To show when neutralization has been reached, an indicator is used. A suitable indicator is phenolphthalein. This indicator is colourless in acid and pink in an alkaline solution. The end-point of this reaction is determined when the colour changes from pink to colourless.
DIAGRAM
HYPOTHESIS
Neutralization reaction will occur when the acid reacts with the base. The chemical equation shown below determines the reaction and the molar ratio.
H2SO4(aq) + 2NAOH(aq) → Na2SO4(aq) + 2H2O(l)
1 : 2
The above reaction shows that the molar ratio between sulphuric acid and sodium hydroxide is 1:2. Hence the number of moles of sodium hydroxide reacting will be twice that of sulphuric acid.
If the acid is added drop wise to the base with phenolphthalein solution then, when it reaches the end point the colour will change from pink to colourless.
DEPENDENT VARIABLE
The dependent variable in this experiment is the volume of sulphuric acid from the burette that was used to neutralize the base.
CONTROLLED VARIABLES
-
The volume of sodium hydroxide in the conical flask was kept constant at a volume of 25cm3
-
The molarity of sodium hydroxide used was kept constant at a value of 0.1mol/dm3.
- The number of drops of phenolphthalein indicator that was added in the conical flask was kept constant at 3-4 drops.
APPARATUSES AND CHEMICALS USED
- 2 beakers of 250ml each
- 1 pipette of 25ml with pump
- 1 burette of 50ml
- 1 metal retort stand
- 1 white tile
- 1 conical flask of 250ml
- 0.1M of Sodium Hydroxide ( NaOH )
-
Unknown molarity of Sulphuric acid ( H2SO4 )
- Phenolphthalein indicator
METHOD
- First, I rinsed out the burette with a little distilled water followed by a little distilled water followed by a little dilute sulphuric acid. Then I filled the burette with dilute sulphuric acid and recorded the initial burette reading in a suitable results table.
-
Then I rinsed out the pipette with some 0.1M sodium hydroxide and then carefully transferred 25cm3 of the NaOH solution to a conical flask
- Then I added 3-4 drops of phenolphthalein solution to the flask. Shake the flask and stand it on a white tile under the burette.
- Then I added a few drops of dilute sulphuric acid from the burette to the flask while swirling the contents of the flask at the same time. I continued this process until the solution in the flask just turned colourless. I recorded the reading on the burette which showed me the approximate end-point. I repeated the experiment four times in order to get an average which is a more accurate reading.
- After I obtained my reading I emptied the burette and other apparatuses and rinsed them thoroughly with distilled water.
RESULTS
DATA COLLECTION
DATA PROCESSING AND PRESENTATION
Chemical equation:
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
Molar Ratio: 1 : 2
For every 1 mole of sulphuric acid it will react with 2 moles of sodium hydroxide.
Mean Volume of H2SO4 = (23.8 + 24 + 23.8 + 24) = 23.9cm3
4
No. of moles of NaOH = Concentration (mol/dm3) x Volume (dm3)
= 0.1 M x 0.025 dm3 = 0.0025 moles
No. of moles of Sulphuric acid = 0.5 x No. of moles of NaOH
neutralized = 0.5 x 0.0025 = 0.00125 moles
Concentration of Sulphuric acid = No. of moles of H2SO4
Volume of H2SO4 (dm3)
= 0.00125
0.0239
= 0.052M
CONCLUSION
After conducting the above the experiment carefully and systematically I conclude that the concentration of sulphuric acid that was used in the experiment is 0.05mol/dm3.
I also conclude that from the chemical equation the molar ratio of sulphuric acid to sodium hydroxide is 1: 2. .
EVALUATION
The sources of error that was present when doing the experiment was parallax error. The measurement of the liquid was supposed to be read from the lower meniscus.
The lab should have burettes with a better stopper so as to allow different amount of volumes to pass through. There should be better pumps for sucking the liquid through the pipette.
The method used here in order to perform the titration gave me the concentration of sulphuric acid to be 0.05mol/dm3, hence the method used was effective.
