Question

The solution you will use for this experiment was prepared by making a 0.25 M CoCl2 solution in a solvent that already contains 4.0 M NaCl. Recall that as an ionic compound, CoCl2 will dissociate into Co(II) and chloride ions. What will the concentration of chloride ions be?

2. If all of the Co(II) ions described in the previous problem form the [Co(H2O)6 ] 2+ complex, what will its initial concentration be?

3. Suppose that the initial concentration of [CoCl4 ] 2－ is 0, and that it increases to some value “x” by the time the reaction reaches equilibrium. Write an expression in terms of “x” for the [Co(H2O)6 ] 2+ and chloride ions.

4. Write an equation for the equilibrium constant for the reaction in terms of “x.”

Answer 1

1. First off, you have to remember that both your compounds are ionic compounds. You will assume complete dissociation. Therefore, from the NaCl solution, you would get:

NaCl ---> Na⁺ + Cl^-

As the reaction ratio is 1:1 for the chloride, you get 4M concentration of Cl^-.

For the second compound:

CoCl₂ ---> Co²⁺ + 2Cl^-

The ratio for the chloride in this reaction is 1:2. Given so, you would have twice the concentration of CoCl₂, for Cl^-, which would be 0.5 M.

This gives a total of 4.5 M of chloride ions.

2. Given that you are considering all the Co is used to form the complex ion, you would get a 0.25 M concentration.

3. The reaction that is taking place in this problem is

(Co(H2O)₆)²⁺ + 4Cl⁻ <---> CoCl₄^2- + 6H₂O

0.25 M 4.5M        0 0 initial

0.25 - x 4.5 - 4x x x equilibrium

4. The equilibrium constant for this reaction is obtained by the following expression: