CH530CH533 Experiment4 SolidAcidExperiment Essay

Submitted By kgpearce
Words: 910
Pages: 4

Aim(s):
In this experiment the compound 12-tungstosilicic acid will be formed using a solvent extraction method.
Quantitatively the available protons in this material will be determined.
The material as a solid acid catalyst will be tested.

Introduction:
Heteropolyacids (HPA’s) are a class of solid acid which have vast applications within the science and technical fields of the world.

HPA systems are extremely diverse and created through chemical techniques which allow this diversity to be harnessed, a good example of this is Tungstosilicic Acid which will be made through this experiment; the initial building block of this system is SiO4 a tetrahedral molecule which can connect to other molecules via sharing corners and edges of the molecule; however never the faces, the connections will occur through hydration, protonation and further hydrolysis.
A HPA system can be formed such as shown in Figure.1. below:5

In Figure.1. the octahedral base is visible in the middle of the complex surrounded by octahedral units of MO6, this image helps to represent the structure however Figure.2. below helps to show the actual bonding not just the 3d layout:

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Due to the properties these compounds possess they are of extreme interest to the scientific world; particularly as catalysts and in this experiment the HPA tungstosilicic acid will be made and tested.

Method:
CH530/CH533, 2nd Year, Experiment 4, page 25-28.
HCL was 11m.
NaOH was 0.102M.
When Separating the layers the yellow layer did not completely disappear and when it was mixed with the other layer to check it made two layers and therefore I could not use it; no matter how much ether was added there was still some yellow layer left, which was a crude layer which was a mix of everything.

Results and Discussion:
Titration:
Titration Number
Volume of NaOH used to neutralize (ml)
1
15.5ml
2
15.6ml
3
15.4ml
Average Titration Value
15.5ml

The following reaction occurs:
H4SiW12O40*7H2O + 4NaOH → 11H2O + 4Na(+) + SiW12O40(4-) and from this titration the following values were given or calculated.

H4SiW12O40*7H2O
NaOH
Concentration
0.01M
0.102M
Volume
30ml
15.5ml
Moles
3.95x10-4 mol
1.58x10-3 mol

Below are the calculations used:
0.102 x (15.5ML/1000) = 1.58x10-3 mol
1.58x10-3 mol/4 = 3.95x10-4 mol
3.95x10-4 mol / (30ml/1000) = 0.013M Solid Acid = 0.01M of Solid Acid per mole of compound.

Test of H4SiW12O40*7H2O as a solid acid (Catalyst):
In this part of the experiment two solutions were made up; one containing the Solid Acid formed and the other without, along with Bromine water, cyclohexanol and cyclohexanone; initially they both appeared the same with 2 layers a golden brown layer which was the bromine sitting on the bottom and another layer which was on top however when the solution containing the Solid Acid was shook the solid acid worked as a catalyst and allowed the bromine to halogenate the cyclohexanol into an alkyl halide (1,2-bromocyclohexane) and the two layers in the test-tube switch positions and the yellow looking layer now sits on the top layer and the alternative layer on the bottom; which is not observable in the solution which is not containing the solid acid.
The Following equation occurs:
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Product:
The product formed was 9.89g however I did not weigh anything before the heating of this so I have no value to compare to in order to calculate other than perhaps the amount of sodium tungstate dihydrate put in which was 15g so when opposing these two values the yield obtained is: 65.93%.

Questions:
1. A keggin unit is the best known structural formula of Heteropolyacids (HPA’s) which is a class of solid acid. A keggin unit is a heteroatom (not a