) = 0.0197 – 0.0081
= 0.0116 ml
Renee (Tablet 2) = 0.01945 – 0.0064
= 0.01305 ml

Calculate the number of moles of NaOH used (C=n/v)
Number of moles of NaOH (Bisodol) = Concentration in mol/dm * Volume in dm3
= 0.5 * 16.2 (1/1000)
= 0.5 * 0.0162
= 0.0081 moles of NaOH
Number of moles of NaOH (Renee) = Concentration in mol/dm * Volume in dm3
= 0.5 * 12.8 (1/1000)
= 0.5 * 0.0128
= 0.0064 moles of NaOH
Calculate the number of moles of excess HCL added before titration
Moles of HCL added before titration (Bisodol) = Concentration in mol/dm * Volume in dm3
= 0.5 * 39.4 (1/1000)
= 0.5 * 0.0394
= 0.0197 moles of HCL
Moles of HCL added before titration (Renee) = Concentration in mol/dm * Volume in dm3
= 0.5 * 38.9 (1/1000)
= 0.5 * 0.0389
= 0.0195 moles of HCL
Calculate the number of moles of HCl that was neutralised by the antacid
Moles neutralized by Antacid = Number of moles of excess acid (#2) – number of moles HCl neutralized by NaOH (#1)
Bisodol = 0.0197 – 0.0081
= 0.0116 moles
Renee = 0.0195 – 0.0064
= 0.0131 moles
Calculate the number of moles of HCl neutralised by 1 antacid tablet
Number of Moles neutralized by 1 antacid tablet are same calculated in required 3 above as 1 tablet was used and requirement 3 calculated the moles neutralized by antacid.
Therefore:
Number of moles of HCI neutralised by 1 Bisodol tablet = 0.0116 moles
Number of moles of HCI neutralised by 1 Renee tablet = 0.0131 moles
How many tablets of the antacid will be required to bring the concentration of HCl in the stomach to a more normal level?
HCl concentration in Hyper Acidic Stomach = 0.030M
Volume of Liquid in Stomach = 250 mL or
= 0.25 L
Number of Moles of HCl in hyper acidic stomach = Volume * HCl Concentration (molarity)
= 0.250 * 0.030
= 0.075 moles
Number of moles nuteralized by 1 Bisodol tab. = 0.0116
Number of tablets required = 0.075 / 0.0116
= 0.65 tablets
Number of moles nuteralized by 1 Renee tab. = 0.0131
Number of tablets required = 0.075 / 0.0131
= 0.57 tablets

(Note: The HCl concentration in a hyper acidic stomach is 0.030 M. The volume of liquid in the stomach is 250 mL)

## Laboratory Report

Introduction

Detecting proteinuria in urine is an important biomarker in medicine. Increased levels of proteins may indicate kidney disease. Two samples of urine were delivered marked as ‘patient’ & ‘healthy. The patient complained about fever and tiredness after endurance, and stated that blood was found in the urine.

Aim

To assess the concentration of proteins, to provide scientific reasoning based on data obtained. Comparison of samples by double-tailed T-Test to reveal the significance of results. To assess the patient’s case.

Hypothesis

It is expected to assess the concentration of proteins in both samples. To assess the significance of the results.

Materials and methods

A quantitative Bradford assay was used for the estimation of proteins. Thus, a dye known as Coomassie blue provides a colorimetric time-tested change from red to blue when it binds to proteins under acidic conditions. Spectrometry was used to assess absorbance. Known concentration of Bovine Serum Albumin (BSA) was used as a reference point to assess unknown protein concentrations in both samples. Standard dilutions of Bradford and BSA were prepared.Absorbing readings were normalized with blank. Spectrometer set to 595 nm wavelength. All spectrometry observation was triplicate repeated, all results were recorded. Incubation time at room temperature 5min.

Results

Concentrations were deduced based on the linear equation after setting up the intercept. y=mx where y=absorbance, x=concentration, m=slop. According to Chart No. 1 calibration, standard curve was plotted based on provided concentration of BSA. Unknow protein concentrations were deduced from rearranged linear equations displayed in Chart No. 1. Results with standard deviation are presented in Table No.1 T-test was used as a statistical tool. Significance was marked on Chart No. 2. Urine protein levels have significance resulted for P. Each group n=3 two-tailed t-test. Obtain data are presented as mean ± standard deviation. t(4)=(-34.87), P<0.00001. Therefore, P****.

Chart No. 1

Discussion and conclusion

All objectives of the analysis were reached. According to Aitekenov, (2020), used methodology is similar. Two points on the calibration curve do not match the trend line. There is a probability that the researcher used incorrect volume in those two calibration BSA samples. Comparison of healthy sample versus patient sample reveals that level of protein is abnormal to patient. Statistical verification by a two-tailed t-test revealed the significance factor of obtaining data. Blood in urine and other patient symptoms lead to the conclusion of glomerulonephritis. Additional screenings should proceed to find the cause of the patient’s condition. Accuracy & efficiency may be improved with the adoption of different screening methods according to Schleicher, (1978).

References

• Aitekenov, S. et al. (2020), Detection and quantification of proteins in human urine. Talanta, p.121718. [online] Available at: https://www.sciencedirect.com/science/article/pii/S0039914020310092?via%3Dihub [Accessed 29 Nov. 2021].
• Schleicher, E. and Wieland, O.H. (1978), Evaluation of the Bradford method for protein determination in body fluids. Journal of Clinical Chemistry and Clinical Biochemistry. Zeitschrift Fur Klinische Chemie Und Klinische Biochemie, [online] 16(9), pp.533–534. Available at: https://pubmed.ncbi.nlm.nih.gov/712344/ [Accessed 29 Nov. 2021].