Question

When species combine to produce a coordination complex, the equilibrium constant for the reaction is called is the formation constant, Kf.

For example, the iron(II) ion, Fe2+, can combine with the cyanide ion, CN−, to form the complex [Fe(CN)6]4− according to the equation

Fe2+(aq)+6CN−(aq)⇌[Fe(CN)6]4−(aq)

where Kf=4.21×1045.

This reaction is what makes cyanide so toxic to human beings and other animals. The cyanide ion binds to the iron that red blood cells use to carry oxygen around the body, thus interfering with the blood's ability to deliver oxygen to the tissues. It is this toxicity that has made the use of cyanide in gold mining controversial. Most states now ban the use of cyanide in leaching gold out of low-grade ore.

a) The average human body contains 6.50 L of blood with a Fe2+ concentration of 2.30×10⁻⁵M . If a person ingests 12.0 mL of 23.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Answer 1

The number of moles Fe²⁺ = Molarity X volume

Moles Fe²⁺ = 2.30×10⁻⁵ X 6.5 L = 1.495 X 10^-4

moles CN- = Molarity X volume

Moles of CN- = 0.012 X 23 X 10^-3 =0.000276
The given reaction is
Fe²⁺ 6 CN- ---> [Fe(CN)₆]⁴⁻

As per stoichiometry one mole of Fe+2 will react with 6 moles of cynide ions

So for moles of CN- (0.000276) = 0.000276 / 6 moles =0.000046 = 0.46 X 10^-4

% of Fe+2 sequestered = Moles of Fe+2 reacted X 100 / moles of Fe+2 present = 0.46 X 10^-4 X 100 / 1.495 X10^-4

% of Fe⁺² = 10.70 %