# Chemistry

The formation constant of [M(CN)6]4– is 2.50 × 1017, where M is a generic metal. A 0.170-mole quantity of M(NO3)2 is added to a liter of 1.31 M NaCN solution. What is the concentration of M2 ions at equilibrium?

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asked by Nat
1. You do these by noting that the formation constant is huge so that the complex essentially forms at 100%, then turn the reaction around as if it were a weak acid and calculate the (M^2+).
..........M^2+ + 6CN^- ==> M(CN)6^4-
I.....0.170M....1.31M........0
C....-0.170...-6*0.170.....+0.170
E........0.....0.290........0.170
Of course the M^2+ is not zero but it is a very small number. Also, of course the complex is not quite 0.170 but it is very close to that. So now we say we have a complex of 0.170 with 0.290M CN^- and ask what the (M^2+) is.
-------------------
I.........0......0.290.........0.170
C.........x......+6x............-x
E.........x......6x+0.290......0.170-x

Kf = [M(CN)6]^4-/(M^2+)(CN^-)^6
2.5E17 = (0.170-x)/(x)(0.290+6x)^6
Solve for x = (M^2+)

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posted by DrBob222

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