Understanding Solubility and Common Ion Effect

This problem involves calculating the concentration of silver ions ($Ag^+$) in a solution containing silver oxalate ($Ag_2C_2O_4$), a sparingly soluble salt. The complexity arises because the solution also contains sodium oxalate ($Na_2C_2O_4$), which introduces a common ion, oxalate ($C_2O_4^{2-}$). This situation demonstrates the common ion effect, which influences the solubility of ionic compounds.

We are provided with the solubility product constant ($K_{sp}$) for silver oxalate, $Ag_2C_2O_4$, which is $5.32 \times 10^{-12}$. We also know the concentration of the added sodium oxalate is $0.010$ M.

Dissolution Equilibrium and $K_{sp}$ Expression

The first step is to represent the dissolution equilibrium of silver oxalate in water. When $Ag_2C_2O_4$ dissolves, it dissociates into its constituent ions:

$Ag_2C_2O_4(s) \rightleftharpoons 2Ag^+(aq) + C_2O_4^{2-}(aq)$

The solubility product constant ($K_{sp}$) expression relates the ion concentrations at equilibrium. For this reaction, it is defined as:

$K_{sp} = [Ag^+]^2 [C_2O_4^{2-}]$

The value of $K_{sp}$ is given as $5.32 \times 10^{-12}$.

Impact of the Common Ion

Sodium oxalate ($Na_2C_2O_4$) is a strong electrolyte and dissociates completely in an aqueous solution:

$Na_2C_2O_4(aq) \rightarrow 2Na^+(aq) + C_2O_4^{2-}(aq)$

This dissociation means that the presence of $0.010$ M $Na_2C_2O_4$ directly adds $0.010$ M of the common ion, $C_2O_4^{2-}$, to the solution. This initial concentration of $C_2O_4^{2-}$ will affect the equilibrium position of the $Ag_2C_2O_4$ dissolution.

Step-by-Step Calculation of $Ag^+$ Concentration

Let '$s$' denote the molar solubility of $Ag_2C_2O_4$ in this specific solution (i.e., the moles of $Ag_2C_2O_4$ that dissolve per liter). Based on the dissolution stoichiometry:

• The concentration of $Ag^+$ ions produced is $2s$.
• The concentration of $C_2O_4^{2-}$ ions produced from the dissolution of $Ag_2C_2O_4$ is $s$.

The total equilibrium concentration of the oxalate ion ($C_2O_4^{2-}$) in the solution is the sum of the contribution from the dissolved $Ag_2C_2O_4$ and the initial concentration from $Na_2C_2O_4$:

$[C_2O_4^{2-}]_{total} = s + [C_2O_4^{2-}]_{from\;Na_2C_2O_4}$

$[C_2O_4^{2-}]_{total} = s + 0.010 \text{ M}$

Now, we plug these equilibrium concentrations into the $K_{sp}$ expression:

$K_{sp} = [Ag^+]^2 [C_2O_4^{2-}]_{total}$

$5.32 \times 10^{-12} = (2s)^2 (s + 0.010)$

$5.32 \times 10^{-12} = 4s^2 (s + 0.010)$

Since the $K_{sp}$ value is extremely small ($5.32 \times 10^{-12}$), it indicates that $Ag_2C_2O_4$ is poorly soluble. Furthermore, the presence of the common ion ($0.010$ M $C_2O_4^{2-}$) further suppresses its solubility. Therefore, we can reasonably assume that the molar solubility '$s$' will be much smaller than $0.010$ M ($s \ll 0.010$).

This approximation allows us to simplify the term $(s + 0.010)$ to just $0.010$. The equation becomes:

$5.32 \times 10^{-12} \approx 4s^2 (0.010)$

$5.32 \times 10^{-12} \approx 0.040 s^2$

Now, we can solve for $s^2$:

$s^2 \approx \frac{5.32 \times 10^{-12}}{0.040}$

$s^2 \approx 1.33 \times 10^{-10}$

To find the molar solubility '$s$', we take the square root:

$s \approx \sqrt{1.33 \times 10^{-10}}$

$s \approx 1.153 \times 10^{-5} \text{ M}$

Determining the Final $Ag^+$ Concentration

The question asks for the concentration of $Ag^+$ ions, which is given by $[Ag^+] = 2s$.

$[Ag^+] = 2 \times s$

$[Ag^+] \approx 2 \times (1.153 \times 10^{-5} \text{ M})$

$[Ag^+] \approx 2.306 \times 10^{-5} \text{ M}$

Checking the Approximation

Our assumption was $s \ll 0.010$. The calculated value $s \approx 1.153 \times 10^{-5}$ M is indeed significantly smaller than $0.010$ M, confirming that the approximation was valid and the calculation is reliable.

Conclusion

The calculated concentration of $Ag^+$ ions in the saturated solution of $Ag_2C_2O_4$ containing $0.010$ M $Na_2C_2O_4$ is approximately $2.306 \times 10^{-5}$ mol L$^{-1}$. This result aligns with one of the provided options.