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Title: Pontiometric Acid-Base Titrations
Description: Lab results, discussion, conclusion, calculations
Description: Lab results, discussion, conclusion, calculations
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Experiment 3: Pontiometric Acid-Base Titrations
I
...
0024x3 + 0
...
6564x + 4
...
83206
12
pH
10
8
6
4
2
0
0
5
10
15
20
25
30
Volume (mL)
14
12
y = -‐0
...
596x2 -‐ 239
...
1
R² = 0
...
33
0
...
76 g/mol
12
14
16
Unknown acid
pKa
pKa of unknown / pH
18
Anilinium hydrochloride
4
...
06
unknown
Molar Mass of
unknown
Vol
¼
½
¾
pH
7
...
3019
7
...
5590g/mol
Vol
13
...
47
13
...
6028
7
...
Discussion
In the identification of the unknown the molecular weight had to be determined by
converting the volume of NaOH to moles of acid, which was divided to the weight of the
unknown acid
...
76g/mol and the nearest molecular weight from the chart
given was anilinium hydrochloride, which was 129
...
The pH was derived from the
equation from the graph as x was substituted by the volume of NaOH, 13
...
06
...
The volume of NaOH, 13
...
The theoretical curve determined by getting ¼, ½, ¾, 5/4, 3/2 of the volume at
equivalence point, as well as the equivalence point itself
...
1012 to get the moles
...
60 as the pKa value
...
The theoretical
curve was relatively close to the actual curve with possible errors, although this could mean
the unknown acid should be different but if the molecular weight is to be followed, anilinium
hydrochloride is the closest to the result
...
990g/mol using the moles of the acid, volume of NaOH at equivalence
point and weight of the unknown acid
...
Conclusion
The unknown acid was determined to be anilinium hydrochloride as it had the closest
molecular weight to the result, which was 128
...
990g/mol
...
This could be due to incorrect unknown acid used, though their molecular
weights are close, which served as the basis for using this unknown acid
...
47mL
...
06, which is also its pH
...
Calculations
M NaOH = 0
...
y = 0
...
514)2 = 0
...
0012)
0
...
0012x2 + 0
...
1997
st
à x = 21
...
3507 1 derivative
2
...
0024x3 + 0
...
6564x + 4
...
0072x^2+0
...
6564
-(0
...
194)2 – 4(-0
...
6564)
2(-0
...
968, 22
...
47mL NaOH
13
...
1092 mol NaOH x
1L
x
1L
1000mL
1 mol acid = 0
...
1755g acid / 0
...
76g/mol à anilinium hydrochloride
3
...
4376 + √ (0
...
0303)(1
...
1214, 4
...
2211mL 2nd derivative
2(0
...
13 - (31
...
0044) y = 3
...
13
...
735mL
-0
...
097x2 – 0
...
8108
-0
...
735)3 + 0
...
735)2 – 0
...
735) + 4
...
057 pH at equivalence point
5
...
47mL endpoint
¼
3
...
1012 = 0
...
7350 mL x 0
...
6816 mol
¾
10
...
1012 = 1
...
1755g ÷ 129
...
3543 x10-3
pKa = 4
...
60 + log (0
...
3543 x10-3) = 7
...
MM = Wt ÷ MV à = 0
...
001363)(13
...
Literature Cited
MM of unknown = 9
Title: Pontiometric Acid-Base Titrations
Description: Lab results, discussion, conclusion, calculations
Description: Lab results, discussion, conclusion, calculations