I calculated mols Fe^2+ as 6.2L x 2.4E-5 = 1.49E-4
CN^- 0.008L x 0.013 M = 1.04E-4
With such a large Kf I assumed equilibrium conditions would be FAR FAR to the right.
..........Fe^2+ + 6CN^- ==> [Fe(CN(6]^-4
initial..1.49E-4 mols
add..............1.04E-4mols
change.1.73E-5...1.04E-4.....1.73E-5
equil...?
% iron complexed = (1.73E-5/1.49E-4)*100 = 11+ %?
The average human body contains 6.20 L of blood with a concentration of 2.40×10−5 M . If a person ingests 8.00 mL of 13.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion?
According to the equation:
Fe2+ + 6CN- ->[Fe(CN)6]4-
kf = 4.21*10^45
3 answers
Thank you! I got it.
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