Need help...not quite clear on this problem

Question

What is the concentration of CN- ion in a 0.100 molar solution of K4Fe(CN)6? Kd for (Fe(CN)6^4- is1.3 x 10^-37

this is what i got..

Fe(CN)6 -> Fe 6 CN

1.3 x 10^-37...x.....6x^6

Answer 1

Since Fe(CN)6---- (aq) ↔ Fe++ (aq) + 6 CN- (aq) [---- is -4 charge],

then Kd = [Fe++] [CN-]^6 / [Fe(CN)6----].

In going to equilibrium Fe(CN)6---- loses a certain amount, call it x. Fe++ gains the same amount, and CN- gains 6x. So at equilibrium, there is 0.100-x of Fe(CN)6----, x of Fe++, and 6x of CN-. Subtituting into the equilibrium expression:

1.3x10^-37 = (x) (6x)^6 / (0.100-x).

To simplify solving for x, assume 0.100>>x, so we can ignore x in the denominator. After simplification, the equation becomes:

1.3x10^-37 = 46,656 x^7 / 0.100.

Solving,

x = [Fe++] = 8.3x10^-7 M

6x = [CN-] = 5.0x10^-6 M (the answer).