[THC]

0.050 mol FeSO₄ is added to 0.500 L 1.00 M NaCN (aq) and the following reaction occurs
Fe²⁺ + 6 CN^- -> [Fe(CN)₆]4-
K = 7.7*10^36.

Calculate [Fe²⁺].

Okay, so
K = [[Fe(CN)₆]4-]/[Fe²⁺]*[CN^-]^6
and c(Fe²⁺) = 0.100 M
and if I insert all the known values
 7.7*10^36 = x/((0.100 M-x)*(1.0 - 6x)^6)

I can't solve that equation, so I guess I should make some assumptions... but I'm not sure what. I'm used to a very low K, not a very high one.

[Borek]

What does a very high K mean? Where is the euilibrium - far to the left, or far to the right?

[THC]

What does a very high K mean? Where is the euilibrium - far to the left, or far to the right?

It means that the equilibrium is shifted far to the right... so should I assume that there is no Fe(II) left?

[Borek]

That's not the best assumption possible, as assuming it you will get directly to (wrong) answer to the original question.

But you are right - equilibrium is shifted to the right. How much complexed iron do you have?

[THC]

Hmm... a little less than 0.100 M ?

[Borek]

Try that

[THC]

I'm not sure I understand yet.

Do you mean like this:
7.7*10^36 = 0.100/((0.100 M-x)*(1.0 - 6x)^6)

[Borek]

That's not different from what you did before.

x is amount of uncomplexed iron left. That means concentration of complex is 0.1-x, but you know that x is much, much smaller then 0.1 - so you can safely ignore x...

[THC]

Ah, solved it

Thank you!