Answer:
[tex]\text{Na}_2\text{C}_2\text{O}_4[/tex]
Explanation:
The common ion effect refers to the decrease in the solubility of a salt when a common ion is added to the solution. This effect can be explained using the principle of Le Chatelier's principle. According to Le Chatelier's principle, if a stress, such as the addition of a common ion, is applied to a system at equilibrium, the system will shift to counteract the stress.
We are given the equilibrium equation:
[tex]\[ \text{SrC}_2\text{O}_4(s) \rightleftharpoons \text{Sr}^{2+}(aq) + \text{C}_2\text{O}_4^{2-}(aq) \][/tex]
Common ion: [tex]\( \text{C}_2\text{O}_4^{2-}[/tex]
When more of this ion is added to the solution, according to Le Chatelier's Principle, the equilibrium will shift to the left to counteract the addition of the common ion. This leads to a decrease in the concentration of the dissolved ions.
Let's test each option to see if they hold true to the common ion effect:
[tex]\( \textbf{Na}_2\textbf{C}_2\textbf{O}_4 \) (\( \textbf{aq} \)): This solution introduces a high concentration of \( \text{C}_2\text{O}_4^{2-} \) ions into the solution, effectively causing the common ion effect.[/tex]
[tex]\( \textbf{SrC}_2\textbf{O}_4 \) (\( \textbf{aq} \)): Although this solution contains the compound \( \text{SrC}_2\text{O}_4 \), it's in a solid state and does not contribute to the ion concentration in solution. Therefore, it doesn't effectively cause the common ion effect.[/tex]
[tex]\( \textbf{Na}_2\textbf{SO}_4 \) (\( \textbf{aq} \)): This solution introduces \( \text{SO}_4^{2-} \) ions, which are not involved in the equilibrium reaction. Thus, it does not cause the common ion effect in this case.[/tex]
Therefore, the correct option is [tex]\text{Na}_2\text{C}_2\text{O}_4[/tex]
