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Acid Base Titration Essay

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Chemistry lab report
28th of Sep 2009/ Monday
Aya Naji Shnawa
Experiment 1.3 An acid base titration
*data collection and processing
*conclusion and evaluation
DATA COLLECTION AND PROCESSING:

Readings 1st trail 2nd trail 3rd trail 4th trail Average
Initial ±0.05cm 3 0.0 0.0 0.0 0.0 0.0
Final   ±0.05cm 3 20.5 21.5 21.0 21.1 21.0

Table 1.1: a presentation of processed data in all the trails.
________________________________________

The experiments The difference
1st to 2nd 1.0
2nd to 3rd 0.5
3rd to 4th 0.1

Table1.2: the difference of the final readings between the experiments
________________________________________

The average Standard deviation
21.0 0.4

Table1.3: the difference between average and standard deviation.
________________________________________
The sum of the difference The average
1.6 0.5

Table 1.4 the average of the difference.
PRESENTATIONN OF DATA:


Chart 1.1: the final readings of each of the four trails.
________________________________________

Chart 1.2: the difference between the readings in each of the trails.


Chart 1.3: the difference between the sum and the average.

CONCLUSION 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 equivalence point during a titration. in a titration the method is about totaling one reactant from the burette (regularly the acid),to a known volume of the other reactant in a conical flask(regularly the base) .
In order to find the concentration of NaOH we need to tag along the following steps:
a- note down the balanced chemical equation for the reaction
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- equate moles ratio from the equation :
  1:1
e- find moles...

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