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).