One of my questions asks, when can you conclude about the relative equillibrium concentrations of CrO4^2- in each of the colutions 0.1 M K2Cr2O7 and 0.1 M K2CrO4?

Im really confused by this question. Could someone clarify it for me? Thanks :)

K2Cr2O7 is red-orange.

K2CrO4 is yellow. The CrO4^-2 is yellow. So there is little CrO4^- in K2Cr2O7 solution.

So is K2CrO4 the equillibrium concentration of CrO4-2 is high whereas in K2CrO7 the CrO4-2 equillibrium concentration is low?

yes

Thank You :)

Certainly! The question is asking about the relative equilibrium concentrations of CrO4^2- in two different solutions: 0.1 M K2Cr2O7 and 0.1 M K2CrO4.

To answer this question, we need to understand the balanced chemical equations for the dissociation of K2Cr2O7 and K2CrO4 in water.

K2Cr2O7 dissociates in water as follows:
K2Cr2O7 → 2K+ + Cr2O7^2-

K2CrO4 dissociates in water as follows:
K2CrO4 → 2K+ + CrO4^2-

From these equations, we can see that both K2Cr2O7 and K2CrO4 produce the CrO4^2- ion. However, the concentration of CrO4^2- in the 0.1 M K2Cr2O7 solution will be influenced by the equilibrium position of the reaction, which is determined by the stoichiometry of the reaction and the initial concentration of the reactants.

In the case of K2Cr2O7, the equilibrium position will be influenced by the concentration of CrO4^2- produced by the dissociation of the compound. Since the reaction produces Cr2O7^2-, there will be a shift in the equilibrium to form more CrO4^2- ions, resulting in a lower relative equilibrium concentration of CrO4^2-.

On the other hand, in the case of K2CrO4, the dissociation directly produces CrO4^2- ions. Therefore, there is no additional shift in equilibrium, and the relative equilibrium concentration of CrO4^2- will be higher compared to the K2Cr2O7 solution.

In conclusion, based on the stoichiometry and equilibrium position of the reactions, we can conclude that the relative equilibrium concentration of CrO4^2- will be higher in the 0.1 M K2CrO4 solution compared to the 0.1 M K2Cr2O7 solution.