Question

A 115.0 −mL sample of a solution that is 2.6×10⁻³ M in AgNO3 is mixed with a 230.0 −mL sample of a solution that is 0.14 M in NaCN .

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

Answer 1

Moles of AgNO3 taken = MxV = 2.6x10^-3 M x 0.115 L = 2.99x10^-4  mol

Moles of NaCN taken = MxV = 0.14 M x 0.230 L = 3.22x10^-2 mol  

AgNO3 and NaCN are completely dissociated in water and when they are mixed Ag+ and CN- react with eachother to form [Ag(CN)₂]^-. The balanced chemical reaction is

Ag+(aq) + 2CN-(aq) ------ > [Ag(CN)₂]^-(aq) , Kf = 3.0 x 10²⁰

1 mol, ----- 2 mol,--------------- 1 mol

1 mol of Ag+(aq) reacts with 2 mol of CN-(aq).

Hence 2.99x10^-4  mol that will react with the moles of CN-(aq) = (2 mol CN- / 1 mol Ag+) x 2.99x10^-4 mol Ag+

=  5.98x10^-4 mol CN-(aq)

Hence Ag+(aq) is the limiting reactant and will react almost completely to form [Ag(CN)₂]^-(aq) as the formation constant of [Ag(CN)₂]^-(aq) is very high.

Also moles of [Ag(CN)₂]^-(aq) produced after the reaction = 2.99x10^-4  mol

Total volume after mixing, Vt = 0.115 L + 0.230 L = 0.345 L

Hence concentration of [Ag(CN)₂]^-(aq) in the solution = moles of  [Ag(CN)₂]^-(aq) / Vt = 2.99x10^-4 mol /  0.345 L

= 8.67x10^-4 M

Moles of CN-(aq) remained after the reaction = 3.22x10^-2 mol - 5.98x10^-4 mol = 3.16x10^-2 mol  CN-(aq)

Hence concentration of  CN-(aq) after the reaction, [CN-(aq)] = 3.16x10^-2 mol  CN-(aq) / Vt

= 3.16x10^-2 mol / 0.345L = 9.16x10^-2 M

Now for the chemical reaction

Ag+(aq) + 2CN-(aq) ------ > [Ag(CN)₂]^-(aq) , Kf = 3.0 x 10²⁰

Kf =  3.0 x 10²⁰ =  [[Ag(CN)₂]^-(aq)] / [Ag+(aq)]x[CN-(aq)] =  8.67x10^-4 M / ( [Ag+(aq)] x(9.16x10^-2 M)²)

=>  [Ag+(aq)] = 8.67x10^-4 M / { 3.0 x 10²⁰ x (9.16x10^-2 M)² } = 3.44 x 10^-22 M (answer)