Determine pH values of salts solutions by using different indicators Prepare acetic acid-sodium acetate buffer
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These are notes and practice questions on chapter 11 of the student's book of IGCSE Chemistry. This will give you a clear understanding on the concept of making salts for example details on precipi...
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This is an excellent and concise formal report
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Chemistry Lab Report Experiment Number 4
CHEMY 102
Rafiuddin Mohammed Mohiuddin ( ( 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 ACT ION OF A BUFFER SOLUTION
REPORT SHEET
A. Hydrolysis of Salts Table I
Solution (0.1 M )
NaCl
Na2CO3
NaC2H3O2
Ion expected to hydrolyze
Spectator ion(s)
---
Na , Cl
+
CO32 ‾
Na
C2H3O2
-
+
-
+
Na
+
Cl
2+
Cl
NH4Cl
NH4
ZnCl2
Zn
KAl(SO4)2
Al
3+
-
-
+
K , SO42 ‾
Table III Net-ionic equation for hydrolysis
Solution (0.1 M )
Na2CO3
NaC2H3O2
NH4Cl
ZnCl2
-
-
Value of
constant ( K a or K b)
K a or K b
-
-
-4
CO3 (aq) + H2O HCO3 (aq) + OH‾ (aq)
K b = [HCO3 ] [OH‾] / [CO3 ]
K b = 5.0 x 10
C2H3O2 ‾ (aq) + H2O HC2H3O2 (aq) + OH‾ (aq)
K b = [HC2H3O2] [OH‾] / [C2H3O2 ‾]
K b = 1.1 x 10 -12
+
+
NH4 (aq) + H2O NH3 (aq) + H (aq)
Zn
2+
+
+
(aq) + H2O Zn(OH) (aq) + H (aq) 3+
KAl(SO4)2
Expression for equilibrium
Al[H2O]6
- 2+
(aq) Al[H2O]5[OH ] + H (aq)
(aq) +
K a = [NH3] [H ] / [NH4 ]
+
+
K a = 1.9 x 10
-11
+
K a = 9.0 x 10
-11
+
2+
- 2+
+
K a = [Zn(OH) ] [H ] / [Zn ] K a = [Al[H2O]5[OH ] ] [H ] / 3+ [Al[H2O]6 ]
K a = 1.2 x 10
-5
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.
