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.

Part A The average human body contains 5.80 L of blood with a Fe2+ concentration of 2.10×10−5 M . If a person ingests 9.00 mL of 10.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion? Express the percentage numerically.

Answer 1

number of moles of Fe2+ = 2.10×10⁻⁵ M x 5.80 L

                                          = 1.218 x 10⁻⁴ mol

volume of NaCN = 9.00 mL convert in to liters = 0.009L

10.0 milli molar convert in to molar = 0.010M

moles of NaCN = 0.010 M x 0.009 L = 9 x 10^-5 mole

NaCN === Na+ + CN-

means one moles of NaCN = moles of CN-

now let see the reaction between Fe2+ and CN-

Fe²⁺ + 6 CN^- = [Fe(CN)₆] ^4-

moles Fe2+ sequestered = 9 x 10^-5 /6= 1.5 x 10^-5

% =(1.5 x 10^-5/ 1.218 x 10⁻⁴) 100

percentage of iron(II) =12.32 %