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 6.00 L of blood with a Fe2+ concentration of 3.10×10⁻⁵ M . If a person ingests 6.00 mL of 18.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Answer 1

Given that:

The average human body contains 6.00 L of blood with 3.10×10⁻⁵ M Fe²⁺ concentration.

Volume and concentration of NaCN to ingest is 6.00 mL 18.0 mM respectively.

moles of Fe²⁺ = 6.00 L x 3.1 x 10^-5 M= 1.86 x 10^-4

moles of CN^- = 6.00 x 10^-3 L x 0.0180 M= 0.000108

Fe²⁺ + 6 CN^- = [Fe(CN)₆]^4-

moles of Fe²⁺ sequestered = 0.000108 /6=0.000018

% of Fe²⁺ sequestered = (0.000018 x 100)/ 1.86 x 10^-4 = 9.67

9.67% of iron(II) in the blood would be sequestered by the cyanide ion.