Answer
Please see the work below.
Work Step by Step
We know that
$Cu^{+2}+2CN^{-}\rightarrow \leftarrow Cu(CN)_2^{-}$
Thus the expression for $K_f$ is
$K_f=\frac{[Cu(CN)_2^-]}{[Cu^{+}][CN^-]^2}=1.0\times 10^{16}$
General Chemistry 10th Edition
by Ebbing, Darrell; Gammon, Steven D.
Published by
Cengage Learning
ISBN 10:
1-28505-137-8
ISBN 13:
978-1-28505-137-6
Chapter 17 - Solubility and Complex-Ion Equilibria - Questions and Problems - Page 736: 17.61
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