Question

A 115.0 −mL sample of a solution that is 3.0×10−3 M in AgNO3 is mixed with a 225.0 −mL sample of a solution that is 0.10 M in NaCN. After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

Answer 1

moles of Ag+ in the solution = M1xV1 = 3.0x10^-3 mol/L x 0.115 L = 3.45x10^-4 mol

moles of CN- in the slution = M2xV2 = 0.1 mol/L x 0.225 L = 0.0225 mol

Total volume of the solution after mixing = 0.115 L + 0.225 L = 0.340 L

Hence concentration of Ag+ after mixing = 3.45x10^-4 mol / 0.340 L = 0.001015 M

concentration of CN- after mixing = 0.0225 mol / 0.340 L = 0.0662 M

After mixing Ag+ reacts with CN- to form [Ag(CN)₂]^-(aq). The balanced reaction is

------------Ag+(aq) + 2CN-(aq) -------- > [Ag(CN)₂]^-(aq) ; Kf = 1.0x10²¹  

init.mol: 3.45x10^-4, 0.0225 ----------------- 0

change: - 3.45x10^-4, - 2x3.45x10^-4, ----- +3.45x10^-4

eqm mol: 0, ------- 0.02181 mol, -------- 3.45x10^-4 mol

Since the Kf value is very high almost all of the Ag+(aq) is converted to  [Ag(CN)₂]^-(aq).

At equilibrium, [CN-(aq)] = 0.02181 mol / 0.340 L = 0.06415 M

[[Ag(CN)₂]^-(aq)] = 3.45x10^-4 mol / 0.340 L = 0.001015 M

Also

Kf = 1.0x10²¹ =   [[Ag(CN)₂]^-(aq)] / ([CN-(aq)]² x [Ag+(aq)] = (0.001015 M) / (0.06415 M)²x [Ag+(aq)]

=>  [Ag+(aq)] =  (0.001015 M) / [(0.06415 M)²x1.0x10²¹] = 2.47x10^-22 M (answer)