Question

A 110.0 −mL sample of a solution that is 2.8×10⁻³ M  in AgNO3  is mixed with a 230.0 −mL sample of a solution that is 0.14 M in NaCN. A complex ion forms.

Question:   After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

Answer 1

Ag^+ + 2CN^- ? [Ag(CN)2]^-
Kform = [[Ag(CN)2]^-]/[Ag^+][CN^-]^2 = 5.3×10^18
Before rxn (using M1V1 = M2V2)

[Ag^+] = (110.0)/(340)×(2.8×10^-3) = 9.0588 ×10^-3
[CN^-] = (230/340)×0.14 = 0.0947 M
At equilibrium let [Ag^+] = x [[Ag(CN)2]^-]

= (9.0588 ×10^-3 -x) ? 9.0588 ×10^-3 this is because x is so small which with experience you can tell from Kform.

To form 9.0588 ×10^-3 M of Ag(CN)2]^- requires
2× 9.0588 ×10^-3 or 0.0181 M of CN^-.

Uncoordinated [CN^-] = 0.0947 - 0.0181 = 0.0766M
Put these values into K form = (9.0588 ×10^-3)/x(0.0766)^2 = 6.93×10^18
x = (9.0588 ×10^-3)/[(0.0766)^2 × (6.93×10^18)] = 8.19×10^-16M
Please Check my mathematic.