Question

The average human body contains 5.10 L of blood with a Fe2+ concentration of 3.20×10−5 M . If a person ingests 11.0 mL of 25.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Answer 1

Ans. Fe²⁺ + 6 CN^- ---------> [Fe(CN)₆]^4-    

Assuming the above reaction holds true in blood.

Give, [Fe²⁺] = 3.20 x 10^-5 M = 3.20 x 10^-5 moles per L

Total amount of Fe²⁺ in blood = molarity of Fe²⁺ x volume of blood

                                    = 3.20 x 10^-5 moles L^-1 x 5.10 L = 16.32 x 10^-5 moles

Amount of CN^- in 11.0 mL of 25.0 mM NaCN = vol. of solution in L x molarity of NaCN

                                    = (0.011 L) x 25.0 x 10^-3 moles L^-1         

                                    = 0.275 x 10^-3 moles.

According to the stoichiometry of reaction, 6 moles of NaCN reacts with 1 mole of Fe²⁺. Thus, maximum amount of Fe²⁺ bound with CN^- = (1/6) x moles of NaCN

                                                = (1/6) x 0.275 x 10^-3 moles

                                                = 0.0458 x 10^-3 moles = 4.58 x 10^-5 moles.

Thus, the given amount of NaCN binds to a maximum of 4.58 x 10^-5 moles of Fe²⁺. That is amount of Fe²⁺ sequestered by CN^- = 4.58 x 10^-5

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

                                    = [ (4.58 x 10^-5) / (16.32 x 10^-5 moles) ] x 100

                                    = 28.06 %