Question

A 130.0 −mL sample of a solution that is 2.6×10−3 M in AgNO3 is mixed with a 220.0 −mL sample of a solution that is 0.13 M in NaCN. After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

Answer 1

The product of the reaction is NaCN(aq) + AgNO3(aq) = AgCN(s) + NaNO3(aq)

[NB: AgCN can dissolve in presence of excess CN- to form complex [Ag(CN)₂]^- and [Ag(CN)₃]^2-. The possibility of formation of complex is not considered here. To consider the complex ion formation we need to know the equilibrium constant value for the formation of complex ions.]

K_sp of AgCN = 1.6* 10^-14

No of moles of AgNO3 in solution = 0.0026*130/1000 = 0.000338 moles

No of moles of NaCN in solution = 0.13*220/1000 = 0.0286 moles

After complete precipitation,

No of moles of CN- in solution = 0.0286 - (2*0.000338) = 0.0279 moles

[CN-] = 0.0279*1000/(130+220) = 0.0797 M

AgCN dissolves to some extent to give Ag+ in solution.

[Ag+] = Ksp/[CN] = 1.6* 10^-14 / 0.0797 =2.0075*10^-13 M