# The purity of aminosulphonic acid

Titration number

1

2

3

4

23.30

22.50

45.10

23.50

0.00

0.10

22.40

1.00

Amount of acid used for each titration:

1- 23.00-0.00=23.00 cm3

2- 22.50-0.10=22.40 cm3

3- 45.10-22.40=22.70 cm3

4- 23.50-1.00=22.50 cm3

The amount of acid used for the first titration is a little higher than it should be so to make the calculation much more accuracy I will ignore the first one and I will work out the average of the other three values:

22.40+22.70+22.50=67.6 cm3

67.6/3=22.53 cm3 (2 decimals place)

22.53 cm3 is average amount of acid (H2NSO3H) reacted with 25.00 cm3 of alkali (NaOH) with concentration of 0.200 moldm-3.

Now I am going to work out the number of moles of NaOH in 25.00 cm3 of that with concentration of 0.200 moldm-3:

CV=M

C=0.200 moldm-3 M=0.200*25.00/1000=0.005 number of moles of NaOH

V=25.00 cm3 reacted with acid.

M=number of moles

As the ratio of the acid to the alkali in the reaction is 1-1 so number of moles of the acid reacted with the alkali are as same as number of moles of the alkali reacted with the acid because of ratio of 1-1.

Now I can say that the number of moles of acid in 22.53 cm3 of that (which is the average amount of the acid I worked that out before) is equal to 0.005. So I can work out the number of moles of acid in 250cm3 of that.

Number of moles in cm3 X=250*0.005/22.53

0.005 22.53 X=0.0555 moles in 250 cm3 acid

X 250

Now I can work out the number of grams of the acid (H2NSO3H) by using its molecular number (Mr=97):

Number of moles of H2NSO3H is equal to this number of grams

1 97

0.0555 Y

Y=0.0555*97=5.3835 grams

Now I can work out the value of percentage purity which is :

Percentage purity = 5.3835/5.6=0.9613=96.13%

Also I can work out the value of percentage error in the result which is:

% Error in the result = (99.1-96.13)*(100/99.1)= 2.99=2.99%

Now I am going to estimate the maximum percentage error in using each piece of apparatus, and the overall percentage error.

Balance= +/-0.1g = (0.1/5.6)

250cm3 volumetric flask = +/-0.25cm3=(0.25/250)

25cm3 pipette = +/-0.05cm3 = (0.05/25)

Burette = +/- 0.10cm3 = (0.10/22.53)

The overall value of error is sum of the above numbers which is:

Overall error = (0.1/5.6) + (0.25/250) + (0.05/25) + (0.10/22.53) = 0.0253=2.53%

Reasons for the values of errors and differences in each step could be:

* Rounding the answers.

* Using unsuitable scales which could be improve by using smaller scale(for example using mg instead if g and also much more sensitive apparatus)

* Using too much indicator (phenolphthalein) which makes it to take more time to be colourless.

* Do not staring the substance during the experiment.

* Do not dissolving the substances completely in the water.

* The acid may have different concentration in different level of that for example higher parts may have lower concentration and lower parts may have higher consequently the samples taken from that may have a different concentration each time. Also this case may apply for alkali as well.

* Subside of the substances.