Fe 2aq 6cn- aq --gtlt-- fecn6 4- aq where kf 4.21 x 10 , Ask an Expert

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Fe 2+(aq) + 6CN- (aq) -->/<-- [Fe(CN)6] 4- (aq) where Kf = 4.21 x 10 ^ 45 The average human body contains 5.30 L of blood with a Fe 2+concentration of 2.70×10 -5 M. If a person ingests 9.00 mL of 14.0mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?

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