Question

The average human body contains 5.70 L of blood with a Fe2+ concentration of 1.50×10−5 M . If a person ingests 12.0 mL of 13.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

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

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

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

where Kf=4.21×10^45

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.

Answer 1

% of Fe²⁺ sequestered = 30.4 %

Explanation

Concentration of Fe²⁺ = 1.50 x 10^-5 M

volume of blood = 5.70 L

moles of Fe²⁺ = (Concentration of Fe²⁺) * (volume of blood)

moles of Fe²⁺ = (1.50 x 10^-5 M) * (5.70 L)

moles of Fe²⁺ = 8.55 x 10^-5 mol

Concentration of NaCN = 13.0 mM = 13.0 x 10^-3 M

volume of NaCN = 12.0 mL = 12.0 x 10^-3 L

moles of NaCN = (Concentration of NaCN) * (volume of NaCN)

moles of NaCN = (13.0 x 10^-3 M) * (12.0 x 10^-3 L)

moles of NaCN = 1.56 x 10^-4 mol

moles of Fe²⁺ sequestered = (moles of NaCN) * (1 mole Fe²⁺ / 6 moles CN^-)

moles of Fe²⁺ sequestered = (1.56 x 10^-4 mol) * (1 / 6)

moles of Fe²⁺ sequestered = 2.6 x 10^-5 mol

% of Fe²⁺ sequestered = (moles of Fe²⁺ sequestered / initial moles Fe²⁺) * 100

% of Fe²⁺ sequestered = (2.6 x 10^-5 mol / 8.55 x 10^-5 mol) * 100

% of Fe²⁺ sequestered = (0.304) * 100

% of Fe²⁺ sequestered = 30.4 %