Chemistry question on Ionic equations

Question

A mixture is first made of 150 cm3 of a 0.200 mol.dm–3 solution of silver nitrate and ycm3 of a 2.50 mol.dm–3 solution of potassium cyanide. Thereafter, 100 cm3 of a 0.120 mol.dm─3 solution of potassium chloride is added to this mixture and the resulting solution is diluted to 2.00 dm3 . Calculate the value of y that will just prevent the precipitation of silver chloride.

Answer 1

Two table constants are needed and neither is specific to this problem, so here they are: Ksp(AgCl) = 1.8 x 10^-10 and Kf for [Ag(CN)2]- = 1.0 x 10^21. As you will see at the end, the answer barely depends on either.

The two competing reactions:

Ag+(aq) + 2 CN-(aq) <==> [Ag(CN)2]-(aq)   (very strong, ties silver up)
Ag+(aq) + Cl-(aq) <==> AgCl(s)   (the precipitation you must prevent)

Cyanide wins the competition. The job is to add just enough of it that the free silver left over is too low to reach the AgCl solubility product.

Step 1 - moles of everything, and the final chloride concentration

Ag+: 0.150 dm^3 x 0.200 mol/dm^3 = 0.0300 mol
Cl-: 0.100 dm^3 x 0.120 mol/dm^3 = 0.0120 mol
CN-: (y/1000) dm^3 x 2.50 mol/dm^3 = 0.00250y mol

Everything ends up in 2.00 dm^3, so:

[Cl-] = 0.0120 / 2.00 = 6.00 x 10^-3 mol/dm^3

Step 2 - the maximum free silver allowed

Precipitation just begins when the ion product equals Ksp:

[Ag+][Cl-] = Ksp
[Ag+] = 1.8 x 10^-10 / 6.00 x 10^-3 = 3.0 x 10^-8 mol/dm^3

Step 3 - the free cyanide that holds silver down to that level

Kf is enormous, so essentially all the silver is in the complex:

[Ag(CN)2-] = 0.0300 / 2.00 = 0.0150 mol/dm^3

Kf = [Ag(CN)2-] / ([Ag+][CN-]^2), so

[CN-]^2 = 0.0150 / ((1.0 x 10^21)(3.0 x 10^-8)) = 5.0 x 10^-16

[CN-] = 2.2 x 10^-8 mol/dm^3

Step 4 - total cyanide, and y

Cyanide is needed for two things: complexing the silver, and the free excess above.

complexed: 2 x 0.0300 = 0.0600 mol
free: 2.2 x 10^-8 x 2.00 = 4.5 x 10^-8 mol

The free amount is about a billion times smaller than the complexed amount, so it disappears in the rounding:

total CN- = 0.0600 mol

y = 0.0600 mol / 2.50 mol/dm^3 = 0.0240 dm^3

y = 24.0 cm^3

The point of the problem

Look at what actually determined the answer: the 2:1 stoichiometry of the complex, and nothing else. The equilibrium constants only served to prove that the extra cyanide needed beyond the stoichiometric amount is negligible.

That is also why the missing Ksp was not fatal. Redo Step 3 with the other commonly tabulated value, Kf = 5.6 x 10^18, and the free cyanide comes out as 3.0 x 10^-7 mol/dm^3 - ten times larger, still a millionth of a percent of the total, and y is still 24.0 cm^3. Whenever a formation constant is this large, check whether the equilibrium detail actually moves the answer before assuming you are stuck without it.