Experiment Anwer

Only available on StudyMode
  • Download(s) : 93
  • Published : May 11, 2013
Open Document
Text Preview
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|...
tracking img