The average human body contains 5.70 L of blood with a Fe2+ concentration of 3.2

Question

The average human body contains 5.70 L of blood with a Fe2+ concentration of 3.20×105 M . If a person ingests 7.00 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, 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.)

Explanation / Answer

The moles of Fe+2 in the body = 5.7 L*3.2*10^-5 = 1.82*10^-4 moles.

The moles of CN- ion = 0.007*1.3*10^-2 = 9.1*10^-5 moles/L

It takes 6 CN- ions to combine with one Fe+2 ion so the number of Fe+2 ions tied up is as follows

= 9.1*10^-5 moles/ 6 = 1.52*10^-5

The percentage of Fe+2 ions tied up would be = (1.52*10^-5)/(1.82*10^-4)

                                                                           = 0.0835

                                                                           = 8.35 %

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