Help me solve this!!!!

Question

You are looking at the formation of the dicyanoargentum complex ion, [Ag(CN)2]^-1(aq). You combined 100.0mL of 0.1358M Ag⁺¹(aq) with 150.0mL of 0.1816M CN^-1(aq) and you allowed the reaction reach equilibrium. If the formation constant K_f, for this reaction is 1.935x10¹⁶, what are the concentrations of all reactants and products at equilibrium?

Answer 1

First, write down the reaction taking place for the formation of the complex ion

Ag⁺(aq) + 2CN^-(aq) <==> [Ag(CN)₂]^-1

Now, calculate the initial concentrations of Ag⁺ and CN^- in the final volume of 0.250 L

[Ag⁺] = 100.0 ml x 1 L / 1000 ml x 0.1358 mol / L = 0.01358 mol / 0.250 L = 0.05432 M

[ CN^-] = 150.0 ml x 1 L / 1000 ml x 0.1816 mol / L = 0.02724 mol / 0.250 L = 0.10896 M

The [CN^-] is twice that of [Ag⁺] which is what the stoichiometry of the balanced equation requires. So, for all practical purposes, there is no limiting reactant. And since the K_f is so large (10¹⁶), we can assume the reaction goes all the way to the right to completion. This will form 0.05432 M [Ag(CN)₂]^-1

Now we can write an ICE table for dissociation of [Ag(CN)₂]^-1

[Ag(CN)₂]^-1 <==> Ag⁺ + 2CN^-

0.05432 ...............0...........0...........Initial

-x........................+x..........+2x........Change

0.05432-x...........x.............2x.........Equilibrium

Writing the K expression, we have

K = [Ag+][CN-]² / [[Ag(CN)₂]^-1 and the K will be 1 / 1.935x10¹⁶ = 5.168x10^-17

5.168x10^-17 = (x)(2x)² / 0.05432-x and x will be very small so ignore it in the denominator

5.168x10^-17 = (x)(2x)² / 0.05432

4x³ = 2.807x10¹⁸

x³ = 7.108x10^-19

x = 8.924x10^-7 M

So, if the math is correct, and you should check it carefully, the equilibrium concentrations will be ...

[Ag⁺] = 8.924x10^-7 M

[CN^-] = 2 x 8.924x10^-7 M = 1.785x10^-6 M

[Ag(CN)₂]^-1 = 0.05432 M