Finding ion concentration after determining whether precipitates will form

In summary, the calculation shows that a precipitate will form when 50.0ml of 0.010M AgNO3 is added to 50ml of 0.0010M Na2CO3. The carbonate ion concentration after equilibrium is established is 0.0059 M.
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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!
 
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  • #2


To find the carbonate ion concentration after equilibrium is established, we can use the equilibrium constant expression for the reaction Ag2CO3 <--> 2Ag+ + CO3 (2-).

Ksp = [Ag+]^2 x [CO3 (2-)]

We know that at equilibrium, the concentration of Ag+ ions will be equal to the solubility of Ag2CO3 in pure water at 25 degrees Celsius, which is 1.3x10^-4. So we can plug this value in for [Ag+] in the equilibrium constant expression:

Ksp = (1.3x10^-4)^2 x [CO3 (2-)]

We also know that the initial concentration of carbonate ions, [CO3 (2-)], was 0.0010M. So we can rearrange the equation to solve for [CO3 (2-)]:

[CO3 (2-)] = Ksp / (1.3x10^-4)^2
= 0.0059 M

Therefore, the carbonate ion concentration after equilibrium is established is 0.0059 M.
 

Related to Finding ion concentration after determining whether precipitates will form

1. How do I determine if precipitates will form in a solution?

The formation of precipitates in a solution depends on the solubility product constant (Ksp) of the compounds involved. If the product of the concentrations of the ions in the solution exceeds the Ksp value, then precipitates will form.

2. What is the relationship between ion concentration and precipitation?

The higher the concentration of ions in a solution, the more likely it is for precipitates to form. This is because there is a higher chance for the product of the ion concentrations to exceed the Ksp value.

3. How do I calculate the ion concentration in a solution?

The ion concentration in a solution can be calculated using the formula: concentration (M) = moles of ions / volume of solution (L). The moles of ions can be determined from the chemical formula and molar mass of the compound.

4. Can I predict the ion concentration without knowing the Ksp value?

No, the Ksp value is necessary to determine if precipitates will form. Without knowing the Ksp value, it is impossible to accurately predict the ion concentration in a solution.

5. How does temperature affect ion concentration and precipitation?

In general, increasing temperature increases the solubility of compounds, resulting in higher ion concentrations. However, this may not always hold true and depends on the specific compounds involved in the solution.

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