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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
Added (ml) Measured pH
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
Errors with additional substances & reading burette.
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).