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Lab Report 10

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Norayr Kloumian

Kevork boyadjian
Ayya Yehia
Chemistry 206 Lab
Exp.10: PH Titration
April 7th, 2022
Rania Mansour
I. Purposes:
 To learn the technique of potentiometric pH titration
 To compare different methods of determining equivalence points
 To determine the dissociation constants for an unknown polyprotic acid
(malonic acid)
 To determine the molecular weight of the unknown acid

II. Rxns:

 H 2 X + KOH ↔ KHX + H 2 O
 H X −¿+ KOH ↔ K X + H O ¿
2 2

III. Tables:

Table 1: Preparation of diprotic acid (malonic acid H 2 X ¿

Mass of Acid (± 0.0001 g ¿ Volume of acid (± 0.03 ml)

1.0530g 250.0ml

Table 2: Titration of malonic acid H 2 X with KOH

V of KOH added in drops pH measurement

0 2.13

3 2.23

6 2.36

9 2.51

12 2.65

15 2.82

18 2.97

21 3.14
24 3.29

27 3.54

30 3.95

33 4.34

36 4.70

39 4.90

42 5.11

45 5.31

48 5.44

51 5.61

54 5.78

57 5.96

60 6.20

63 6.68

66 11.26

69 12.10

72 12.42

75 12.50

78 12.67

81 12.79
Table 3: 1st Derivative table

V of KOH in Drops ∆ pH
∆V

3 0.0333

6 0.0433

9 0.0500

12 0.0467

15 0.0567

18 0.0500

21 0.0567

24 0.0500

27 0.0833

30 0.137

33 0.120

36 0.130

39 0.0667

42 0.0700

45 0.0667

48 0.0433

51 0.0567

54 0.0567

57 0.0600

60 0.0800
63 0.160

66 1.53

69 0.280

72 0.107

75 0.0267

78 0.0567

81 0.0400

Table 4: Summary of Results:

Equivalence point 1 Equivalence point 2

pH V in drops pH V in drops

Indicator 3.95 30 11.26 66

Titration curve 3.90 30 9.40 65

1st Derivative -- 30 -- 66

Ka1 1.51 ×10


−3

Ka2 −6
4.90 × 10

IV. Calculations and Results:

 Visual method of determining the equivalnce point:


 Methyl orange changed the color from red to yellow
 Phenolphthalein changed the color from yellow to pink

V 1 30
= =15 drops p Ka1=2.82
2 2
3V 1 90
= =45drops p Ka2=5.31
2 2

 Calibration of a dropper:

1ml of KOH 26 drops

 Calculation of the molar mass of the malonic acid H 2 X :

1 ml of KOH →26 drops


x → 30 drops
30 ×1
ml of KOH at the first equivalnce point=x= =1.15 ml
26

At the first equivalence point:

n KOH n H X
= 2

1 1

C KOH ×V KOH −3 −3
nH X= =1.000 ×1.15 ×10 =1.15 ×10 moles (¿ 25.0 ml)
2
1

−2
n H X ( ¿ 250.0 ml ) =1.15× 10 moles
2

mH X mH X 1.0530
nH X= 2
→ M H X= 2
= =91.6 g /mol
2
MH 2 X
2
nH 2 X 1.15 ×10−2

V. Possible Errors:

 Parallax errors
 Not fully immersing the pH electrode
 Uncalibrated balance (preparation of primary standard)
 Not protecting the primary standard KOH from CO 2 (which makes the concentration
decreases because of acid-base rxn)
 Wrong number of added drops (excess)
 Uncalibrated pH meter
 Not using an indicator
 Errors in equivalence point detection

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