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The Ka & Molar Mass of a Monoprotic Weak Acid By Elina Melan Professor Dr Paul Gilletti November 5, 2020

Abstract

The purpose of this experiment was to determine the and of a weak acid. ThispK a

K a

was accomplished by graphing a titration curve mL by mL as a mystery solute was titrated by a .1 M (experimentally 0.098 M) solution of NaOH. Through the derivatives of this curve, it was found that the and of this weak acid were 4.62 and 2.399�-5, respectively. By use of thepK a

K a

molarity of NaOH and the chemical equation 2

,X(aq) aOH (aq) H + N → H O aX(aq) + N along with the mass of the solute HX, the molar mass of HX was calculated to be 63.78 g/mol. Given the actual molar mass, 69.49 g/mol, it can

be concluded that this weak acid is Hydroxylamine Hydrochloride, with molecular formula ClH 4

NO The purpose of this experiment was to determine the and of a weak acid. ThispK a

K a

was accomplished by graphing a titration curve

mL by mL as a mystery solute was titrated by a .1 M (experimentally 0.098 M) solution of NaOH. Through the derivatives of this curve, it was found that the and of this weak acid were 4.62 and 2.399�-5, respectively. By use of thepK a

K a

molarity of NaOH and the chemical equation 2

,X(aq) aOH (aq) H + N → H O aX(aq) + N along with the mass of the solute HX, the molar mass of HX was calculated to be 63.78 g/mol. Given the actual molar mass, 69.49 g/mol, it can be concluded that this weak acid is

Hydroxylamine Hydrochloride, with molecular formula ClH 4

NO The purpose of this experiment was to determine the and of a weak acid. ThispK a

K a

was accomplished by graphing a titration curve mL by mL as a mystery solute was titrated by a

.1 M (experimentally 0.098 M) solution of NaOH. Through the derivatives of this curve, it was found that the and of this weak acid were 4.62 and 2.399�-5, respectively. By use of thepK a

K a

molarity of NaOH and the chemical equation 2

,X(aq) aOH (aq) H + N → H O aX(aq) + N along with the mass of the solute HX, the molar mass of HX was calculated to be 63.78 g/mol. Given the actual molar mass, 69.49 g/mol, it can be concluded that this weak acid is Hydroxylamine Hydrochloride, with molecular formula ClH 4

NO The purpose of this experiment was to determine the and of a weak acid. ThispK a

K a

was accomplished by graphing a titration curve mL by mL as a mystery solute was titrated by a .1 M (experimentally 0.098 M) solution of NaOH. Through the

derivatives of this curve, it was found that the and of this weak acid were 4.62 and 2.399�-5, respectively. By use of thepK a

K a

molarity of NaOH and the chemical equation 2

,X(aq) aOH (aq) H + N → H O aX(aq) + N

along with the mass of the solute HX, the molar mass of HX was calculated to be 63.78 g/mol. Given the actual molar mass, 69.49 g/mol, it can be concluded that this weak acid is Hydroxylamine Hydrochloride, with molecular formula ClH 4

NO The purpose of this experiment was to determine the Ka and Pka and molar mass of a weak acid. To find the acid dissociation constant, Ka, we used the equation (1). Our experiment was accomplished by preparing and standardize 0.1 molarity NaOH and graphing a titration curve. Through the derivatives of this curve, it was found that the volume at equivalence= 23.27 ml, PH at equivalence point = PKa = 4.025 and Ka=9.4 x 10-5. By use of the molarity of NaOH=0.0993 M

and the volume at the equivalence point, the molar mass of unknown acid was calculated to be 174.5 g/mol. −¿ ( aq ) +¿ ( aq )+ A ¿ ¿ HA ( aq ) +H 2 O ( l ) → H 3 O

(1)

Introduction The objective of this experiment is to determine the molar mass, pKa, and Ka of a weak acid based on a titration curve. An acid is a compound that can donate a proton, and the strength of an acid can be determined by the compound’s ability to donate a proton. The stronger the acid, the farther the equilibrium of reaction (1). The acid dissociation constant, Ka, is the acidity constant and equals the molar concentration of hydronium times the molar concentration of the conjugate base divided by the molar concentration of the weak acid according to equation (2). Taking the negative antilog of the pKa to the tenth power, results in the Ka of the unknown weak acid (3). The strength of an acid is determined by the pH or negative log of the hydronium ion concentration according to equation (4) and (5). −¿ ¿ A ¿ +¿ H 3 O¿ ¿ ¿ k a=¿

(2)

pKa = -log Ka +¿ ¿ H3O ¿ ¿ PH =−log 10 ¿ +¿ H 3 O¿ ¿ ¿ pKa = -log Ka

(3)

(4)

(5) (5)

A titration curve is a graphical representation of the pH of a solution during a titration. At the equivalence point of the titration, the moles of the weak acid and the base added are equal and it corresponds with a color change of the indicator. At half of the equivalence point, the pH is equivalent to the pKa because the molar concentration of the resulting conjugate base is equal to the molar concentration of the acid. By multiplying the volume of NaOH used to titrate to

equivalence by the molarity of NaOH we found the moles of NaOH equal 2.31. At equivalenced point, the moles of acid equal the moles of base. By dividing the mass of the unknown acid by this number we found the molar mass of acid equal 174.5 g/mole.

Method In the first step of this experiment, the needed mass of NaOH was calculated on a lowprecision balance and dissolve it in a 250-mL beaker with DI water. It was poured the dissolved NaOH into a 500-mL polyethylene bottle and dilute with more DI water to the final volume. It was used a plastic bottle because hydroxide solutions react with glass on prolonged contact, etching the glass and changing the [OH –] value. It was weighed out 204.233 g/mol the of KHP on the analytical balance and record the mass to the full available precision. It was dissolved the KHP sample in an Erlenmeyer flask with about 50 mL of deionized water. Then added 2 drops of phenolphthalein indicator and titrate with our NaOH solution to a pale pink endpoint. It was Set up the buret for a pH meter titration while Keeping the solution gently stirred with the magnetic stirrer. It was added enough base solution from the buret to produce a change of 0.2 units in the pH meter reading. It was repeated the addition of base solution until it well past the equivalence point. At the very end of your titration, the pH values will change only slightly with the addition of 2 mL or more of base solution. Finally, we used the excel to graph the titration curve to find the Ka and Pka. https://drive.google.com/file/d/1xc5nJboYXp3eQ4CggR7AJW3lssUkleNH/view?ts=5e7a9984 (date of access 11/05/2020)

Data Table 1. Burette reading mL of 0.0993 M NaOH and PH value

#

1 2 3 4 5 6

Burette reading mL of 0.0993 M NaOH

PH readin g

#

Burette reading mL of 0.0993 M NaOH

PH reading

1.62 1.04 1.99 3.00 3.88 5.00

2.83 3.00 3.17 3.32 3.43 3.55

26 27 28 29 30 31

21.88 22.38 22.58 22.79 22.88 22.97

5.50 5.80 5.96 6.52 6.80 7.16

7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

6.01 6.98 8.06 8.99 9.98 10.98 12.00 12.98 13.97 15.07 16.04 16.98 18.01 19.01 19.37 19.82 20.37 20.88 21.40

3.66 3.75 3.84 3.93 4.02 4.09 4.18 4.25 4.32 4.41 4.50 4.59 4.71 4.84 4.88 4.96 5.04 5.17 5.30

32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

23.02 23.09 23.17 23.18 23.27 23.30 23.38 23.47 23.61 23.80 24.23 24.46 25.29 25.51 26.47 27.50 28.64 30.69 34.48

7.56 7.69 7.86 7.83 8.05 8.86 9.01 9.15 9.47 9.68 9.97 10.06 10.20 10.38 10.57 10.70 10.81 10.98 11.18

Delta y

Delta x 0 -0.58 0.95 1.01 0.88 1.12 1.01 0.97

# 26 27 28 29 30 31 32 33

Delta y

Delta x 0.48 0.50 0.20 0.21 0.09 0.09 0.05 0.07

Table 2. first derivative

# 1 2 3 4 5 6 7 8

0 0.17 0.17 0.15 0.11 0.12 0.11 0.09

0.20 0.30 0.16 0.56 0.28 0.36 0.40 0.13

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

0.09 0.09 0.09 0.07 0.09 0.07 0.07 0.09 0.09 0.09 0.12 0.13 0.04 0.08 0.08 0.13 0.13

1.08 0.93 0.99 1.00 1.02 0.98 0.99 1.10 0.97 0.94 1.03 1.00 0.36 0.45 0.55 0.51 0.52

34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

0.17 -0.03 0.22 0.81 0.15 0.14 0.32 0.21 0.29 0.09 0.14 0.18 0.19 0.13 0.11 0.17 0.20

0.08 0.01 0.09 0.03 0.08 0.09 0.14 0.19 0.43 0.23 0.83 0.22 0.96 1.03 1.14 2.05 3.79

Table 3. plot volume and first derivative Plot volume

first derivative

#

Plot volume

first derivative

1 2 3

0 1.33 1.515

26 27 28

21.64 22.13 22.48

0.416667 0.6 0.8

4

2.495

29

22.685

2.666667

5 6

3.44 4.44

0 -0.2931 0.17894 7 0.14851 5 0.125 0.10714 3

30 31

22.835 22.925

3.111111 4

#

7 8

5.505 6.495

9

7.52

10

8.525

11

9.485

12 13

10.48 11.49

14

12.49

15

13.475

16

14.52

17

15.555

18

16.51

19 20 21 22

17.495 18.51 19.19 19.595

23

20.095

24

20.625

25

21.14

0.108911 0.09278 4 0.08333 3 0.09677 4 0.09090 9 0.07 0.08823 5 0.07142 9 0.07070 7 0.08181 8 0.09278 4 0.09574 5 0.116505 0.13 0.111111 0.17777 8 0.14545 5 0.25490 2 0.25

32 33

22.995 23.055

8 1.857143

34

23.13

2.125

35

23.175

-3

36

23.225

2.444444

37 38

23.285 23.34

27 1.875

39

23.425

1.555556

40

23.54

2.85714

41

23.705

1.105263

42

24.015

0.674419

43

24.345

0.391304

44 45 46 47

24.875 25.4 25.99 26.985

0.168675 0.818182 0.197917 0.126214

48

28.07

0.096491

49

29.665

0.082977

50

32.85

0.05277

Graphs Graph1. Volume vs. PH

Graph2. First derivative

Calculations 1. Volume of equivalence point = 23.27 ml 2. M of NaOH = 0.0993 mol/l 3. Mole of NaOH = volume at equivalence point x M of NaOH Mole of NaOH = (23.27 x 10-3 L) x 0.0993 M = 0.0023 mol 4. At equivalence point 1.23

Moles of base = moles of acid at

Mole NaOH = 0.0023 mole acid

Molar mass of acid = grams of acid / moles of acid Molar mass of acid = 0.4030 g / 0.0023 mole = 174.4 g/mol 5. PH at equivalence point = 8.05 PH at halfway to the equivalence point = 8.05 / 2 = 4.025 6.

7.

Pka = PH = 4.025

+¿ ¿ H3O ¿ ¿

8. Ka =

+¿ ¿ H3O x 10-5 ¿ ¿

+¿ ¿ H3O x 10-5 ¿ ¿

Conclusion Through this experiment we determined the Ka and Pka and molar mass of a weak acid. To find the acid dissociation constant, Ka, we used the equation (1). Our experiment was accomplished by preparing and standardize 0.1 molarity NaOH and graphing a titration curve. Through the derivatives of this curve, it was found that the volume at equivalence= 23.27 ml, PH at equivalence point = PKa = 4.025 and Ka=9.4 x 10-5. By use of the molarity of NaOH=0.0993 M and the volume at the equivalence point, the molar mass of unknown acid was calculated to be 174.5 g/mol.

References 1. Gilletti, Paul. CHM152LL Lab Manual the Ka & Molar Mass of a Monoprotic Weak Acid; Mesa Community College: Mesa, AZ, 2020

Atkins, P.; Jones, L.; Laverman, L. Chemistry Principles: The Quest for Insight  , 7th ed.; W.H. Freeman and Company: New York, NY, 2016; pp 264. Atkins, P.; Jones, L.; Laverman, L. Chemistry Principles: The Quest for Insight

 , 7th ed.; W.H. Freeman and Company: New York, NY, 2016; pp 264. 2. Atkins, P.; Jones, L.; Laverman, L. Chemistry Principles: The Quest for Insight, 7th ed.; W.H. Freeman and Company: New York, NY, 2016; pp 264....