Volumetric Analysis is a quantitative method and which deals with the volume of a particular substance used in the analysis. However in volumetric analysis most important factor is the volume. Here the volumetric analysis is proceed with known amount of solutions. But, concentrations of both the solution has not been mentioned since it is done to find out the concentration of known acid or a base. When adding known concentration solution to unknown concentration solution drop wise will result a color change at a point and this procedure called Titration. In volumetric analysis there are 5 main types as acid base titration, redox titration, precipitation titration, complexometric titration and iodometric titration (Datta and Johar, 2014).
Titration is the process of observing the color change of mixed solution when known concentration solution is added to unknown concentration solution.
Titrant is the solution in volumetric analysis whose concentration is known.
Titrand is the solution in volumetric analysis whose concentration is to be found. (Solution with unknown concentration).
Indicator is the substance used to observe the color change of the resulting solution.
Equivalence point is the point which will show the color change of the resulting solution for the first time when adding solutions from burette drop wise.
End point is the point at which the color change will not be disappeared and stay permanently.
Phenolphthalein is used because it is slightly acidic and it can detect the pH range of 8.3 to 10.0 resulting color change colorless to pink. Phenolphthalein is insoluble in water where it is soluble is alcohols. (Matthews, 2003).
Color change of Indicators (Phenolphthalein)
(Pradeep and Dave, 2013)
Preparation of 0.1moldm-3 of NaOH
Took 10.0ml of NaOH (1moldm-3) to the graduated pipette using pipette filler and introduced it into the volumetric flask whose volume is 100ml. Then diluted the solution up to 100ml by adding distilled water until the lower meniscus of the diluted NaOH solution to tally with the ring mark present on the volumetric flask.
Acid Base Titration is carried out to find the concentration of an acid or a base by neutralizing the acid or the base of which the concentration is unknown.
The burette should be filled with the Base (known concentration solution) and the titration flask should contain the acid which the concentration is to be determined. Then three drops of indicator, phenolphthalein should be added to the titration flask, because the phenolphthalein will leads towards the color change in the resulting solution.
In the titration above, the acid which has the unknown concentration is HCl and the base which has the known concentration is NaOH. When NaOH added with HCl, it results NaCl with H2O.
Sodium hydroxide + Hydrochloric acid Sodium Chloride + Water
NaOH + HCl NaCl + H2O
After setting up the equipment, the titration is carried out by allowing base from burette to fall into the acid drop wise and should observed the resulting solution very carefully until it occurs the color change. At some point, when the NaOH drops falls into the HCl solution will result pink color due to phenolphthalein and that pink color is not permanent since it disappears. When observing the pink color for the first time, it will be the equivalence point.
Then the test should be carried out so on until pale pink color will remain without disappearing and it is called the endpoint of the titration. Once endpoint is reached, tap of the burette should close soon and get stop the titration. Then tabulate the reading by repeating the experiment at least three times to get three readings. Finally calculate the concentration of the unknown solution by putting the reading into the basic equations of chemistry.
The objective of this practical is to find out the unknown concentration of an acid or a base by Volumetric Analysis which is a quantitative analytical method.
First the graduated pipette was washed with a small amount of Sodium Hydroxide (NaOH). Then the filled sodium hydroxide was discarded and 10.0 ml of NaOH was filled to the graduated pipette using pipette filler and transferred to a volumetric flask. A little amount of distilled water was poured into the volumetric flask and swirled. Then more distilled water was added up to the volumetric flask till it rises up to the ring mark and lower meniscus balanced.
Cleaned the burette with small amount of NaOH. After that the burette was fixed vertically to the burette holder and the burette was filled up to 0.00ml mark. Then the graduated pipette was washed with some (Hydrochloric) HCl acid to clean up. After cleaning the graduated pipette by HCl acid, it was filled with HCl acid whose concentration is to be found, exactly up to 10.0 ml with the help of pipette filler.
Then 10.0ml of HCl acid was introduced to the titration flask and three drops of phenolphthalein was directly put into HCl acid in the titration flask without let it touching the glass of the titration flask. After that, placed the titration flask under the burette which was fixed to the holder.
The tap of the burette was slowly opened to release the NaOH drop wise into the titration flask, then at one place, the NaOH drops when fall into the HCl acid will show some pink color dot for the very first time, and that point will call as the equivalence point of the titration. Then the tap was adjusted slowly to reduce the rate at which it release NaOH drops.
Few seconds later at a point the pale pink color can be seen which is permanent, and that point is called the end point. Then closed the tap of the burette and the volume of NaOH which has been used to neutralize the acid is recorded. After recording the first reading continued the practical two more times to get another two readings. Results were tabulated.
The experiment is carried out for three times to get three readings and reading taken at each test has been tabulated below.
Test Number |
Initial volume/cm3 |
Final volume/cm3 |
Volume used/cm3 |
1 |
0.00 |
32.10 |
32.10 |
2 |
0.00 |
34.00 |
34.00 |
3 |
0.00 |
35.20 |
35.20 |
Average volume = 32.10cm3 + 34.00cm3 + 35.20cm3 3
= 33.76 cm3
Titration is the process of determining the concentration of unknown solution when two solutions are put into to neutralize. A chemical reaction has been used for this purpose and the reaction should be fast and should be a complete reaction. The titration should have a determinable equivalence point and an endpoint.
We used HCl and diluted NaOH (0.1moldm-3) in this titration process and the indicator used is phenolphthalein which can cover a range of 8.3 to 10.0 pH value. Here, the indicator is acting like a signal showing where to stop the titration at endpoint.
When NaOH added with HCl, it results NaCl with H2O.
Sodium hydroxide + Hydrochloric acid Sodium Chloride + Water
NaOH + HCl NaCl + H2O
Actually the endpoint is the neutral point where acid and base get neutralized. For this reaction, Ionic equation is:
H+ + OH– H2O
Here at the endpoint in first titration, the whole resulting solution turned into pale pink color, but other two reading were too far apart due to some errors.
The errors which may lead to deviate the readings may be:
Sometimes it can be due to the usage of contaminated solutions because taking the pipette for the next test without washing. It can be due to the way which cleaned the pipette after using once.
Sometimes the titration results can alter due to human errors such as difficulty with reading the burette value or sometimes hand glows can have some drops of some other solutions and get dropped to the test which is going on.
As the first reading 32.10cm3 was a successful effort, but still the test 02 (34.00cm3) and test 03 (35.20cm3) was unsuccessful due to those errors. Sometimes the color change was so rapid change and unable to stop the tap of the burette on time.
The concentration of the unknown was determined by using titration process and by the known value for the other solution. Calculation should be carried out in order to get the final answer to determine the unknown concentration of HCl.
Calculation:
Average volume = 32.10cm3 + 34.00cm3 + 35.20cm3 3
= 33.76cm3
Assumption: Number of moles remain unchanged
Therefore by using, C1V1 = C2V2
0.100moldm-3 x 0.100dm-3 = C2 x 0.033dm-3
C2 = 0.100moldm-3 x 0.100dm-3 0.033dm-3
C2 = 0.30moldm-3
THE UNKNOWN CONCENTRATION OF HCl IS 0.30moldm-3
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