Sample Undergraduate Chemistry Report

Chemistry Experiments Lab Report

The Technique of Volumetric Analysis


To determine the unknown concentration of NaOH using titration technique


A volumetric analysis’s primary goal is to determine the quantity of a particular compound present in a sample, whether liquid or solid. If there are two aqueous solutions used, the concentration of one solution has to be known, and the other solution’s concentration has to be quantified using titration method (Gonzalez, Jimenez and Asuero, 1990). This report involved the determination of the concentration of 20 mL NaOH via titration with 0.1M HCl.

Materials and Methods

The burette was filled up to the 0 marks with 0.1 M HCl. After that, a 20 mL volume of NaOH has dispensed onto a conical flask three drops of indicator added to the solution. The 0.1 M HCl solution was then titrated into the conical flask contents while swirling it constantly until the endpoint was reached. The volume of the endpoint volume of HCl was recorded to two decimal places. The whole procedure was repeated twice times to obtain three replicates.


Standard titration of HCl and NaOH/3
We already know that the volume of NaOH used is a constant 20 mL. Now we need to determine the average volume of the HCl used.
Average Vol. = 20.20 + 19.20 +18.60 = 19.33 mL./3

Place Table Here

For the Replication 1, σ = (20.10-19.33)2 = 0.59

Replication 2, σ = (19.20-19.33)2 = 0.017

Replication 3, σ = (18.60-19.33)2 = 0.53

Determination of the concentration of NaOH

NaOH (aq)+ HCl (aq) ————–> NaCl (aq)+ H2O (l)

The concentration of HCl is 0.1M

No of moles of HCl = Con. X Vol in L

Convert 19.33mL to L—————> 19.33/1000 = 0.019

Moles of HCl = 0.1 x 0.019 = 0.0019 moles = No. of moles of NaOH (1:1 ratio in the reaction equation)

Therefore, the concentration of

NaOH = 0.0019 moles/0.020L = 0.095 M


Titration can be perceived as the method used to determine either the number of moles in a given compound or the unknown concentration of a substance in a particular sample. Consequently, a chemical reaction is normally used to achieve this goal, and the reaction has to be rapid and has a quantifiable endpoint.

The chemical reactions involving strong bases and acids meet these requirements, and acid-base titrations comprise the most fundamental examples of this method.

In this study, the sample is HCl as the acid and NaOH as the base where sodium hydroxide is not known. Thus, the concentration of the acid is 0.1 M, and that of the NaOH is unknown. An indicator is usually used to indicate the point which endpoint is reached. An acid-base indicator is believed to be a weak acid or base different in colour from its salt.

However, at least one of them, either the indicator or it’s salt needs to be strongly coloured. It is to ensure that it so that it can be seen even in strongly dilute solutions.

The solution colour is, therefore, differently based on the basicity or acidity of the solution diluted in. When the acidity of acid changes significantly, a colour change is bound to take place.
In acid-base reaction equation for the given solution can be given;

HA + MOH  H2O + MA

Acid base water salt

The endpoint can be seen as the neutral point. The endpoint about the acid-base titration leads to salt and water production with a pH of 7. The net ionic reaction equation can be written as:

H+ + OH-  H2O

In the above experiment, the indicator that was used is phenolphthalein. If the base was to be used as the titrant, and acid as a solution in the Erlenmeyer flask, the solution would have turned to light pink after the endpoint had been reached (Janos, 2007). However, in this case, since an acid was used as the titrant, and base NaOH, was contained in the Erlenmeyer flask, the solution in the flask turns colourless after the endpoint had been reached. In this particular experiment, the chemical equation can be written as;

HCl + NaOH →NaCl + H2O

In this study, there are some possibilities of errors, for instance, using more than the required acid volume to titrate the NaOH solution. The volume has gone past the endpoint. Therefore the volume used would be much more than required. To address this issue, we have to do the titration slowly while firmly shaking the volumetric flask for nearly 30 seconds or until when the solutions indicate a change to colourless. The other important reasons are utilising volumetric flask that had been used with other solutions. Thus, the newest solution’s concentration will most likely affect the results (Gonzalez, Jimenez and Asuero, 1990). To deal with this issue, we have to ensure that volumetric flask is spotlessly clean and properly dried. It could be the reason our result was not that precise and accurate as it should have been as can be confirmed from the standard deviation. Nevertheless, found out that the distribution of result is ranged significantly.


The concentration and the number of moles of an acid or a base can be quantified using volumetric analysis, particularly the titration process. As long as one of the compounds’ concentration is known, the other substance’s concentration and moles can always be determined. The concentration of NaOH was found to be 0.095 M.


Gonzalez, G. G., Jimenez, A. M. and Asuero, A. G. (1990) ‘Titration errors in acid-base
titrations’, Microchemical Journal, 41(1), pp. 113–120. doi: 10.1016/0026-

Janos, P. (2007) ‘Acid-base titration curves of solid humic acids’, Reactive and Functional
Polymers, 68(1), pp. 242–247. DOI: 10.1016/ARTICLE.

Quantitative Determination of Proteins


This experiment aimed to determine the unknown protein concentrations using the Lowry assay test (Lee et al., 2015). This entails a colourimetric analysis that quantifies the concentration based on light absorption figures detected using a spectrometer. The absorbance of a compound depends on wavelength and concentration and type of the substance.

A standard curve plotted based on the absorption OD is then used to determine unknown proteins’ concentration. In most cases, a series of standards stock solutions are prepared and then the graph of concentration vs OD values is plotted (Janairo et al., 2015).

After determining the linear trend, an equation is generated that can be used to determine the concentration depending on the OD values.

The Lowry Protein Test largely depends on amino acids’ presence, particularly the aromatic ones in the protein. A peptide bond complex is created, followed by improvement of a phosphomolybdate complex with subsequent aromatic amino acids.

Therefore, this essay is often not perfectly linear and bound to be interfered by many compounds. When running the UV analysis, a wavelength of 750 nm is normally, and the sensitivity range is between 20-250 μg of assayed protein.

Materials and Methods


Solution D

Folies Reagent

BSA stock 200 μg/mL

Unknown protein sample.


A series of BSA dilutions from the stock solution was made, and a triplicate of the unknown protein was set. After that, 5 ml of solution D was added to each tube, vortexed and mixed well. The solution was then allowed to stand at room temperature for 10 minutes. Consequently, 0.5 ml of diluted Folin’s reagent was added to all tubes and immediately vortexed.

The samples were then left at room temperature for 30minutes, and the absorbance for each sample read at 750 nm. Finally, a standard curve of the standards was plotted and used to determine the unknown proteins’ concentration in μg/mL.


Dilution table for the preparation of standard curve for BSA using Lowry assay
NB: The final BSA concentration was determined using the formula

C1V1= C2V2

For example for 1st test tube,

C1 = 200 ug/mL, V1 = 0, V2 =1 and C2=?
C2 = C1V1/V2 = 200 x 0/1= 0

Place Table here

The same applies to the rest of the samples/test tubes

Absorbance Values

Place Table Here

We need to calculate the average absorbance values for the three replicates for each of the samples.

The formula for calculating mean is given by; μ = ( Σ Xi ) / N

The standard deviation is given by the formula; σ = sqrt [ Σ ( Xi – μ )2 / N ]

So for;

Test tube 1,

Average abs = 0.013 +0.015 +0.016 / 3 = 0.015

Test tube 2

Average abs = 0.255 +0.248 +0.261 / 3 = 0.255

Test tube 3

Average abs = 0.600 +0.412 +0.423  = 0.478

Test tube 4

Average abs = 0.527 +0.545 +0.549 / 3= 0.540

Test tube 5

Average abs = 0.697 +0.670 +0.706 / 3 = 0.691

Test tube 6

Average abs = 0.255 +0.261 +0.261 / 3 = 0.259

Test tube 7

Average abs = 0.526 +0.504 +0.520 / 3 = 0.517

Standard Curve

Place image here

The concentration of the unknown A and unknown B

From the standard curve,

Y =0.0033x + 0.0684

Where Y is the absorbance and X the concentration

Thus for unknown A;

Y=absorbance = 0.259 = 0.0033x +0.00684
0.0033x= 0.259-0.00684 =0.252

X = 0.252/0.0033 = 76.364 ~ 76 /mL

Therefore the concentration of the unknown A is 76 m/mL

For the unknown B;

Y=absorbance = 0.517 = 0.0033x +0.00684
0.0033x= 0.517-0.00684 =0.510

X = 0.510/0.0033 = 154.594 ~ 155 mg/mL

Therefore the concentration of the unknown B is 155 mg/mL


The purpose of this practical was to quantify the concentration of unknown proteins in a sample by utilising the Lowry Protein Assay techniques. A linear relationship was developed by plotting the standard curve, and the equation y = 0.0033x + 0.0684 was produced.

By fixing the absorbance element as y, the x, the unknown protein concentration, can be computed and quantified. The absorbance figures for the unknowns A and B were 0.259 and 0.517 respectively. The protein concentrations for the A and B were determined to be 76 and 155 /mL, respectively.

Since the R² value was is 0.9587, the calculated protein concentrations may be close to the exact value (Lee et al., 2015). Nevertheless, the concentrations can still be perceived as relatively good predictions. To sum up, a different standard curve with at least a 0.99 R² value needs to be applied to obtain the best results. It will help in obtaining the most accurate protein concentration value.


This experiment aimed to assay the protein concentration of two unknown protein samples through the Lowry Protein test technique. A standard curve was plotted by using the absorbance figures of the standard protein samples. A linear relationship is was developed from the curve where the equation was found as y = 0.033x + 0.0687.


Janairo, G., Linley, M.S., Yap, L., Llanos-Lazaro, N. and Robles, J., 2015. Determination of the the sensitivity range of biuret test for undergraduate biochemistry experiments. Lee, N., Shin, S., Chung, H.J., Kim, D.K., Lim, J.M., Park, H. and Oh, H.J., 2015.

Improved quantification of protein in vaccines containing aluminium hydroxide by a simple modification of the Lowry method. Vaccine, 33(39), pp.5031-5034.

Colourimetric Assay for Paracetamol


Different organic compounds absorb different wavelengths of radiations. This kind of property characteristic of organic substances can be used to pinpoint the substance. The presence of various metal ions, or certain functional groups, for example, inorganic compounds, influences how particular substances absorb the diverse wavelengths of light, therefore causing these compounds to tend towards specific colours (Wilkinson 1976).

The magnitude of absorption is also largely dependent on the concentration of the material if it is in solution form. Quantification of the amount of absorption can be utilised in determining concentrations of strongly dilute solutions (Shihana et al., 2010). Consequently, an absorption spectrometer is used to quantify how the radiation absorbed by a compound differs across the UV spectrum.

In this experiment, the popular medicinal drug paracetamol concentration was examined in a solution whereby it was reacted with nitrous acid before the solution being analysed using UV-spectrophotometer.

Materials and Methods

Precisely 100 mg of paracetamol was weighed and dissolved in about 10 mL of distilled water in a universal flask. The solution was then vortexed until it completely dissolved. It was transferred to a 100 mL volumetric flask and topped up with distilled water to the top and labelled as stock solution A. 1 mL aliquots of each diluted series were prepared in Eppendorf tubes from the stock solution.

100 uL of each of the standards and unknown samples were pipetted into Eppendorf tubes, and 1 mL of 3% trichloroacetic acid added the solution and vortexed. Accordingly, 300 uL of 0.07M sodium nitrite solution was added to each tube, vortexed and left to stand at room temperature for 20 minutes. %0 uLof 8M NaOH solution was then added to each tube, capped and vortexed for 10 minutes.200 uL from each of the tubes was transferred in quadruplicates to the UV spectrophotometer, and the absorbance was read.


Place Table & here

Computation of the mean, the SD and % CV
The formula for determining mean is given by;
Mean = ( Σ Xi ) / N
The SD is given by the formula;
σ = sqrt [ Σ ( Xi – μ )2 / N ]
While %CV is given by;
σ / * 100%
Mean absorbances
Flask 1
´Mean = 0.038 +0.185 +0.039 +0.186 = 0.112
σ =
1 1
1 1 n n
X i i
i i
n n

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   
     
   
  = Sqrt (0.112-0.038)2 +(0.185-0.112)2 +(0.112-

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