implementation
10 g of potassium permanganate is dissolved in 250 ml of demineralized water. The solution is a concentrated potassium iodide solution. The mixture is
heated in a 500 ml three-necked flask equipped with a mantle heater for 30 minutes. Subsequently, the excess of permanganate is eliminated by the
addition of alcohol (decoloration of the solution). The solution is filtered with suction flask and suction filter. The filtrate thus obtained is
acidified with acetic acid and evaporated to the burner on a ceramic hob to the crystallization point. So that the crystallization is complete, the
solution after the first crystals have failed, cooled with ice. KIO3 the resulting crystals are filtered off and washed with ethanol.
observation
-It Had to be filtered several times since always came with brown stone in the filtrate.
- After adding acetic acid to give only a weak acidic solution, which was not acidified.

chemical
solubility

mp
KI
1000 g H2O at 20 ° C: 143 g
686 ° C
KMnO4
1000 g H2O at 20 ° C: 64 g
50 ° C
ETOH
Acetic acid dil.


- The evaporation was slow and was under constant stirring.
- When the first crystals were precipitated, the beaker was allowed to stand still on the ceramic hob for a few minutes, until the solution was cooled
with ice.
- It must be re-filtered several times to collect all KIO3 crystals in the filter cake. It was then rinsed with EtOH.
-The Filter cake was scraped from the filter into a crystallizing dish. -The Crystals were dried in the crystallizing dish. calculations
Weighing KMnO4: 10.05 g
Weighing CI: 5.05 g
+ → conc. KI solution weighing H2O: 35.39 g
1000g H2O >> 143 g KI
x g H2O >> 5 g KI
x = 34.97 g
Weighing crystallizing: 89.30 g
Weighing Kristallisiersschale + KIO3 crystals: 91.67 g amount of KIO3: 2,37g
yield calculation
M (KI) = 165.91 g / mol M (KIO3) = 214 g / mol
Based on AI
Of 1 mol of KI produced 1 mol of KIO3
165.91 g → 1 mol 5.05 g → 0.03 mol
From 0.03 mol KI KIO3 arise 0.03 mol

theoretical yield
0.03 mol * 214 g / mole = 6.51 g
Useful yield
2.37 g
(2.37 g / 6.51 g) * 100 = 36.4% = yield evaluation / Questions
Were obtained 2,37g potassium iodate. This corresponds to a yield of 36.4%. Some loss is einzuberechnen during filtration and the scraping of the
filter. During the final filtration product could be reached in the mother liquor and not have been completely scraped from the filter. In addition,
secondary reactions occur, consume potassium, eg arises KOH.
The addition of ethanol to the reaction mixture results in a redox reaction with potassium permanganate and the formation of CO2 and H2, as well as
manganese dioxide, which causes a discoloration of the solution.
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I was
using ethanol denatured with methanol and it seemed to work fine. I boiled down the filtrate from 150ml to 25ml and cooled it in an ice bath. One
point I need to emphasize is to cool the solution slowly. I just threw mine straight into the ice bath and lost probably half the yield due to
formation of crystals which could pass through my medium grit filter paper, even after refiltering twice. I can't say anything about the yield
though, since I lost more than half of my yield to the funnel
Product identity
after washing twice with denatured ethanol and once with distilled water was confirmed by adding a small amount of the product to concentrated
hydrochloric, and concentrated sulfuric acid. On addition of HCl, the solid fizzed and dissolved, turning the solution yellow and giving it a smell
of chlorine. On addition of H2SO4, a clear solution resulted indicating no iodide was present, though addition of a small amount of KI resulted in a
deep brown color instantly.