I already know what Le Chatliers Principle is, but I need help describing what is going on in specific in the following reactions. For example, how is the chemical equation changing to minimize change.
1. Bromine water is initially yellow/brown but changes to yellow when sodium hydroxide is added. Then when sulfuric acid is added the solution turns yellow/brown again.
What is happening? Note: Br2(aq) + H2O(l) <----> HOBr2(aq) + Br2 -(aq) + H+
2. Potassium chromate is initially yellow. When sulfuric acid is added the solution turns orange. Then when sodium hydroxide is added the solution turned back yellow.
What is happening? Note: CrO4 2-(aq) + 2H+(aq) <-----> Cr2O7 2-(aq) + H2O(aq)
3. Ethyl acetoacetate and water are clear initially, then when iron(III) chloride is added the solution turned deep purple. Then when bromine water was added the solution turned clear. Over time the deep purple color reappeared in the solution.
What is happening? Note: Ethyl acetate exists as an equilibrium mixture of two forms(called tautomers) -when iron(III) chloride is added to ethylaccetoacetate the solution becomes coloured because of a reaction between the enol form and the iron(III) chloride. The enol form of ethylacetoacetate also reacts with bromine water.
For questions 1 and 2 you need to know that sodium hydroxide (NaOH) is a strong base (produced OH- ions) while sulfuric acid is a strong acid (produces H+ ions).
Q1) When NaOH is added, the OH- derived from this base reacts with the H+ that already exist in the equation to form water (H2O). Therefore, all of a sudden it is as if you are adding more water to the reaction and according to Le Chatelier's principle the reaction is shifted to the right, which results in production of more aqueous bromine and hence the yellow color. On the other hand when a strong acid is added it is as if you are ...
The solution discusses Le Chateliers Principle.
Le Chateliers Principle Lab
1 1mL Ba(NO3)2 + 0.5mL K2CrO4 + drops HCl? + drops NaOH? "Ba(NO3)2=clear and colourless
Ba(NO3)2 + K2CrO4= yellow turbid solution (precipitate)
HCl= clear yellow solution (precipitate dissolved)
NaOH= yellowish turbid solution (precipitate returns by neutralizing acid) "
2 1mL Ba(NO3)2 + 0.5mL K2Cr2O7 + 10 drops HCl "Ba(NO3)2=clear and colourless
Ba(NO3)2 + K2Cr2O7= orange-yellow slightly cloudy solution (cloudy d/t reaction with distilled water))
HCl= clear yellowish solution (no precipitate) "
3 1mL Ba(NO3)2 + 5 drops HCl + 5 drops K2CrO4 "Ba(NO3)2=clear and colourless
Ba(NO3)2 + HCl= clear and colourless (acidified Barium Nitrate)
+K2CrO4= yellow clear solution (No precipitate d/t rapid conversion to dichromate) "
4 1mL Ba(NO3)2 + 5 drops CH3CO2H + 5 drops K2CrO4 "Ba(NO3)2=clear and colourless
Ba(NO3)2 + CH3CO2H (acetic acid)= clear and colourless
+K2CrO4= yellow cloudy solution (precipitate formed b/c acetic acid is a weak acid) "
1) Explain what happened in each test regarding Le Chateliers principle
2) When few drops if sodium fluoride solution are added to a solution containing iron (III) ions a colourless solution is obtained. However, when a few drops if sodium chloride are added to a solution
Fe³⁺(aq) + 6F‾ (aq) ⇋ FeF₆³‾(aq)
(pale yellow) (colourless)
containing iron (III) ions a pale yellow solution is obtained, with the yellow colour deepening as more sodium chloride is added.
Fe³⁺(aq) + 4Cl‾ (aq) ⇋ FeCl₄‾(aq)
(pale yellow) (Intense yellow)
Predict what will occur, if a solution of sodium fluoride is added to the solution prepared from the iron (III) and sodium chloride solutions. Justify your prediction in terms of Le Chatelier's principle.