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=4.21×10^45.

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.

1. The average human body contains 6.10 L of blood with a Fe2+ concentration of 1.20×10⁻⁵M . If a person ingests 12.0 mL of 16.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Answer 1

Solution :-

Fe2+(aq)+6CN−(aq)⇌[Fe(CN)6]4−(aq)

Lets calculate the moles of Fe^2+ and CN^- using the volume and molarity of the each

Moles of Fe^2+ = 1.20*10^-5 mol per L * 6.10 L = 7.32*10^-5 mol Fe^2+

Moles of CN^- = (16.0 mM * 1 mol per L / 1000 mM)* 0.012 L = 1.92*10^-4 mol CN-1

Now lets calculate the moles of Fe^2+ reacts with CN-

1.92*10^-4 mol CN- * 1 mol Fe^2+ / 6 mol CN^- = 3.2*10^-5 mol Fe^2+

Now lets find the moles of Fe^2+ sequestered

% Fe^2+ sequestered = (3.2*10^-5 mol Fe^2+/7.32*10^-5mol Fe^2+)*100%

                                       = 43.72 %

So it causes 43.72 % of Fe^2+ to get sequestered