Question

A 110.0 −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

from the given data,

[Ag⁺] = 110 x 0.0026 / 1000 = 0.000286 M

[CN^-] = 220 x 0.13 / 1000 = 0.0286

When 0.000286 moles of Ag⁺ is reacted with 0.0286 moles of CN^-, essentially 0.000286 moles of [Ag(CN)₂]^- ion should be obtained.

It is diluted to 330 mL of solution

So, [Ag(CN)₂]^- = 0.000286 mol/0.330 L= 0.000867 M

Remaining number of moles o CN^- = 0.0286 - 2 (0.000286) = 0.028028 mol

New, [CN^-] = 0.028028 / 0.330 = 0.08493 M

Formation constant of [Ag(CN)₂]^- is = 1 x 10²¹

Ag⁺ (aq.) + 2 CN ^- (aq.) <---------> [Ag(CN)₂] ^- (aq.)

K_f = [Ag(CN)₂] ^- / [Ag⁺][CN^-]²

1 x 10²¹ = 0.000867 / [Ag⁺](0.08493)²

[Ag⁺] = remaining concentration at equilibrium = 1.2 x 10 ^-22 M