How much Silver Nitrate to produce a precipitate?

[tobywashere]

Homework Statement

What is the minimum amount of AgNO₃, in grams, that has to be added to 2.50L of 0.020M K₂CrO₄ to produce a precipitate?

Homework Equations

K_sp Ag₂CrO₄ = 9.0 x 10^-12
[Ag⁺]²[CrO₄^-2] = 9.0x10^-12

The Attempt at a Solution

2AgNO₃(aq) + K₂CrO₄(aq) --> Ag₂CrO₄(s) + 2KNO₃(aq)
Ag₂CrO₄(s) is the precipitate
The net ionic equation is:
Ag₂CrO₄(s) --> 2Ag⁺(aq) + CrO₄^-2(aq)
Let y represent the concentration of CrO₄^-2 at equilibrium. The concentration of Ag⁺ will be 2y due to the 2:1 stoichiometry. Plugging into the solubility equilibrium expression:
[Ag⁺]²[CrO₄^-2] = 9.0x10^-12
(2y)²(y) = 9.0x10^-12
Therefore, y = 1.31 x 10^-4 M
So the equilibrium concentration of CrO₄^-2 is 1.31 x 10^-4 M, and the equilibrium concentration of Ag⁺ is double that, or 2y, or 2.62x10^-4M
The initial concentration of CrO₄^-2 is 0.020M (from the 0.020M K₂CrO₄). The decrease in concentration of CrO₄^-2 to form the precipitate is:
0.020M - 1.31 x 10^-4M = 0.01987M
The initial concentration of Ag⁺ (what we're trying to find) decreases by twice this amount, 0.03974M, to reach an equilibrium concentration of 2.62x10^-4M (the 2y found earlier).
Initial concentration of Ag⁺ - 0.03974M = 2.62x10^-4M.
Initial concentration = 2.62x10^-4M + 0.0397M = 0.040M
Find mass:
0.040M x 2.5L x 170g / mol = 17g
Therefore, 17 g of AgNO₃ needs to be added to form a precipitate.
Did I do this correctly? I'm really not sure.<br>

 

[Borek]

No, it is wrong. You add silver to the solution that contains chromate. Nothing consumes chromate, so its concentration stays constant, and concentration of Ag⁺ goes up - till you reach solubility product.

Technically when adding silver nitrate as a solution you would dilute chromate a little bit, but we can assume you add it as a solid, not changing solution volume.

 

[tobywashere]

Oh ok. So you just need 2.62x10^-4M of AgNO₃.
2.62x10^-4M x 2.50L x 170g/mol = 0.11135 grams?

 

[Borek]

No, concentration of silver is much lower than that.

Ksp defines saturated solution concentrations, it doesn't mean ratio of concentrations is given by precipitate stoichiometry. Ksp for AgCl is about 10^-10. If you put solid AgCl into water, saturated solution will be 10^-5M both in [Ag⁺] and [Cl^-]. However, saturated solution can be also 1M in [Ag⁺] and 10^-10M in [Cl^-], or - say - 10^-3M in [Ag⁺] and 10^-7M in [Cl^-] - if it was prepared by mixing solutions containing these ions in correct quantities.

 

[tobywashere]

Oh so I made this problem unnecessarily complicated.
[0.020M][Ag⁺]² = 9.0x10^-12
[Ag⁺]² = 2.12x10^-5M
2.12x10^-5M x 2.5L x 170g/mol = 0.0090 g

 

[tobywashere]

Is that right?
10 char

 

[Borek]

Yes.