Question

Using the dissociation constant,Kd=2.2×10−34, calculate the equilibrium concentrations of Co³+ and NH₃ in a 0.500 M solution of Co(NH₃)₆³⁺.

a) [Co3+]=2.2×10−12 M, [NH3]=0.500 M
b) [Co3+]=0.500 M, [NH3]=2.2×10−12 M
c) [Co3+]=4.4×10−23 M, [NH3]=0.500 M
d) [Co3+]=0.500 M, [NH3]=4.4×10−23 M

Answer 1

To calculate the equilibrium concentrations of Co³+ and NH₃ in a 0.500 M solution of Co(NH₃)₆³⁺, use the dissociation constant Kd. The correct answer is [Co³+] = 2.2 × 10⁻¹² M and [NH₃] = 0.500 M (a).

Step-by-step explanation:

To calculate the equilibrium concentrations of Co³+ and NH₃ in a 0.500 M solution of Co(NH₃)₆³⁺, we can use the dissociation constant Kd. The dissociation constant is defined as the ratio of products to reactants in a dissociation reaction. In this case, the reaction is:

Co(NH₃)₆³⁺ ⇌ Co³+ + 6 NH₃

The equilibrium constant (Kd) is equal to the concentration of the products divided by the concentration of the reactants. So, we can write:

Kd = [Co³+] / ([NH₃]⁶)

Given Kd = 2.2 × 10⁻³⁴ and [NH₃] = 0.500 M, we can solve for [Co³+]. Let's assume x is the equilibrium concentration of Co³+:

2.2 × 10⁻³⁴ = x / (0.500⁶)

Calculating x, we find that [Co³+] = 2.2 × 10⁻¹² M and [NH₃] = 0.500 M. Therefore, the correct answer is option a) [Co³+] = 2.2 × 10⁻¹² M and [NH₃] = 0.500 M.