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, Kf. For example, the iron(ll) 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)64-(aq) where K1 = 4.21 times 1045 This reaction is what makes cyanide so toxic to human beings and other animals. The cyanide ion binds to the iron that red blood cells use to carry oxygen around the body, thus interfering 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 leaching gold out of low-grade ore. The average human body contains 6.10L of blood with a Fe2+ concentration of 1.80 times 10-5M . If a person ingests 9.00mL. of 23.0mM NaCN. what percentage of iron(ll) in the blood would be sequestered by the cyanide ion? Express the percentage numerically.

Explanation / Answer

The average human body contains 5.60L of blood with a Fe2+ concentration of 2.40×10-5 M. If a person ingests 11.0mL of 13.0 mM NaCN , what percentage of iron(II) in the blood would be sequestered by the cyanide ion? assuming that 1 CN- attaches to 1 Fe2+ and that "sequesters" the iron... moles Fe attached to CN-..... 11.0 mL NaCN x (1 L / 1000 mL) x (13.0 mmole / L) x (1 mole / 1000 mMole) x (1 mole CN- / 1 mole NaCN) x (1 mole Fe2+ / 1 mole CN-) = 1.43x10^-4 moles Fe2+ total moles Fe2+ in your blood... 5.60 L x (2.40x10^-5 moles / L) = 1.34x10^-4 moles Fe2+ % Fe attached to CN- = 100%.. there is more CN- than Fe2+ ************* Now that I've read Joshua's answer, I guess it depends on how you intrepret this problem... if you mean this... Fe2+ + 2 CN- ------> Fe(CN)2, then by all means... moles Fe attached to CN- goes by this... 11.0 mL NaCN x (1 L / 1000 mL) x (13.0 mmole / L) x (1 mole / 1000 mMole) x (1 mole CN- / 1 mole NaCN) x (1 mole Fe2+ / 2 mole CN-) = 0.72x10^-4 moles Fe2+ and you end up with 0.72 / 1.34 x 100% = 54% bound Fe.. ********* however.... Fe2+ in blood is predominantly found in hemoglobin. The chemistry of hemoglobin binding with O2 and CN- is not a chemical reaction but rather a coordination complexation. Each heme group contains 1 Fe2+ which can reversibly bind with 1 O2. It does NOT react with O2. Same goes for carbon monoxide and CN-. 1 CN- ion "binding" to that Fe2+, which it will because the Fe2+ / CN- complexation is preferred over Fe2+ / O2, will prevent the complexation of the heme Fe2+ with O2 thus "sequestering" the Fe2+ *********** so my advice to you is this..... submit either answer or both and explain your rationale then ask your instructor whether 1 CN- sequesters 1 Fe2+ or whether 2 CN- sequesters 1 Fe2+... ************ what class is this for anyway? We covered hemoglobin complexation in inorganic chemistry in college. That was a senior level course fyi. The average human body contains 6.50 L of blood with a [Fe2+] concentration of 1.40×10-5 M........? If a person ingests 8.00 mL of 15.0 mM NaCN , what percentage of iron(II) in the blood would be sequestered by the cyanide ion? moles Fe3+ = 6.50 L x 1.40 x 10^-5 M=0.0000910 moles CN- = 0.00800 L x 0.0150 M = 0.000120 Fe2+ + 6 CN- = [Fe(CN)6]4- moles Fe2+ sequestered = 0.000120/6 = 0.0000200 % = 0.0000200 x 100/ 0.0000910=22.0

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.