Methyl Orange

Topics: Titration, Chemistry, Acid-base titration Pages: 5 (1152 words) Published: February 23, 2008
In this experiment the initial mass of the sodium carbonate used was 2.69g. In each titration, 3 drops of methyl orange was added to the sodium carbonate solution. With this information the titration can begin, and the results obtained are shown below:

Titration readings

Titre (cm3)4.3017.7016.6016.6017.6016.30

Therefore, the average titre would be calculated as follows; all the titre measurements added together (not including the rough titre), and divide this amount by the total number of titrations taken. In this case, I am only going to find the average titre of 3 of the results because to have successive measurements, the titrations have to be within 0.1 of each other, and in this case it was for only 3 of the results shown above. The average titre is shown below:

16.60 + 16.60 + 16.30 = 49.50 / 3 = 16.5cm3

Going back to the beginning the equation for the reaction that is taking place is as follows:

Na2CO3 (aq) + H2SO4 (aq)  Na2SO4 (aq) + CO2 (g) + H2O (l)

Concentration of sodium carbonate solution

In this experiment, a solution carbonate had to be made. This was done by mixing sodium carbonate with distilled water, to get the desired solution. The mass of the sodium carbonate used here was 2.69g. In order to find out the concentration of sodium carbonate used, the number of moles used need to be worked out first.

1.The equation for finding the number of moles is as follows: •Amount of Na2CO3 (mol) = mass (g) / molar mass (g mol-1)
•The mass is 2.69g and the molar mass of Na2CO3 is [(26 x 2) + 12 + (16 x 3)] •= 106 g mol-1
•Therefore, 2.69 / 106 = 0.0253mol

We can now use this number of moles to find the concentration of sodium carbonate used in solution.

2.The equation for finding the concentration of a solution is as follows: •Concentration (moldm-3) = amount of solute (mol) / volume of solution (dm-3) •In this reaction, the volume of solution used was 250cm3 and the amount of solute used was 0.0253mol (as we found out earlier). •As you can see from the equation, the volume is measured in dm3, therefore we have to convert 250cm3 into dm3. This is done by 250 / 1000, because there are 1000cm3 in one dm3. •So, 250 / 1000 = 0.25dm3

•Concentration = 0.0253 / 0.25 = 0.1012moldm-3
•Therefore the concentration of sodium carbonate used in this experiment is 0.1012moldm-3

As we have found out the concentration of the sodium carbonate solution, we can now go on to finding the concentration of the acid used in the experiment.

Concentration of sulphuric acid

Follow the same procedures as above.

Due to this being an acid base titration, we are trying to find out how much acid is used. In this case the number of moles of acid used is the same as the number of moles in the sodium carbonate solution, which is 0.0253mol. We can now find the concentration of the acid.

The equation for finding the concentration of a solution is as follows: •Concentration (moldm-3) = amount of solute (mol) / volume of solution (dm-3) •The volume of the solution used here is the average titre, which was found to be 16.5cm3. This has to be converted into dm3. Dividing 16.5 by 1000 to equal 0.0165dm3 does this. •Therefore the concentration of acid is as follows 0.0253 / 0.0165 = 1.53moldm-3



The procedure and the experiment itself carried quite a few limitations.

•Top balance could only read to 2 decimal places. Therefore, a very precise measurement of sodium carbonate couldn't be taken because the reading doesn't go up that high. •The burette wasn't large enough to accommodate all of the acid used in the experiment. So, between successive titres, the acid had to keep on being added to the burette, and this wasn't always filled up to the same amount as the previous initial...
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