[125wewe]

Here is my question:
A sample of AgCl was treated with 5ml of 2M Na₂CO₃ solution to produce Ag₂CO₃. The remaining solution contained 0.003 g of Cl^- (Chloride Ion) per litre. Calculate the solubility product of AgCl.(Ksp of Ag₂CO₃ = 8.2 *10_-12).

[125wewe]

Please help.

[Borek]

You have to show your attempts at solving the question to receive help. This is a forum policy.

Please list all related equations that can be used when solving this type of the problem.

[125wewe]

I calculated the concentration by dividing gm/litre by 35.5 to give molarity. But, I am now stuck at calucating the solubility product of AgCl. This is all I could do.
Relevent Equations:
ksp of AgCl=[Ag⁺]*[Cl^-]

[AWK]

You have two solubility equilibriums in your solution: AgCl and Ag₂CO₃. Concentration of Ag+ is the same in both equilibriums.

[125wewe]

So, 0.01 moles of Na₂CO₃ produces 0.01 moles of Ag₂CO₃ in 5ml solution. I think the solubility of Ag₂CO₃ must be same as that of Na₂CO₃.
Then,
ksp of Ag₂CO₃ = [2xSolubility]²x[Solubility]
or, Solubility = 1.270334x10^-4 == Concentration of [Ag⁺]
Now,
[Cl^-]=0.003/35.5 M = 8.45 x 10^-5
Therefore, I think Ksp of AgCl should be:
Ksp = 1.270334x10^-4 x 8.45 x 10^-5 = 1.073 X 10^-8.

Is this correct?
I wonder why concentration and volume of Na₂CO₃ was given.

[Borek]

So, 0.01 moles of Na₂CO₃ produces 0.01 moles of Ag₂CO₃ in 5ml solution.

That would mean much higher concentration of Cl^-.

I think the solubility of Ag₂CO₃ must be same as that of Na₂CO₃.

Impossible - one is well soluble, the other is weakly soluble, you should know that.

Try to describe - in words - what is happening when you add the carbonate solution to the solid AgCl.

[125wewe]

I am little confused with Solubility product.
To describe what is happening with it, AgCl must have been excess. So, when Na2CO3 is added Ag2Co3 is formed along with excess AgCl that remained in solution.

That's all I got. Please show the solution and I will try to understand and write my understanding here.

[AWK]

AgCl must have been excess

Solid AgCl is in the excess.
And CO₃^2- is in the large excess.

[125wewe]

And CO32- is in the large excess.

How did you get that? If AgCl was excess, shouldn't all Na2CO3 be consumed?

[Borek]

AgCl is solid, the only way for the Ag₂CO₃ to be created is to first dissolve the AgCl, then to precipitate Ag₂CO₃. Whether it may happen or not depends on the combination of solubility products, but even if it happens, it is a very slow process. It doesn't happen here, which is obvious when you look at the concentration of chlorides - it tells you only a very tiny amount of AgCl dissolved.

In this particular case you are adding relatively concentrated carbonate to the saturated solution of the AgCl. What is happening is that some carbonate precipitates (removing part of Ag⁺ from the solution). When Ag⁺ is removed, a little bit of additional AgCl dissolves, till you reach a new equilibrium with two solids present - AgCl and Ag₂CO₃. When both solids are present, both K_sp are at work. In general this leads to a rather nasty set of equations, however, we know that the amount of AgCl dissolved was small - so we also know amount of precipitated carbonate was small. If so, concentration of CO₃^2- is practically unchanged.

[125wewe]

I think I got it:

Ksp=[Ag⁺]*[Cl^-]

From given Cl^- concentration,
[Cl^-]=0.003/35.5 M = 8.45 x 10^-5 mol/litre

As,concentration of CO3^-- is practically unchanged,
Ksp of Ag2CO3=[Ag⁺]² x [CO₃^--]
or,  8.2 x 10^-12 = x² x 2     (2M Na₂CO₃ => [CO₃^--] = 2M)
=>x= 2.025 x 10^-6 = [Ag⁺]
So,
Ksp of AgCl = 2.025x10^-6 x 8.45 x 10^-5
=1.7*10^-10

Did I get right?

[Borek]

Looks OK

[125wewe]

Thank you for helping.