± Cyanide Poisoning For example, the iron(II) ion, Fe2+, can combine with the cy

Question

± Cyanide Poisoning

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)

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.

Part A

The average human body contains 6.20 L of blood with a Fe2+ concentration of 2.20×105M . If a person ingests 12.0 mL of 22.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Express the percentage numerically.

SubmitHintsMy AnswersGive UpReview Part

Provide FeedbackContinue

Chemistry 106 Summer 2016

Help

Close

Homework Set #6

± Cyanide Poisoning

ResourcesConstantsPeriodic Table

« previous 3 of 22 next »

± Cyanide Poisoning

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.

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.

Part A

The average human body contains 6.20 L of blood with a Fe2+ concentration of 2.20×105M . If a person ingests 12.0 mL of 22.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

Express the percentage numerically.

  % Fe2+  

SubmitHintsMy AnswersGive UpReview Part

Provide FeedbackContinue

Explanation / Answer

Moles of NaCN = 12*22*10^-6 = 2.64*10^-4

Moles of Fe+2 present in blood = MV = 6.2 * 2.20×105 = 1.364*10^-4

6 moles of CN- ion reacts with 1 mole of Fe+2

2.64*10^-4 moles of CN- ion reacts with ====>  2.64*10^-4 / 6 = 4.4*10^-5 moles of Fe+2

Moles of Fe+2 sequesterd are 4.4*10^-5

percentage of iron(II) in the blood sequesterd =100 * 4.4*10^-5 / 1.364*10^-4 = 32.25

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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