A mixture of two salts is used to prepare a solution S, which gives the following results : White pr

Q: Why doesn't $\text{Zn}^{2+}$ precipitate with dilute HCl?

Zinc belongs to Group IV in qualitative analysis. Its chloride, $\text{ZnCl}_2$, is highly soluble in water. It only precipitates as a sulfide in a basic medium.

Q: Does $\text{Bi}^{3+}$ react with dilute HCl?

Bismuth chloride ($\text{BiCl}_3$) is soluble in acidic solutions. However, if the solution is heavily diluted with water, it can hydrolyze to form a white precipitate of bismuth oxychloride ($\text{BiOCl}$). Since the reagent is dilute HCl, the acidic medium prevents this hydrolysis, keeping it soluble.

Q: Why does $\text{Ag}^{+}$ form a brown precipitate with NaOH instead of a white hydroxide?

Silver hydroxide ($\text{AgOH}$) is highly unstable at room temperature. As soon as it forms, it spontaneously dehydrates to form silver oxide ($\text{Ag}_2\text{O}$), which is a dark brown precipitate.

Q: What happens to $\text{Hg}^{2+}$ with NaOH?

Similar to silver, mercury(II) hydroxide ($\text{Hg(OH)}_2$) is unstable and immediately decomposes into mercury(II) oxide ($\text{HgO}$). Depending on the particle size, this precipitate appears yellow.

Q: Can $\text{Pb}^{2+}$ precipitate in both tests?

Yes! Lead(II) is unique. It precipitates as $\text{PbCl}_2$ (white) with dilute HCl (Group I) and also forms an amphoteric white precipitate of $\text{Pb(OH)}_2$ with dilute NaOH.

Question

A mixture of two salts is used to prepare a solution $	ext{S}$, which gives the following results :

White precipitate(s) only $\xleftarrow[	ext{Room temperature}]{	ext{Dilute NaOH(aq.)}}$ $	ext{S}$ (aq. solution of the salts) $\xrightarrow[	ext{Room temperature}]{	ext{Dilute HCl(aq.)}}$ White precipitate(s) only

The correct option(s) for the salt mixture is(are)

Options:

(A) $	ext{Pb(NO}_3	ext{)}_2$ and $	ext{Zn(NO}_3	ext{)}_2$

(B) $	ext{Pb(NO}_3	ext{)}_2$ and $	ext{Bi(NO}_3	ext{)}_3$

(C) $	ext{AgNO}_3$ and $	ext{Bi(NO}_3	ext{)}_3$

(D) $	ext{Pb(NO}_3	ext{)}_2$ and $	ext{Hg(NO}_3	ext{)}_2$

Sigma Percentile · Accepted Answer

The world of qualitative salt analysis is like being a chemical detective. You are given a mysterious mixture, and by adding specific reagents, you observe clues—colors, precipitates, and gases—to deduce the culprits (the ions) hiding within.

In this fascinating JEE Advanced problem, we are tasked with identifying a mixture of two salts that behaves in a very specific way.

### Decoding the Problem Statement

We have a solution $	ext{S}$ containing a mixture of two salts. The problem provides us with two crucial pieces of evidence:
1. When we add **dilute $	ext{HCl}$** at room temperature, we get **only white precipitate(s)**.
2. When we add **dilute $	ext{NaOH}$** at room temperature, we again get **only white precipitate(s)**.

The word "only" is the trap here. It means that if even a single drop of a colored precipitate forms, or if no precipitate forms at all when one is expected, that option is instantly disqualified. Let's break down the chemistry behind these two reagents.

### The First Test: Dilute HCl

Dilute hydrochloric acid ($	ext{HCl}$) is the classic group reagent for **Group I cations** in qualitative analysis. The cations in this group are Lead ($	ext{Pb}^{2+}$), Silver ($	ext{Ag}^{+}$), and Mercury(I) ($	ext{Hg}_2^{2+}$).

When dilute $	ext{HCl}$ is added to a solution containing these ions, they form insoluble chlorides, which all happen to be white precipitates:
$$ 	ext{Pb}^{2+} + 2	ext{Cl}^{-} ightarrow 	ext{PbCl}_2 \downarrow 	ext{ (White)} $$
$$ 	ext{Ag}^{+} + 	ext{Cl}^{-} ightarrow 	ext{AgCl} \downarrow 	ext{ (White)} $$

Most other common metal cations, such as $	ext{Zn}^{2+}$, $	ext{Bi}^{3+}$, and $	ext{Hg}^{2+}$, form soluble chlorides and will remain in the aqueous solution without forming any precipitate.

### The Second Test: Dilute NaOH

Dilute sodium hydroxide ($	ext{NaOH}$) is a strong base that provides hydroxide ($	ext{OH}^{-}$) ions. Many heavy metal cations form insoluble hydroxides when reacted with $	ext{NaOH}$. However, the colors of these precipitates vary wildly, which is exactly what this question tests.

- **Lead ($	ext{Pb}^{2+}$)** forms a white precipitate of lead(II) hydroxide:
$$ 	ext{Pb}^{2+} + 2	ext{OH}^{-} ightarrow 	ext{Pb(OH)}_2 \downarrow 	ext{ (White)} $$
- **Zinc ($	ext{Zn}^{2+}$)** forms a white precipitate of zinc hydroxide:
$$ 	ext{Zn}^{2+} + 2	ext{OH}^{-} ightarrow 	ext{Zn(OH)}_2 \downarrow 	ext{ (White)} $$
- **Bismuth ($	ext{Bi}^{3+}$)** forms a white precipitate of bismuth(III) hydroxide:
$$ 	ext{Bi}^{3+} + 3	ext{OH}^{-} ightarrow 	ext{Bi(OH)}_3 \downarrow 	ext{ (White)} $$

But here is where things get colorful:
- **Silver ($	ext{Ag}^{+}$)** initially forms silver hydroxide, but it is highly unstable at room temperature and instantly dehydrates to form a **brown** precipitate of silver oxide:
$$ 2	ext{Ag}^{+} + 2	ext{OH}^{-} ightarrow 	ext{Ag}_2	ext{O} \downarrow 	ext{ (Brown)} + 	ext{H}_2	ext{O} $$
- **Mercury(II) ($	ext{Hg}^{2+}$)** similarly forms an unstable hydroxide that decomposes into a **yellow** precipitate of mercury(II) oxide:
$$ 	ext{Hg}^{2+} + 2	ext{OH}^{-} ightarrow 	ext{HgO} \downarrow 	ext{ (Yellow)} + 	ext{H}_2	ext{O} $$

Armed with this knowledge, let's interrogate our suspects (the options).

### Evaluating the Suspects

**Option A: $	ext{Pb(NO}_3	ext{)}_2$ and $	ext{Zn(NO}_3	ext{)}_2$**
- With dilute $	ext{HCl}$: $	ext{Pb}^{2+}$ forms white $	ext{PbCl}_2 \downarrow$, while $	ext{Zn}^{2+}$ remains soluble as $	ext{ZnCl}_2$. The result is a white precipitate.
- With dilute $	ext{NaOH}$: Both $	ext{Pb}^{2+}$ and $	ext{Zn}^{2+}$ form white precipitates ($	ext{Pb(OH)}_2 \downarrow$ and $	ext{Zn(OH)}_2 \downarrow$). The result is only white precipitates.
- **Verdict:** Option A perfectly satisfies both conditions!

**Option B: $	ext{Pb(NO}_3	ext{)}_2$ and $	ext{Bi(NO}_3	ext{)}_3$**
- With dilute $	ext{HCl}$: $	ext{Pb}^{2+}$ forms white $	ext{PbCl}_2 \downarrow$, while $	ext{Bi}^{3+}$ remains soluble as $	ext{BiCl}_3$. The result is a white precipitate.
- With dilute $	ext{NaOH}$: Both $	ext{Pb}^{2+}$ and $	ext{Bi}^{3+}$ form white precipitates ($	ext{Pb(OH)}_2 \downarrow$ and $	ext{Bi(OH)}_3 \downarrow$). The result is only white precipitates.
- **Verdict:** Option B also perfectly satisfies both conditions!

**Option C: $	ext{AgNO}_3$ and $	ext{Bi(NO}_3	ext{)}_3$**
- With dilute $	ext{HCl}$: $	ext{Ag}^{+}$ forms white $	ext{AgCl} \downarrow$, while $	ext{Bi}^{3+}$ remains soluble. So far, so good.
- With dilute $	ext{NaOH}$: $	ext{Bi}^{3+}$ forms white $	ext{Bi(OH)}_3 \downarrow$, but $	ext{Ag}^{+}$ forms the **brown** precipitate of $	ext{Ag}_2	ext{O} \downarrow$. 
- **Verdict:** The presence of the brown precipitate violates the "only white precipitate" rule. Option C is incorrect.

**Option D: $	ext{Pb(NO}_3	ext{)}_2$ and $	ext{Hg(NO}_3	ext{)}_2$**
- With dilute $	ext{HCl}$: $	ext{Pb}^{2+}$ forms white $	ext{PbCl}_2 \downarrow$, while $	ext{Hg}^{2+}$ remains soluble as $	ext{HgCl}_2$. 
- With dilute $	ext{NaOH}$: $	ext{Pb}^{2+}$ forms white $	ext{Pb(OH)}_2 \downarrow$, but $	ext{Hg}^{2+}$ forms the **yellow** precipitate of $	ext{HgO} \downarrow$.
- **Verdict:** The yellow precipitate ruins it. Option D is incorrect.

### The Final Verdict

By systematically applying our knowledge of group reagents and precipitate colors, we have successfully deduced that the mixtures in **Option A** and **Option B** are the only ones that yield exclusively white precipitates under both given conditions.

This problem is a beautiful reminder of why memorizing the characteristic colors of precipitates in salt analysis is absolutely non-negotiable for JEE Advanced. Keep practicing, and soon these colors will become second nature to you!

Select Answer:

Visualized Solution

$Problem Analysis$

$Reaction with Dilute HCl$

$Reaction with Dilute NaOH$

$Evaluating Option A$

$Evaluating Option B$

$Evaluating Option C$

$Evaluating Option D$

$Conclusion$

The Sigma Insight: Salt Analysis

Decoding the Problem Statement

The First Test

The Second Test

Evaluating the Suspects

The Final Verdict

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Problem Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reaction with Dilute HCl

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reaction with Dilute NaOH

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Option A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Option B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Option C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Option D

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion