Question

A sufficient amount of NaCN was added to 0.015M AgNO₃ to give a solution that was initially 0.100M CN^-. What is the concentration of Ag⁺ in this solution after Ag(CN)₂^- forms? The formation constant K_f for the complex ion Ag(CN)₂^- is 5.6x10¹⁸.

Answer 1

The formation constant is the equilibrium constant for the reaction to form the complex, which for this complex is:

Ag+ + 2CN- --> Ag(CN)2-

Kf = [ Ag(CN)2- ] / { [Ag+] [CN-}^2 }

Once you mix the two solutions, the volume is doubled, so each starting concentration is half in the mixed solution. This means [Ag+] initial = 0.01 M and [CN-] initial = 0.10 M. Starting there, you can make a few assumptions.

Since Kf is very large and Ag+ is the limiting reactant, essentially all of it is converted to complex. Since 1 mole of Ag+ that reacts forms one mole of complex, the concentration of complex will be essentially 0.010 M after the reaction reaches equilibrium.

Since nearly all the Ag+ is converted to complex, and two moles of CN- react for every mole of Ag+ that reacts, the ending concentration of CN- will be 0.1-2(0.01) = 0.08 M.

Set the ending concentration of Ag+ to the variable z, plug in the values for each species in the equation for Kf, then solve for z:

Kf = 5.3 x 10^18 = 0.01 / z(0.08)^2

z = 2.95 x 10^-19.