When species combine to produce a coordination complex, the equilibrium constant

Question

When species combine to produce a coordination complex, the equilibrium constant for the reaction is called is the formation constant K_f. For example, the iron (II) ion, Fe^2+, can combine with the cyanide ion, CN^-, to form the complex [Fe(CN)_6]^6- according to the equation. Fe^2+ (aq) + 6CN^- (aq) [Fe(CN)_4]^4- (aq) where K_1 = 4.21 times 10^46. This reaction is what makes cyanide so toxic to human beings and other animals. The cyanide ion binds to the ion, that red blood cells use to carry oxygen around the body, thus interacting 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 teaching gold out of low-grade ore. Part A The average human body contains 6.40 L of blood with a Fe^2+ concentration 2.50 times 10^-5 mol L^-1. If a person ingests 10.0 mL of 15.0 mmol L^-1 NaCN, what percentage of ion(II) is the blood would be sequestered by the cyanide ion? At what pH will precipitation of AI(OH)_3 begin if 2.00 kg of aluminum sulfate, Al_2(SO_4)_1 is added to 5150 L of water (with a negligible change in volume)?

Explanation / Answer

Part A)

1 mole of Fe2+ reacts with 6 moles of CN-

moles of Fe2+ present = 2.5 x 10^-3 M x 6.40 L = 0.016 mol

moles of CN- added = 15 x 10^-3 M x 0.01 L = 1.5 x 10^-4 mol

moles of Fe2+ reacted = 1.5 x 10^-4/6 = 2.5 x 10^-5 mol

%Fe sequestered = 2.5 x 10^-5 x 100/0.016 = 0.156%

Part B) Ksp for Al(OH)3 = 4 x 10^-15

4 x 10^-15 = [Al3+][OH-]^3

[OH-] = cube rt.(4 x 10^-15/(2.90 x 10^3/342.151 x 5150))

         = 1.34 x 10^-4 M

[H+] = 1 x 10^-14/1.34 x 10^-4 = 7.46 x 10^-11 M

pH = -log[H+] = 10.13 all of Al(OH)3 would precipitate out of solution

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