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## 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!