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Offline the_guitarist

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Analysis [Long]
« on: March 22, 2008, 05:13:53 PM »
THIS IS THE ANALYSIS OF MY TITRATION RESULTS.IF SOMEONE COULD CHECK IT I WOULD APPRECIATE IT, THIS IS ALSO THE LAST I WILL POST REGARDING THIS PARTICULAR TOPIC.

The reaction:
Na2CO3(aq)+H2 SO4(aq)---->H2O(l)+Na2SO4(aq)+CO2(g)

Starting reading/cm3    End reading/cm3    Total acid used/cm3
0                                   20.65                     20.65
0                                   20.15                     20.15
0                                   20.00                     20.00
0                                   20.00                     20.00

 
 To find the concentration of my Na2CO3 solution I used the following formulae:
Mass=Moles x Relative Molecular Mass
Moles=Concentration x Volume

Firstly, I worked out that Na2CO3 has a relative molecular mass (RMM)  of 106:
2Na=23 x2=46
C=12
3O=16 x 3=48
46+12+48=106

After this I calculated the number of moles in the solution by diving 2.67 by the RMM of sodium carbonate, 106.
2.67g/106
=0.025mol.

Concentration=Moles/volume.
= 0.025mol./0.25dm=0.1moldm-3


Because one molecule of H2SO4 reacted with one molecule of Na2CO3 (see equation) the molar ratio in this equation is 1:1 the volume ratio will be 25:20, meaning that the calculation to work out the concentration of the H2SO4 will be this:
Moles=Concentration x Volume
1:1 moles means that the volume ratio is  20dm3:25dm3

Concentration H2SO4=   0.025dm3/0.2dm3  =0.125 moldm-3
                                             
                                       
The solution of the H2|SO4 is 0.125 moldm-3

  My results for the 4 titrations show that, as I said in my plan, the first titration was a rough measurement to gain an approximate end point volume (the volume of acid needed to neutralise the alkali) and I found that this was a volume of 20.65cm3. This volume, however, created a very vibrant colour change, indicating that the end point had long since past; from this though we know that the volume of sulphuric acid needed to reach the end point. From here I can continue to perform the second titration; because I already knew that 20.65cm3 is too large of a volume to reach the exact end point so I simply added around 15cm3 of acid from the burette in one go and agitated the solution. After the 15ml were released I added acid at a slower rate, but not quite drop-by-drop. Once I started to see small,short lasting colour changes in the solution I started adding the acid drop by drop from the burette. I did in fact go too far past the end point on the second titration as I saw a much more vibrant colour change that I was actually expecting to see, this was at 20.15cm3 of acid.
  Due to not already having concordant results I hate to perform some more titrations until I gathered results that were, this took 2 more attempts which left me with a total of 4 titrations.
The third and fourth titrations are the ones in which I got concordant results of 20cm3 for both of the titrations. I got this as the second titration was much closer to the precise end point; I added about 15cm3  and then added more at a slower rate until I saw momentary colour changes I began to add the acid drop by drop. Eventually I noticed a definite colour change, although be it slight, and concluded that this was my end point.
 
  The average titre for this experiment is worked out as follows:
(Sum of titres)/number of titrations =average titre.

So it would be:
(20.65+20.15+20+20)/4= 20.2

  This shows that the average titre was very close to the actual titre which I got through performing the practical,20cm3.. Because the value was so close to the actual volume it shows that my results were actually valid, rendering my interpretation of the end point as a perfectly acceptable interpretation. I will talk more about the problems associated with this in my evaluation.

Offline Guitarmaniac86

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Re: Analysis [Long]
« Reply #1 on: March 22, 2008, 06:09:26 PM »
The only thing Id change is the average titre.

Take the concordant ones, ie titres that differ by no more than 0.2cm3 therefore your actual average is (20 + 20 + 20.15)/3 = 20.05cm3

Don't believe atoms, they make up everything!

Offline merkl

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Re: Analysis [Long]
« Reply #2 on: March 23, 2008, 07:36:09 AM »
Guitarist, are you by chance british, with an AS plan due in soon?

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