What Is the Initial Concentration of Na2CrO4 in a Mixed Ag2CrO4 Solution?

In summary, when 50.0 mL of a 2.0x10-4 M Ag2CrO4 solution was added to 50.0 mL of a Na2CrO4 solution a ppt formed (Ag2Cro4). Ksp Ag2CrO4 = 1.1 × 10-12.
  • #1
j3llzang
9
0

Homework Statement


When 50.0 mL of a 2.0 × 10-4 M Ag2CrO4 solution was added to 50.0 mL of a Na2CrO4 solution a ppt formed (Ag2Cro4). What was the initial concentration of this Na2CrO4 solution?
Ksp Ag2CrO4 = 1.1 × 10-12

Homework Equations


Given A2 (aq) + B(aq) -> AB(s)
Ksp = [A]n

The Attempt at a Solution


I tried using the "ICE" Table, but it didn't help:
Ag2CrO -> 2Ag + CrO4
I 2.0x10-4 2.0x10-4 ?
C +x(?) -2x(?) -x(?)
E 2.0x10-4(unchanged) 2.75x10-5

I found [CrO4] at eqb using Ksp equation. (2.75x10-5M)
The initial [CrO4] was the [CrO4] from Ag2CrO4 plus Na2CrO4, right?
but why is it that [CrO4] at equilibrium is smaller than that of the initial [CrO4] in the initial Ag2CrO4 concentration?
Can anyone solve this problem?
Explanation would be nice :)

Thanks!~
 
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  • #2
What was the final concentration of Ag+ after solutions has been mixed? (Don't think about Ksp, it was just diluted).

Can you calculate concentration of chromate now?
 
  • #3
Borek said:
What was the final concentration of Ag+ after solutions has been mixed? (Don't think about Ksp, it was just diluted).

Can you calculate concentration of chromate now?

I like how everyone answers the questions with another question. lol

As I stated above, I did find the final concentration of Ag+, which came out to be 2.0x10-4M
 
  • #4
I like how everyone concentrates on the first part of the answer but ignores the other part, one that leads to the solution.

So you have a 2.0x10-4M solution of Ag[sup+[/sup] and you know concentration of CrO42- is high enough to start precipitation. What is concentration of CrO42-? Where did it came from?
 
  • #5
Hmm.. concentration of CrO4 comes from Na2CrO4...
Well, I used algebra to find the initial concentration of CrO4, and ended up with a negative value.
Here's what I did:
I declared x as [CrO4]init,
so [CrO4] @ eqb = 1.0x10-4 + x

Since Ksp = [Ag]^2[CrO4],
[CrO4] (at eqb?) = 1.1x10-12 / (2.0x10-4)^2
x = -7.25x10^5 M = [CrO4] initial.

I have no clue as to what this value actually means...
Any idea?
Thanks for your help :D
 
  • #6
LOL, unless I am missing something that's not your fault. Question is wrong. There is no such thing as 2x10-4M solution of Ag2CrO4, no wonder you are getting negative concentrations.

Check what is silver chromate solubility.

On the second thought, there is a slight chance that question is OK and you were expected to take into account chromate protonation. But I strongly doubt it.
 
  • #7
oh my gosh, that's so funny! I thought my chemistry teacher was evil enough to give us a question that had a negative answer :)
Well, solubility of Silver Chromate is 1.1x10-12, so I guess the question had it right.
But how do you know that 2.0x10-4 M solution doesn't exist?
Thanks!
(ps: I don't think we know what protonation is yet...)
 
  • #8
j3llzang said:
Well, solubility of Silver Chromate is 1.1x10-12, so I guess the question had it right.

This is not solubility, this is Ksp. Solubility is a concentration of a saturated solution.
 

FAQ: What Is the Initial Concentration of Na2CrO4 in a Mixed Ag2CrO4 Solution?

1. What is the common ion effect?

The common ion effect is a phenomenon where the solubility of a salt is decreased when a common ion is added to the solution. This is due to the reduction of the concentration of the ions in the solution, which leads to a decrease in the solubility product.

2. How does the common ion effect affect the pH of a solution?

The common ion effect can affect the pH of a solution by shifting the equilibrium of a weak acid or base reaction. For example, when a strong acid is added to a solution containing a weak acid and its conjugate base, the common ion (the strong acid) will react with the weak base, shifting the equilibrium to the left and decreasing the pH.

3. How do you calculate the common ion effect?

The common ion effect can be calculated by using the solubility product constant (Ksp) and the concentration of the common ion in the solution. The equation for calculating the common ion effect is: [ion] = √(Ksp/[common ion])

4. Can the common ion effect be used to increase the solubility of a salt?

Yes, the common ion effect can be used to increase the solubility of a salt by adding a common ion to a solution containing a slightly soluble salt. This will shift the equilibrium of the reaction to the right, increasing the concentration of ions and therefore increasing the solubility of the salt.

5. How does temperature affect the common ion effect?

Temperature does not have a significant effect on the common ion effect. However, it can affect the solubility product constant (Ksp) which is used in the calculation of the common ion effect. As temperature increases, the solubility of a salt may also increase, which can affect the concentration of the common ion and therefore the common ion effect.

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