Dichromate Ion Equilibrium Lab Report

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PART A
I. The Chromate Ion – Dichromate Ion Equilibrium.
1.) As a result of the reaction, the equilibrium had shifted in the response to the addition of acid (H2SO4), toward the formation of orange dichromate ion. Presumably, this shift was to the right due to the fact that when hydrogen ions were added to the reactant side of the reaction, the concentration of H+ ions in the reaction had increased, resulting in a shift to the right; hence, a larger production of dichromate ion found as a product (due to the visible color change found in table 1). Ultimately, in this case the equilibrium system was subjected to a change in concentration of a reacting species and therefore the system responded by attaining a new equilibrium shift to …show more content…

Applications of the Law of Chemical Equilibrium to Copper (II) Sulfate.
1.) A. After the addition of sodium hydroxide to the copper sulfate solution, a precipitate formed, it was identified as copper (II) hydroxide.
B. Referring to table 7 trial #2 it was identified that two precipitates were formed based on the chemical reaction, CuSO4(aq) + 2NaOH(aq) ⇌ Cu(OH)2(s)+ Na2SO4(s). Based on the qualitative observations when NaOH was added the solution contained a dark blue precipitate which matches the identification of Cu(OH)2 whereas, Na2SO4(s) would appear as a white solid precipitate.

2.) A. After the second addition of NaOH, the reaction established equilibrium
B. With reference to table 7 found under results, it was visibly noted that when NaOH was added, the solution had reversed its color change after turning into a light blue liquid solution due to the presence of hydrochloric acid (favouring the reactants). This demonstrates a change in concentration of reactants and products in order to alter equilibrium. Moreover, when NaOH was added the equilibrium shifted towards the right (favouring the products). Ultimately, the reaction went forward then reverse and lastly forward again, which demonstrates an established equilibrium after the second addition of

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