what is the concentration of CN-1(aq) that is still in solution when it reaches eq??

Question

Iron(II) nitrate and hydrocyanic acid react to form the hexacyanoferrate(II) ion, [Fe(CN)₆]^-4(aq), with a formation constant of 3.284 x 10³¹.

A.) If you add 134.787g of iron(II) nitrate solid to 760.0mL of 0.8434M CN^-1(aq) solution, what is the concentration of CN^-1(aq) thats still in solution when the reaction reaches equilibrium?

Answer 1

Iron(II) nitrate = Fe(NO₃)₂

Hydrocyanic acid = HCN

Formation of complex:

Fe²⁺_(aq) + 6CN^-_(aq) ==> [Fe(CN)₆]^4-_(aq)

Kf = [Fe(CN)₆]^4- / [Fe²⁺][CN-]⁶

molar mass Fe(NO₃)₂ = 179.85 g / mol

mols Fe²⁺ = 134.787 g Fe(NO₃)₂ x 1 mol / 179.85 g = 0.7494 mols Fe(NO₃)₂ = 0.7494 mols Fe²⁺

Concentration of Fe²⁺ = 0.7494 mol / 0.760 L = 0.9861

Concentration CN^- = 0.8434

Fe²⁺_(aq) + 6CN^-_(aq) ==> [Fe(CN)₆]^4-_(aq)

0.9861.......0.8434.................0...............Initial

-08434......-08434............+0.8434......Change

0.1427...........0...................0.8434........Equilibrium

Now, looking at the dissociation of the complex, we have...

[Fe(CN)₆]^4-_(aq) <==> Fe²⁺_(aq) + 6CN^-_(aq)

0.8434.......................0.1427.............0............Initial

-x...............................+x..................+6x..............Change

0.8434-x................0.1427+x.............6x.............Equilibrium

Kdissoc = 1/Kf = 1/3.284x10³¹ = 3.045x10^-32 [CN-]⁶[Fe²⁺] / [Fe(CN)₆]⁴

3.045x10^-32 = (6x)⁶(0.1427-x) / 0.8434-x

Solve for x and multiply that value by 6 to get the [CN-]