Question

Given the Ksp = 1.2 x 10-12 for Ag2 CrO4 and K+ = 1.7 x 107 for (Ag(NH3)21, calculate the equilibrium constant for Ag2CrO4(s) + 4 NH3(aq) + 2 Ag(NH3)21+ (aq) + Cr042-aq) and explain whether the reaction favors reactants or products at equilibrium? Show work. TTT Arial 3 (12pt) • T. E. 3. 's Path:p Words:0

Answer 1

We can write the given reaction as the sum of two reactions: the reaction of dissociation of Ag2CrO4 and the reaction of formation of [Ag(NH3)2]+.

Dissociation:

Ag2CrO4 (s) ---------> 2Ag+ (aq) + CrO4 2- (aq) ---------------------------------(1)

When a solid substance dissolves in an aqueous solution, the equilibrium constant is given by the solubility product.

So, for this reaction, Ksp = 1.2 x 10^-12. (Given in the question)

Formation:

Ag+ (aq) + 2NH3 (aq) ----------> [Ag(NH3)2]+ (aq) ---------------------------------(2)

For the formation of a complex in a solution, the equilibrium constant is given by the fomation constant. So, for this reaction, Kf = 1.7 x 10^7. (Given in the question)

To get the final equation, we can add equation (1) and 2 x equation (2).

Here we multiply the second equation by 2 so that we can get the exact final equation as shown below.

Two important rules to be kept in mind are:

• When a reaction is multiplied by a number 'n', its equilibrium constant is raised to the nth power.
• When two reactions at equilibrium are added, then the equilibrium constant of the resulting reaction is the product of the equilibrium constants of those two reactions.

Ag, C904 (8) ► 2 Agt(aq) + C04?" (aq) Kep = 1-2 *1012 24gHaq) + 4Ning (aq) → 2 [ag (artis](aq) that 416 7 x 107 )* Ag, G7Oy (5) +42113 (aq) →6042+(aq) + 2 [Ag (NX),]* Cag) Keq Kop Kg ²

So, the overall reaction obtained here is same as the reaction given in the question. The overall equilibrium constant is

Keq = Ksp x Kf^2 = 346.8

If Keq > 1, then products are favoured, if Keq < 1, then reactants are favoured, and if Keq = 1, then neither reactants nor products are favoured.

As in this case, Keq is much greater than 1, so the reaction is product favoured.