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To a solution containing S2- 0.20 M, Cd2+ 0.020 M and Ni2+ 0.020, Sodium cyanide 2.0...

To a solution containing S2- 0.20 M, Cd2+ 0.020 M and Ni2+ 0.020, Sodium cyanide 2.0 M is added. Find the concentration of Cd2+ and Ni2+ without precipitating in the final solution.
pKs (CdS) = 28.0 pKs (NiS) = 24.0
Cd2+/Cd(CN)42-: log K = 17.1 Ni2+/Ni(CN)42-: log K = 30.0
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Answer #1

I'm pretty sure it's right. conc. of products over conc. of reactant. I reversed the equation b/c they did that in an example problem in the textbook. And since it's reversed Kf becomes 1/kf.

Cd(CN)42- -> Cd2+ + 4CN-

Kf = [Cd2+][CN-]4 / [Cd(CN)42-]

which = 1/6.0 x 1018

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