Question

What is the concentration of free Co3+ in 1.3136e-4 M Co(NO3)3 and 1.2463 M NH3? Co3+ + 6 NH3 → [Co(NH3)6]3+ Kf = 5.000e+33

Answer 1

The concentration of free Co3+ in the given solution is 4.804e-7 M.

Step-by-step explanation:

To determine the concentration of free Co3+ in the solution, we need to consider the equilibrium of the reaction between Co3+ and NH3. The given reaction is:

`Co3+ + 6 NH3 < -- > [Co(NH3)6]3+`

The equilibrium constant (Kf) for this reaction is given as 5.000e+33. We are given the concentrations of Co(NO3)3 and NH3, which are 1.3136e-4 M and 1.2463 M, respectively. We can use the equilibrium expression to determine the concentration of free Co3+.

Since the stoichiometry of the reaction is 1:6 between Co3+ and NH3, the initial concentration of Co3+ is the same as the concentration of
`Co(NO3)3,`36e-4 M.

Let x be the concentration of free Co3+ at equilibrium, then the concentration of
`[Co(NH3)6]3`brium would be (6x)^1/6, and the concentration of NH3 at equilibrium would be 1.2463 - 6x.

Using the equilibrium expression:

`Kf = ([Co(NH3)6]3+)/(Co3+ * NH3^6)`

Substituting the known values:

`5.000e+33 = (6x)^1/6 / (1.3136e-4 * (1.2463 - 6x)^6)`

Simplifying the equation, solving for x and plugging in the given values gives:

x = 4.804e-7 M

Therefore, the concentration of free Co3+ in the given solution is 4.804e-7 M.