Let's calculate the molar solubility of Ag_2CO_3 in a solution that is 0.0200 M in Na_2CO_3.
The Ksp of Ag_2CO_3= 8.1 x 10^{-12}
What is the chemical reaction and represents this solubility? X and Y represent the stoichiometric coefficients. For your chemical species do not use subscripts. For example, Ag_2CO_3 would be written as Ag2CO3. The charges have been included for you.
blank1 - Word Answer
Ag2CO3

You are correct
 \leftrightharpoons   X
blank2 - Word Answer
Ag

You are correct
^+ + Y
blank3 - Word Answer
CO3

You are correct
 ^{-2}
X = 
blank4 - Word Answer
2

You are correct

Y = 
blank5 - Word Answer
1

You are correct
In this problem 
blank6 - Word Answer
carbonate

You are correct
(silver, sodium or carbonate) is the common ion.
Total ​[CO_3^{2-}] = 
blank7 - Numeric Answer

M
​[Ag^+] =

The problem you're working on involves finding the molar solubility of silver carbonate (\( \text{Ag}_2\text{CO}_3 \)) in a solution that already contains sodium carbonate (\( \text{Na}_2\text{CO}_3 \)), which introduces a common ion (\( \text{CO}_3^{2-} \)).

### Step-by-step breakdown:

1. **The dissociation of \( \text{Ag}_2\text{CO}_3 \) in water:**
   The equilibrium reaction for the dissociation of \( \text{Ag}_2\text{CO}_3 \) is as follows:
   \[
   \text{Ag}_2\text{CO}_3 \leftrightharpoons 2\text{Ag}^+ + \text{CO}_3^{2-}
   \]
   In this reaction:
   - \( X = 2 \), representing the stoichiometric coefficient of \( \text{Ag}^+ \)
   - \( Y = 1 \), representing the stoichiometric coefficient of \( \text{CO}_3^{2-} \)

2. **The common ion effect:**
   Since the solution already contains \( 0.0200 \, \text{M} \) \( \text{CO}_3^{2-} \) from sodium carbonate, the common ion effect will suppress the dissociation of \( \text{Ag}_2\text{CO}_3 \), reducing the solubility of \( \text{Ag}^+ \).

3. **Total carbonate concentration:**
   The total concentration of \( \text{CO}_3^{2-} \) in the solution is \( 0.0200 \, \text{M} \), contributed entirely by the sodium carbonate, because sodium carbonate is a strong electrolyte and fully dissociates.

   Therefore, the total \( [\text{CO}_3^{2-}] \) is:
   \[
   \text{blank7: } 0.0200 \, \text{M}
   \]

4. **Using the \( K_\text{sp} \) expression:**
   The solubility product constant \( K_\text{sp} \) for \( \text{Ag}_2\text{CO}_3 \) is given as \( 8.1 \times 10^{-12} \). The expression for the solubility product is:
   \[
   K_\text{sp} = [\text{Ag}^+]^2 [\text{CO}_3^{2-}]
   \]

   Let \( s \) be the molar solubility of \( \text{Ag}_2\text{CO}_3 \) (i.e., the concentration of \( \text{Ag}^+ \) ions). Therefore, the concentration of \( \text{Ag}^+ \) will be \( 2s \), and the concentration of \( \text{CO}_3^{2-} \) is \( 0.0200 \, \text{M} \), as mentioned earlier.

   Substituting into the \( K_\text{sp} \) expression:
   \[
   8.1 \times 10^{-12} = (2s)^2(0.0200)
   \]

   Simplifying:
   \[
   8.1 \times 10^{-12} = 4s^2(0.0200)
   \]
   \[
   8.1 \times 10^{-12} = 0.0800s^2
   \]

   Solving for \( s \):
   \[
   s^2 = \frac{8.1 \times 10^{-12}}{0.0800} = 1.0125 \times 10^{-10}
   \]
   \[
   s = \sqrt{1.0125 \times 10^{-10}} = 1.006 \times 10^{-5} \, \text{M}
   \]

   Therefore, the concentration of \( [\text{Ag}^+] \) is:
   \[
   \text{blank8: } 1.006 \times 10^{-5} \, \text{M}
   \]

The molar solubility of \( \text{Ag}_2\text{CO}_3 \) is \( 1.006 \times 10^{-5} \, \text{M} \) in the presence of \( 0.0200 \, \text{M} \) \( \text{Na}_2\text{CO}_3 \).

Would you like further details or clarifications on this calculation?

### Related Questions:
1. How would the solubility change if the concentration of \( \text{Na}_2\text{CO}_3 \) were increased to \( 0.0500 \, \text{M} \)?
2. What would the molar solubility of \( \text{Ag}_2\text{CO}_3 \) be in pure water?
3. How does the common ion effect influence solubility in this case?
4. What is the solubility of \( \text{Ag}^+ \) if the solution also contained \( 0.0100 \, \text{M} \, \text{AgNO}_3 \)?
5. What is the relationship between \( K_\text{sp} \) and solubility for a compound with different stoichiometry, like \( \text{CaSO}_4 \)?

### Tip:
Remember, the common ion effect decreases the solubility of salts by adding an ion already present in the equilibrium, shifting the reaction to the left (less dissociation).

Let's calculate the molar solubility of Ag2CO3 in a solution that is 0.0200 M in Na2CO3. The Ksp of Ag2CO3 = 8.1 x 10^{-12}. What is the chemical reaction and represents this solubility? X and Y represent the stoichiometric coefficients. For your chemical species do not use subscripts. For example, Ag2CO3 would be written as Ag2CO3. The charges have been included for you. Ag2CO3 ⇌ 2Ag^+ + CO3^{2-}.

Learn how to calculate the molar solubility of Ag2CO3 in a solution that is 0.0200 M in Na2CO3 using the solubility product constant (Ksp). This detailed solution covers chemical equilibrium concepts, the common ion effect, and relevant calculations for Grades 11-12.

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