Experiment Anwer
KF 1(a) SOLUBILITY OF AN SALT BY TITRATION
Date : __________________________
Name : _____________________________________________________
Matric no. : __________________________
1. Objective(s) of Experiment
To measure the change in solubility product of potassium periodate (KIO4), when an inert salt (NaNO3) is added to the solution.
2. Why important to determine solubility?
The solubility product expression can be used for predicting whether or not precipitation will occur upon mixing two solutions and whether or not a precipitate will dissolve when in contact with a given solution.
3. Experimental Results
Table 1: Preparation of Sodium Nitrate Solutions
Chemical formula : NaNO3
Molecular weight : 23(1) + 14(1) + 16(3) = 85 g/mol
Weight of Sodium Nitrate : 0.20M x 0.5L x 85g/mol = 8.5 g
C, mol/L | Volume of 0.2M solution, mL | Total Volume, mL | 0.10 | 50 | 100 | 0.05 | 25 | 100 | 0.02 | 10 | 100 | 0.02 | 5 | 100 |
Table 2: Effect of Ionic Strength on the Solubility of KIO4
C(mol/L) | Volume of Na2S2O3 (mL) | [IO4-](mol/L) | I(mol/L) | Ks’(mol/L)2 | Solubility, S(mol/L) | 0.20 | 23.7 | 0.0148 | 0.2148 | 2.19 x 10-4 | 0.0148 | 0.10 | 23.3 | 0.0146 | 0.1146 | 2.13 x 10-4 | 0.0146 | 0.05 | 22.8 | 0.0143 | 0.0643 | 2.04 x 10-4 | 0.0143 | 0.02 | 22.3 | 0.0139 | 0.0339 | 1.93 x 10-4 | 0.0139 | 0.01 | 21.9 | 0.0137 | 0.0237 | 1.88 x 10-4 | 0.0137 | 0.00 | 21.4 | 0.0134 | 0.0134 | 1.80 x 10-4 | 0.0134 | 4. Discussion a) Give an example of calculation of Ks’ and S.
Ks’= [K+] [IO4+]
[K+]= [IO4+]
Ks’= [IO4+]2 = (0.0148mol/L)2 = 2.19 x 10-4 mol2/L2 S=Ks' = 2.19 x 10-4mol2L2 = 0.0148 mol/L
b) On a separate sheet of graph paper, plot log Ks’ against I. c) From the plot obtain Ks
Appendix:
IO4+ 8H++ 7I-→4I2+ 4H2O
2S2O32-+ I2→S4O62-+ 2I-
MV(IO4)MV(S2O62-)=18
[IO4]=MV(S2O62-)8
I=C x [IO4]
Log Ks’ | I | -3.660 | 0.2148 | -3.672 |
Date : __________________________
Name : _____________________________________________________
Matric no. : __________________________
1. Objective(s) of Experiment
To measure the change in solubility product of potassium periodate (KIO4), when an inert salt (NaNO3) is added to the solution.
2. Why important to determine solubility?
The solubility product expression can be used for predicting whether or not precipitation will occur upon mixing two solutions and whether or not a precipitate will dissolve when in contact with a given solution.
3. Experimental Results
Table 1: Preparation of Sodium Nitrate Solutions
Chemical formula : NaNO3
Molecular weight : 23(1) + 14(1) + 16(3) = 85 g/mol
Weight of Sodium Nitrate : 0.20M x 0.5L x 85g/mol = 8.5 g
C, mol/L | Volume of 0.2M solution, mL | Total Volume, mL | 0.10 | 50 | 100 | 0.05 | 25 | 100 | 0.02 | 10 | 100 | 0.02 | 5 | 100 |
Table 2: Effect of Ionic Strength on the Solubility of KIO4
C(mol/L) | Volume of Na2S2O3 (mL) | [IO4-](mol/L) | I(mol/L) | Ks’(mol/L)2 | Solubility, S(mol/L) | 0.20 | 23.7 | 0.0148 | 0.2148 | 2.19 x 10-4 | 0.0148 | 0.10 | 23.3 | 0.0146 | 0.1146 | 2.13 x 10-4 | 0.0146 | 0.05 | 22.8 | 0.0143 | 0.0643 | 2.04 x 10-4 | 0.0143 | 0.02 | 22.3 | 0.0139 | 0.0339 | 1.93 x 10-4 | 0.0139 | 0.01 | 21.9 | 0.0137 | 0.0237 | 1.88 x 10-4 | 0.0137 | 0.00 | 21.4 | 0.0134 | 0.0134 | 1.80 x 10-4 | 0.0134 | 4. Discussion a) Give an example of calculation of Ks’ and S.
Ks’= [K+] [IO4+]
[K+]= [IO4+]
Ks’= [IO4+]2 = (0.0148mol/L)2 = 2.19 x 10-4 mol2/L2 S=Ks' = 2.19 x 10-4mol2L2 = 0.0148 mol/L
b) On a separate sheet of graph paper, plot log Ks’ against I. c) From the plot obtain Ks
Appendix:
IO4+ 8H++ 7I-→4I2+ 4H2O
2S2O32-+ I2→S4O62-+ 2I-
MV(IO4)MV(S2O62-)=18
[IO4]=MV(S2O62-)8
I=C x [IO4]
Log Ks’ | I | -3.660 | 0.2148 | -3.672 |