Aim:
To
determine the percentage of chloride ions in average sea water.
Procedure:
1. Collect the provided sample of
seawater which is to be analyzed.
2. Requires titration of the solution.
3. In the Burrete contains the silver nitrate.
4. In the conical flask below, take 25
ml of seawater with a pipette and add few drops of indicator.
5. Add the silver nitrate until the
indicator changes color slightly. (this case: changes yellow to red)
6. Note the difference and analyses.
DATA
COLLECTION & PROCESSING
During the
experiment, there was a visible change, which is a part of the qualitative
data, and there was a numerical change also. Therefore the qualitative analysis
is listed below:
Qualitative
Analysis:
|
Before Reaction:
|
During Reaction:
|
After Reaction:
|
|
The burrete was colorless
with only silver nitrate in it.
The conical flask contained a yellowish solution which was a mixture of colorless seawater and yellowish indicator of potassium chromate. |
During the addition of the silver nitrate
into the flask, slight reddish change would occur but soon would dissolve
back to yellowish color. This usually changed back due to shaking of the
flask.
|
The burrete was closed as
soon as there was a permanent light reddish precipitate in the flask. This
meant that all the reaction is over and there was a mere excess of the silver
chloride added.
|
Quantitative
analysis:
The data
recorded during the practical is as follows:
Raw Data:
|
Trial
number
|
Initial
meniscus(±0.5)
|
Final
Meniscus(±0.5)
|
|
1
|
50
|
23.1
|
|
2
|
30
|
2.7
|
|
3
|
30
|
3.3
|
Processed
Data:
|
Trial
number
|
Initial
meniscus (±0.5ml)
|
Final
Meniscus(±0.5ml)
|
volume
of silver nitrate(±1ml)
|
|
1
|
50
|
23.1
|
26.9
|
|
2
|
30
|
2.7
|
27.3
|
|
3
|
30
|
3.3
|
26.7
|
|
Average
|
26.97
|
||
Calculations:
The
difference to find the volume of silver nitrate used: initial volume – final volume =
50
ml – 23.1 ml = 26.9 ml (first trial data)
Uncertainty
of the difference:
∆V1+∆V2 =
0.5 ml +0.5 ml = 1ml
Therefor
the volume of the silver nitrate used in the first trial is: 26.9 ml ± 1ml
Average:
(sum of data of all the trials)/ number of trials = (26.9+27.3+26.7)/3 =
26.9666≈26.97
Uncertainty:
(∆V1+∆V2+∆V3)/3(mean
deviation)= (1+1+1)/3 = 1 ml
Average volume used: 26.97ml ± 1ml
________________________ ______________________________
Thus now we
have the volume of the silver nitrate, so now we will deduce the main aim of
determining the percentage of chlorine present in the seawater.
Conversion
factor:
1 ml ≈ 1cm3
Thus, 26.97ml ± 1ml ≈ 26.97 cm3 ± 1 cm3
Uncertainty
: 1/26.97 *100 =3.7%
Number of
moles of silver nitrate calculation:
Concentration =
0.05M
C=n/V
Thus, n = CV
so, n= (0.05Mx26.97dm3)/1000
n=0.0013485
moles
Number of
moles of silver in 1 moles of silver nitrate: 1 (Agno3 has 1 Ag)
Number of moles of silver in 0.0013485 moles
of silver nitrate = 0.0013485 moles
The reaction between Chlorine and silver
nitrate:
Ag+ + Cl- >>AgCl
Thus ratio is 1:1
Therefore moles of chlorine ions used:
0.0013485 moles
And therefore the mass of chlorine used:
Mr *n = 35.5g/moles x 0.0013485 moles = 0.04787 g
Therefore thee
concentration of chlorine in seawater: 0.04787g/0.025 dm3 = 1.91487g/dm3≈1.91gdm-3
Uncertainty: 3.7% of 1.91 = 0.07 g dm-3
Thus, the
concentration of chlorine is : 1.91gdm-3
The average concentration of the seawater is: 1.025 g/ml = 1025g dm-3
(taken from: http://en.wikipedia.org/wiki/Seawater)
(taken from: http://en.wikipedia.org/wiki/Seawater)
Thus the percentage of chlorine = 1.91gdm-3/1025g
dm-3x 100= 0.186% = 0.19%
Therefore the chlorine percentage in seawater is 0.19 %
_____________________________________________________________
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