A colorless aqueous solution contains nitrates of two metals, X and Y. When it was added to an aqueo

Q: Why doesn't silver chloride dissolve in hot water like lead chloride does?

Silver chloride has a much higher lattice enthalpy compared to its hydration enthalpy. Even at elevated temperatures, the thermal energy is insufficient to overcome the strong ionic bonds in the AgCl crystal lattice, whereas lead chloride's lattice is more easily disrupted by hot water.

Q: What exactly is formed when the residue dissolves in aqueous ammonia?

When silver chloride reacts with aqueous ammonia, it forms a soluble coordination complex called diamminesilver(I) chloride, with the formula $[\text{Ag}(\text{NH}_3)_2]\text{Cl}$.

Q: Why do we use excess sodium thiosulfate?

Excess sodium thiosulfate ensures that the silver ions are fully complexed to form the highly soluble dithiosulfatoargentate(I) complex, $\text{Na}_3[\text{Ag}(\text{S}_2\text{O}_3)_2]$. If not in excess, the reaction might not go to completion, leaving some residue.

Q: Could the yellow precipitate be something else?

In the context of this specific separation scheme, a yellow precipitate formed by adding KI to the hot water soluble fraction is definitively lead(II) iodide ($\text{PbI}_2$). Other yellow precipitates exist, but not from Group 1 cations under these conditions.

Q: What if the initial white precipitate turned black with ammonia?

If the white precipitate turned black upon adding ammonia, it would indicate the presence of mercurous chloride ($\text{Hg}_2\text{Cl}_2$). The black color is due to the formation of finely divided metallic mercury alongside white mercuric amino chloride.

Question

A colorless aqueous solution contains nitrates of two metals, $\mathbf{X}$ and $\mathbf{Y}$. When it was added to an aqueous solution of $	ext{NaCl}$, a white precipitate was formed. This precipitate was found to be partly soluble in hot water to give a residue $\mathbf{P}$ and a solution $\mathbf{Q}$. The residue $\mathbf{P}$ was soluble in aq. $	ext{NH}_3$ and also in excess sodium thiosulfate. The hot solution $\mathbf{Q}$ gave a yellow precipitate with $	ext{KI}$. The metals $\mathbf{X}$ and $\mathbf{Y}$, respectively, are

Options:

(A) $	ext{Ag}$ and $	ext{Pb}$

(B) $	ext{Ag}$ and $	ext{Cd}$

(C) $	ext{Cd}$ and $	ext{Pb}$

(D) $	ext{Cd}$ and $	ext{Zn}$

Sigma Percentile · Accepted Answer

### The Thrill of Chemical Deduction

Qualitative analysis is essentially chemical detective work. You are presented with an unknown mixture, and through a series of logical, systematic interrogations using chemical reagents, you deduce the identities of the hidden culprits. In this problem, we are tasked with identifying two unknown metals, $\mathbf{X}$ and $\mathbf{Y}$, starting from a colorless aqueous solution of their nitrates. Let's walk through the clues step by step and unravel this mystery.

### The First Clue: The White Precipitate

The first piece of evidence is the addition of an aqueous solution of sodium chloride ($	ext{NaCl}$), which results in the formation of a white precipitate. This is a massive hint! In the standard scheme of qualitative salt analysis, the addition of dilute chloride ions is the test for **Group 1 cations**.

The cations that form insoluble chlorides in cold water are Silver ($Ag^+$), Lead ($Pb^{2+}$), and Mercurous ($Hg_2^{2+}$). Therefore, our unknown metals $\mathbf{X}$ and $\mathbf{Y}$ must be chosen from this exclusive trio. The chemical reactions occurring at this stage are:

$$Ag^+_{(aq)} + Cl^-_{(aq)} ightarrow AgCl_{(s)} 	ext{ (White)}$$

$$Pb^{2+}_{(aq)} + 2Cl^-_{(aq)} ightarrow PbCl_2_{(s)} 	ext{ (White)}$$

### The Thermal Interrogation: Hot Water Treatment

Our next clue states that the white precipitate is *partly* soluble in hot water, yielding a residue $\mathbf{P}$ and a solution $\mathbf{Q}$. This is the classic method used to separate Lead from Silver and Mercury in Group 1 analysis.

Lead(II) chloride ($PbCl_2$) has a relatively low lattice enthalpy compared to its hydration enthalpy at elevated temperatures. Consequently, its solubility increases dramatically in hot water. On the other hand, Silver chloride ($AgCl$) and Mercurous chloride ($Hg_2Cl_2$) remain stubbornly insoluble even when heated.

Because the precipitate is only partly soluble, we know we have a mixture. The part that dissolved into the hot solution $\mathbf{Q}$ must be $PbCl_2$, and the insoluble residue $\mathbf{P}$ must be either $AgCl$ or $Hg_2Cl_2$.

### Interrogating Residue P: The Silver Confirmation

To identify residue $\mathbf{P}$, we look at its chemical behavior. The problem tells us that $\mathbf{P}$ is soluble in both aqueous ammonia ($	ext{NH}_3$) and excess sodium thiosulfate ($	ext{Na}_2	ext{S}_2	ext{O}_3$).

This is the definitive signature of Silver chloride. When $AgCl$ is treated with aqueous ammonia, it undergoes a complexation reaction to form the soluble diamminesilver(I) complex:

$$AgCl_{(s)} + 2NH_{3(aq)} ightarrow [Ag(NH_3)_2]^+_{(aq)} + Cl^-_{(aq)}$$

Similarly, it dissolves in excess sodium thiosulfate to form the highly stable and soluble dithiosulfatoargentate(I) complex:

$$AgCl_{(s)} + 2S_2O_3^{2-}_{(aq)} ightarrow [Ag(S_2O_3)_2]^{3-}_{(aq)} + Cl^-_{(aq)}$$

If the residue had been Mercurous chloride ($Hg_2Cl_2$), adding ammonia would have caused a disproportionation reaction, turning the precipitate black due to the formation of finely divided metallic mercury. Since our residue dissolved completely to form a clear solution, we can confidently conclude that residue $\mathbf{P}$ is $AgCl$, meaning **Metal X is Silver (Ag)**.

### Interrogating Solution Q: The Golden Lead

Finally, we must confirm the identity of the metal in the hot solution $\mathbf{Q}$, which we suspect is Lead. The problem states that adding potassium iodide ($	ext{KI}$) to solution $\mathbf{Q}$ produces a yellow precipitate.

This is the classic confirmatory test for Lead(II) ions. The reaction produces Lead(II) iodide, which is famous for its brilliant, golden-yellow color:

$$Pb^{2+}_{(aq)} + 2I^-_{(aq)} ightarrow PbI_{2(s)} 	ext{ (Yellow)}$$

This perfectly aligns with our earlier deduction. The hot solution $\mathbf{Q}$ indeed contains $Pb^{2+}$ ions, confirming that **Metal Y is Lead (Pb)**.

### The Final Verdict

By systematically analyzing the solubility rules, thermal properties, and complexation behaviors, we have successfully unmasked the unknown metals. The initial colorless solution contained the nitrates of Silver and Lead. Therefore, the metals $\mathbf{X}$ and $\mathbf{Y}$ are $	ext{Ag}$ and $	ext{Pb}$, respectively. This elegant sequence of reactions highlights the logical beauty of inorganic qualitative analysis!

Select Answer:

Visualized Solution

\text{Initial Reaction with NaCl}

\text{Group I Cations}

\text{Effect of Heat}

\text{Confirming Residue P (AgCl)}

\text{Confirming Solution Q (PbCl}_2\text{)}

\text{Final Conclusion}

\text{The Way Forward}

The Sigma Insight: Salt Analysis

The Thrill of Chemical Deduction

The First Clue

The Thermal Interrogation

Interrogating Residue P

Interrogating Solution Q

The Final Verdict

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