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.

The average human body contains 6.40 L of blood with a Fe2+ concentration of 1.30×10⁻⁵M . If a person ingests 11.0 mL of 11.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Express the percentage numerically.

Answer 1

In blood

Fe⁺² moles = Concentration X volume = 1.3 X 10^-5 X 6.4 = 8.32 X 10^-5 =a

CN- moles added = 11 X 10^-3 X 11 / 1000 = 12.1 X 10^-5 = b

Kf is very high so they will react completely as per the stoichiometry

Let x moles of complex are formed

The reaction is

                 Fe²⁺(aq)         +        6CN⁻(aq)        ⇌     [Fe(CN)₆]⁴⁻(aq)

So each mole of Fe+2 will react with 6 moles of CN-

Moles of CN- = 12.1 X 10^-5 , they will react with 12.1 X 10^-5 / 6 moles of Fe+2 = 2.017 X 10^-5 moles of Fe+2

% of Fe+2 sequestered by CN- = Moles of Fe+2 used X 100 / Initial moles = 2.017 X 10^-5 X 100 / 8.32 X 10^-5

%of Fe+2 sequestered = 24.24 %