Question

Using the dissociation constant, Kd=1×10−44, calculate the equilibrium concentrations of Fe^3+ and CN⁻ in a 0.333 M solution of Fe(CN)₆³⁻.

a) [Fe3+]=1×10−33 M, [CN−]=0.333 M
b) [Fe3+]=0.333 M, [CN−]=1×10−33 M
c) [Fe3+]=1×10−11 M, [CN−]=0.333 M
d) [Fe3+]=0.333 M, [CN−]=1×10−11 M

Answer 1

The correct answer is that the equilibrium concentrations of Fe3+ and CN- in a 0.333 M solution of Fe(CN)6^3- are approximately [Fe3+]=1×10^-11 M and [CN-]=0.333 M, using the given dissociation constant Kd=1×10^-44. The correct option is C.

Step-by-step explanation:

To calculate the equilibrium concentrations of Fe3+ and CN− in a 0.333 M solution of Fe(CN)63−, we can use the dissociation constant (Kd). The reaction for the dissociation of Fe(CN)63− is:

Fe(CN)63− → Fe3+ + 6 CN−

Given that Kd = 1 × 10−44, we can set up the expression:

Kd = [Fe3+][CN−]6

Since the initial concentration of Fe(CN)63− is 0.333 M and assuming complete dissociation, we would have [Fe3+] = x and [CN−] = 6x. However, because Kd is so small, the extent of dissociation is minimal, therefore x is negligible compared to the initial concentration. We can then approximate that [CN−] remains 0.333 M. Plugging these into the equilibrium expression and solving for x gives:

1 × 10−44 = x × (0.333)6

[Fe3+] = x ≈ 1 × 10−11 M

[CN−] ≈ 0.333 M

The correct answer is therefore (c) [Fe3+]=1×10−11 M, [CN−]=0.333 M.