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In: Chemistry

The average human body contains 5.10 L of blood with a Fe2+ concentration of 3.20×10−5 M...

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?

Solutions

Expert Solution

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 %


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