Chemistry HElp? solubility help?

Question

a) Calculate the solubility of silver chloride in 10.0 M ammonia given the following informa$on:

Ksp (AgCl) = 1.6 × 10–10
Ag+ +NH3<-> AgNH3+ K=2.1×103

AgNH3+ + NH3 <->Ag(NH3)2+ K = 8.2 × 103

 

b)Calculate the concentra$on of NH3 in the final equilibrium

mixture.

 

 

Answer 1

(a)

Step 1:

Concentration of silver chloride is [AgCl] = 10.0 M.

The solubility product of the two reactions are as follows.

                 Solubility product .

             Solubility product: .

     Solubility product: .

So the overall concentration of silver [Ag] is sum of the concentration of the silver used in the three reactions.

The overall reaction is .

Solubility product of silver cholride in the overall reaction is .

Solubility product is .

Step 2:

Let x be the concentration and .

Then the concentration of the  is 10-2x.

Solubility product .

The solubility of the silver chloride in 10.0 M ammonia is 0.475 M.

Solution :

The solubility of the silver chloride in 10.0 M ammonia is 0.475 M.

Answer 2

(b)

Step 1:

Concentration of silver chloride is [AgCl] = 10.0 M.

The overall reaction is .

Let x be the concentration and .

Then the concentration of the is 10-2x.

The solubility of the silver chloride in 10.0 M ammonia x is 0.475 M.         (From (a))

Concentration of the is .

Concentration of the is 9.05 M.

Solution :

Concentration of the is 9.05 M.