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tobywashere
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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
Ksp 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>