When a student prepares an aqueous solution containing the five cations Ag+(aq), Hg2 2+(aq), Cu2+(aq — Solutions Chemistry Question

STEPS:

1. Understand the core chemical condition: The problem states that an aqueous solution contains five specific cations: $\text{Ag}^+$, $\text{Hg}_2^{2+}$, $\text{Cu}^{2+}$, $\text{Mn}^{2+}$, and $\text{Ba}^{2+}$. The critical observation is that no precipitates form when the unknown anion is introduced.
2. Apply the concept of solubility rules: For no precipitate to form, the chosen anion must form highly soluble salts with all five of these cations. If the anion forms an insoluble compound with even one of the cations, a precipitate would be observed, violating the student's observation.
3. Recall the fundamental solubility rule for nitrates: In aqueous chemistry, all common nitrate ($\text{NO}_3^-$) salts are highly soluble in water with no exceptions.
4. Evaluate the result of adding nitrate ($\text{NO}_3^-$): Pairing $\text{NO}_3^-$ with each of the cations yields $\text{AgNO}_3$, $\text{Hg}_2(\text{NO}_3)_2$, $\text{Cu}(\text{NO}_3)_2$, $\text{Mn}(\text{NO}_3)_2$, and $\text{Ba}(\text{NO}_3)_2$. Because all five of these compounds are completely soluble in water, they will remain fully dissociated as free aqueous ions, and no precipitate will form. This aligns perfectly with the observation, confirming Option D is the correct answer.

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WHY_OTHERS_WRONG:

• A is incorrect: Chloride ($\text{Cl}^-$) forms insoluble white precipitates with both silver and mercury(I) ions. Adding $\text{Cl}^-$ would immediately precipitate out $\text{AgCl}(s)$ and $\text{Hg}_2\text{Cl}_2(s)$.
• B is incorrect: Carbonates ($\text{CO}_3^{2-}$) are generally insoluble in water, except when paired with Group 1 alkali metal cations or ammonium. Adding $\text{CO}_3^{2-}$ would cause multiple insoluble metal carbonates to precipitate, such as $\text{BaCO}_3(s)$ and $\text{CuCO}_3(s)$.
• C is incorrect: Chromates ($\text{CrO}_4^{2-}$) form highly insoluble precipitates with several metal cations. Adding chromate would result in the precipitation of yellow barium chromate ($\text{BaCrO}_4(s)$) and red-brown silver chromate ($\text{Ag}_2\text{CrO}_4(s)$).
• E is incorrect: While most sulfates ($\text{SO}_4^{2-}$) are soluble, there are well-known exceptions. Sulfate ions form a highly insoluble white precipitate with barium ions ($\text{BaSO}_4(s)$) and mercury(I) ions ($\text{Hg}_2\text{SO}_4(s)$), which would prevent the solution from remaining clear.