Question

The average human body contains 5.70 L of blood with a F e 2+ concentration of 3.10×10−5 M . If a person ingests 6.00 mL of 11.0 mM NaCN , what percentage of iron(II) in the blood would be sequestered by the cyanide ion? Express the percentage numerically.

Answer 1

Molarity = no. of moles / volume of solution in litre

So number of moles of Fe²⁺ = 3.10 x 10^-5 x 5.70

= 1.767 x 10^-4 moles

NaCN = Na⁺ + CN^-

So number of moles of NaCN = number of moles of CN^-

number of moles of NaCN = 0.006 x 11.0 x 10^-3

= 6.6 x 10^-5 moles

Now each Fe²⁺ ion needs 2 CN^- ions to make FeCN2.

Since the CN^- available are 6.6 x 10^-5 moles , so Fe²⁺ ions sequestered by it will be =

( 6.6 x 10^-5) / 2 = 3.3 x 10^-5 Fe²⁺ moles

Percentage = multiplied by 100

= 18.68 %