Chemy102 Lab4 Hydrolysis Of Salts k4j56

Chemistry Lab Report Experiment Number 4

CHEMY 102

Rafiuddin Mohammed Mohiuddin (‫( الدين محي محمد الدين رفيع‬ ID number: 20135673 Section 05

Name: Rafiuddin Mohammed

ID: 20135673 CHEMY 102, Section 05

Chemistry Lab Report Experiment No. 4 HYDROLYSIS OF SALTS AND THE ACTION OF A BUFFER SOLUTION Aim: The purpose of this experiment is to learn about the concept of hydrolysis and to gain familiarity with the behavior of buffer solutions.

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Results and Calculations: Experiment 4: HYDROLYSIS OF SALTS AND THE ACTION OF A BUFFER SOLUTION

REPORT SHEET

A. Hydrolysis of Salts Table I

Ion expected to hydrolyze

Spectator ion(s)

---

Na+, Cl-

CO32‾

Na+

C2H3O2-

Na+

NH4Cl

NH4+

Cl-

ZnCl2

Zn2+

Cl-

KAl(SO4)2

Al3+

K+, SO42‾

Solution (0.1 M)

NaCl

Na2CO3

NaC2H3O2

Table II Solution

pH

[H+]

Color of the Indicator Phenolphthalein

[OH‾]

H2O (unboiled)

6.96

1.1 x 10-7

0.9 x 10-7

Colorless

H2O (boiled)

7.04

0.9 x 10-7

1.1 x 10-7

Colorless

NaCl

7.28

0.5 x 10-7

2 x 10-7

Colorless

Na2CO3

11.85

1.4 x 10-12

7.1 x 10-3

Pink

NaC2H3O2

7.55

0.3 x 10-7

3.3 x 10-7

Colorless

NH4Cl

5.86

13.8 x 10-7

7.2 x 10-9

Colorless

ZnCl2

5.52

3.0 x 10-6

3.3 x 10-9

Colorless

KAl(SO4)2

2.96

1.1 x 10-3

9.1 x 10-12

Colorless

Table III Net-ionic equation for hydrolysis

Expression for equilibrium constant (Ka or Kb)

Value of Ka or Kb

CO3- (aq) + H2O  HCO3- (aq) + OH‾ (aq)

Kb = [HCO3-] [OH‾] / [CO3-]

Kb = 5.0 x 10-4

C2H3O2‾ (aq) + H2O  HC2H3O2 (aq) + OH‾ (aq)

Kb = [HC2H3O2] [OH‾] / [C2H3O2‾]

Kb = 1.1 x 10-12

NH4Cl

NH4+ (aq) + H2O  NH3 (aq) + H+ (aq)

Ka = [NH3] [H+] / [NH4+]

Ka = 1.9 x 10-11

ZnCl2

Zn2+ (aq) + H2O  Zn(OH)+ (aq) + H+ (aq)

Ka = [Zn(OH)+] [H+] / [Zn2+]

Ka = 9.0 x 10-11

KAl(SO4)2

Al[H2O]63+ (aq)  Al[H2O]5[OH-]2+ (aq) + H+ (aq)

Ka = [Al[H2O]5[OH-]2+] [H+] / [Al[H2O]63+]

Ka = 1.2 x 10-5

Solution (0.1 M)

Na2CO3

NaC2H3O2

B- The Action of a Buffer Solution Table IV

pH on addition of 0.1 M NaOH to Volume (total) / mL

pH on addition of 0.1 M HCl to

buffer

pure water

buffer

pure water

0.0

7.46

7.60

7.44

7.20

1 drop

7.48

9.44

7.42

6.83

1.0 mL

7.56

11.36

7.31

2.92

5.0 mL

8.28

11.80

6.85

2.25

Discussion: The experiment enlightens us to the reason behind the acidity or basicity of a solution at the equivalence point. The reason is easily explicable by the concept of hydrolysis. The salt formed by the reaction between a strong acid and a strong base is a neutral salt and does not undergo hydrolysis. This is why titration between a strong acid and a strong base gives a neutral equivalence point. Similarly, titration between a strong acid and a weak base OR a weak acid and a strong base gives an acidic OR basic equivalence point because of the hydrolysis of the cation OR anion of the respective salts. Buffers are interesting solutions because of their resistance to high changes in pH. They are found in various natural systems and are of great value even in our own body.

These experiments, conducted to determine the acidity and basicity of the various salts give values for their dissociation constants that are, at best, gross approximations. This is because of the various errors involved that include the presence of CO2 in the atmosphere which dissolves in the solutions and changes the pH, the inaccuracy of the pH meter in giving readings, the improper cleaning of the containers in which the measurements are taken, the possible incomplete immersion of the electrode in the solution, etc. The experiment could have been better conducted by using freshly prepared solutions of the salts, using a fast and accurate pH meter, carefully making the measurements, and cleaning the apparatus before each reading.

Conclusion: Knowing the pH of a salt solution is an important aspect of various processes and studying hydrolysis is consequently of significant importance.