Finding ion concentration after determining whether precipitates will form

  • Thread starter Porta
  • Start date
  • #1
1
0

Homework Statement


A. Show by calculation whether a precipitate will form when 50.0ml of 0.010M AgNO3 is added to 50ml of 0.0010M Na2CO3.

B. What is the carbonate ion concentration after equilibrium is established in the previous question?

Homework Equations


What I'm working with is:
Ksp of Ag2CO3 = 8x10^-12 at 25 degrees celcius
So the equation is: Ag2CO3 <--> 2Ag+ + CO3 (2-)

Information I figured out from earlier parts of this problem set (may or may not be useful, I'll post my work shown if needed):
[Ag+] in a saturated solution whose [CO3 (2-)] is .01M = 2.8x10^-5
Solubility of Ag2CO3 in pure water at 25 degrees Celsius = 1.3x10^-4
Its solubility in 0.0010M Ag2CO3 = 8x10^-6

The Attempt at a Solution


A. For A, I converted both compounds to their moles using the Molarity formula, then solved for the moles of the ions I was interested that had something to do with Ag2CO3 <--> 2Ag+ + CO3 (2-). Then I divided the moles of each by the added volume to find the new molarity of them when added together, then put them in the formula for Q.

.010M AgNO3 = x/.05L, x = 5x10^-4 moles AgNO3 x (1 mol Ag+/1 mol AgNO3)/(.05L + .05L)
= 5.00x10^-3 M Ag+

.0010M Na2CO3 = x/.05L, x = 5x10^-5 moles Na2CO3 x (1 mol CO3 (2-)/1 mol Na2CO3)/(.05L + .05L)
= 5.00x10^-4 M CO3 (2-)

Q=[Ag+]^2 x [CO3 (2-)]
Q= 1.25x10^-8

Comparing Q to Ksp,
1.2x10^-8 > 8x10^-12
Q > Ksp
So a precipitate forms.

B. B is where I had the most trouble. I'm not sure where to start to be honest, and any help would be appreciated!
 

Answers and Replies

Related Threads on Finding ion concentration after determining whether precipitates will form

Replies
8
Views
1K
  • Last Post
Replies
7
Views
873
  • Last Post
Replies
1
Views
6K
Replies
1
Views
2K
Replies
9
Views
22K
  • Last Post
Replies
10
Views
5K
Replies
3
Views
2K
  • Last Post
Replies
5
Views
6K
Replies
13
Views
2K
Replies
1
Views
2K
Top