Question

The following equilibrium is established when copper ions and bromide ions are placed in solution.

heat+ Cu(H₂O)₆⁺² + 4Br⁻ --> 6H₂O + CuBr₄⁻²

the tube on the left contains only copper sulfate dissolved in solution. the tube on the right is the result of adding some potassium bromide solution. given that the Cu(H₂O)₆⁺² ion is blue and that the CuBr₄⁻² ion is green, what happened to the concentration of each of the ions when the kbr was added?
a. equilibrium shifted to the left, causing [reactants) to increase
b. equilibrium shifted to the right, causing [Br⁻] to increase
c. [Br⁻] increased, causing [products) to decrease
d.[Br⁻] increased, causing [products) to increase

Answer 1

The correct answer is option b.) Equilibrium shifted to the right, causing [Br⁻] to increase.

When potassium bromide (KBr) is added to the solution, it introduces bromide ions (Br⁻) into the system. Looking at the given equilibrium:

`\[ \text{Cu(H}_2\text{O)}_6^(2+) + 4\text{Br}^- \rightleftharpoons 6\text{H}_2\text{O} + \text{CuBr}_4^(2-) \]`

The blue Cu(H₂O)₆²⁺ ion is on the left (reactant side), and the green CuBr₄²⁻ ion is on the right (product side). When KBr is added, it provides additional Br⁻ ions.

The equilibrium will shift to the right, favoring the formation of products (CuBr₄²⁻), as the system tries to counteract the increase in Br⁻ concentration. This results in an increase in the concentration of products and a decrease in the concentration of reactants.

So, the correct answer is:

b. Equilibrium shifted to the right, causing [Br⁻] to increase.