Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: null on July 25, 2019, 11:11:46 AM
-
Here is a problem that I don't know how to get the answer...
In a test tube with water a small quantity of both AgCl and Ag2CrO4 are added. What is the resulting concentration of silver cations? Ksp (AgCl) = 1.8*10-10 and Ksp (Ag2CrO4) = 1.2*10-12
The answer key from my teacher is [Ag+] = 1.35*10-4 mol/L
However, I'm bit confused because both solubility equilibria are going to affect each other, right? The sum of the silver cation concentrations calculated from each equilibrium alone (1.34*10-5 for AgCl and 1.338*10-4 for the chromate) does not give the answer and I think it is because the remaining anions will precipitate some cations. I thought about an ICE table, but the equilibria are simultaneous... Can anyone help me?
-
6.69*10-5 - this is not the silver concentration from chromate itself - this is the concentration of chromate.
Put the correct concentration of silver into the solubility product of AgCl and add chloride concentration (= silver concentration from AgCl in the presence of silver chromate) to this concentration of silver. This simple addition will give you a quite reasonable approximation of total silver concentration in one step.
-
6.69*10-5 - this is not the silver concentration from chromate itself - this is the concentration of chromate. (...)
You're right. I've just fixed my post.
What is the "correct concentration" from your post? I don't get this part. I think I understand the part of adding chloride concentration as silver (because the AgCl is less soluble than the Ag2CrO4, right?) but I don't get what you are calling the "correct concentration".
-
1.338*10-4 - insert this value into solubility product of AgCl