28th of Sep 2009/ Monday
Aya Naji Shnawa
try 1.3 An acid base titration
*data collection and processing
* ending and evaluation
DATA COLLECTION AND PROCESSING:
Readings1st spark advance 2nd trace3rd trailquaternary trailAverage
Initial ±0.05cm 30.00.00.00.00.0
Final ±0.05cm 320.521.521.021.121.0
Table 1.1: a presentation of processed data in all the trails.
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The experimentsThe remnant
1st to 2nd 1.0
2nd to 3rd 0.5
3rd to 4th 0.1
Table1.2: the difference of opinion of the concluding readings between the experiments
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The average Standard excursion
21.0 0.4
Table1.3: the difference between average and standard deviation.
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The sum of the difference The average
1.60.5
Table 1.4 the average of the difference.
PRESENTATIONN OF DATA:
Chart 1.1: the final readings of each of the four trails.
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Chart 1.2: the difference between the readings in each of the trails.
Chart 1.3: the difference between the sum and the average.
end point AND EVALUATION
CONCLUSION:
Equivalence point is the point at which the moles of H+ is equal to the moles of OH+,an indicator is used to show the comparison point during a titration. in a titration the method is around totaling one reactant from the burette (regularly the acid),to a known volume of the some other reactant in a conical flask(regularly the base) .
In order to hap the concentration of NaOH we need to tag along the following stairs:
a-note down the balanced chemical equation for the reply
C8H5O5K+NaOH?C8H4O4KNa+H2O
b-pull out applicable information from the experiment:
C8H5O5K V=0.025 dm3 C=0.2M
NaOH V=0.021 dm3 C= ??
c-calculate number of moles
n(C8H5O5K)=VxC = 0.2x0.025 = 0.005 moles .
d- represent moles ratio from the equation :
1:1
e-find moles...If you demand to get a full essay, order it on our website: Orderessay
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