The reaction Pb(NO3)2 and NaCl in water produces a precipitate that dissolves upon the addition of H

Q: Why doesn't $\text{PbCl}_2$ dissolve in water?

$\text{PbCl}_2$ has a high lattice energy compared to its hydration energy at room temperature, making it sparingly soluble in cold water. However, it is soluble in hot water!

Q: Why does adding $\text{HCl}$ dissolve the precipitate instead of forming more?

While a small amount of $\text{Cl}^-$ would shift the equilibrium to form more precipitate (Common Ion Effect), an *excess* of $\text{Cl}^-$ acts as ligands, forming a soluble coordination complex $[\text{PbCl}_4]^{2-}$.

Q: What is the coordination number of lead in this complex?

The coordination number is 4, as four chloride ligands are attached to the central lead ion.

Q: Can other metal chlorides show similar behavior?

Yes! For example, $\text{AgCl}$ dissolves in excess ammonia to form $[\text{Ag(NH}_3)_2]^+$, and $\text{CuCl}$ dissolves in excess chloride to form $[\text{CuCl}_2]^-$. Complexation is a powerful tool to dissolve precipitates.

Q: Is the oxidation state of lead changing here?

No, the oxidation state of lead remains $+2$ throughout the entire process. It's purely a complexation reaction, not a redox reaction.

Question

The reaction $	ext{Pb(NO}_3)_2$ and $	ext{NaCl}$ in water produces a precipitate that dissolves upon the addition of $	ext{HCl}$ of appropriate concentration. The dissolution of the precipitate is due to the formation of

Options:

(A) $	ext{PbCl}_2$

(B) $	ext{PbCl}_4$

(C) $[	ext{PbCl}_4]^{2-}$

(D) $[	ext{PbCl}_6]^{2-}$

Sigma Percentile · Accepted Answer

### The Initial Encounter: Mixing the Reagents

Imagine you are standing in a chemistry laboratory, holding a test tube filled with a clear, aqueous solution of lead nitrate, $	ext{Pb(NO}_3)_2$.

To this, you slowly add another clear solution of sodium chloride, $	ext{NaCl}$. Both of these are highly soluble salts, completely dissociated into their respective ions in water.

But what happens the moment these two different sets of ions meet in the same container? A chemical reaction is bound to happen.

### The White Precipitate: Lead Chloride

As the solutions mix, the $	ext{Pb}^{2+}$ ions come into contact with the $	ext{Cl}^-$ ions.

According to the solubility rules, lead(II) chloride is sparingly soluble in cold water. So, the moment they combine, they form a solid lattice.

$$ 	ext{Pb(NO}_3)_2	ext{(aq)} + 2	ext{NaCl(aq)} ightarrow 	ext{PbCl}_2\downarrow + 2	ext{NaNO}_3	ext{(aq)} $$

This solid immediately crashes out of the solution, appearing as a dense, milky white precipitate. You can physically see it settling down at the bottom of the test tube.

The formation of this precipitate is the crucial first half of our problem.

### The Magic of Complexation: Dissolving the Insoluble

Now, the question introduces a twist. It states that we add an appropriate concentration of hydrochloric acid ($	ext{HCl}$) to this test tube containing the precipitate.

Hydrochloric acid is a strong acid, providing an abundance of $	ext{H}^+$ and $	ext{Cl}^-$ ions. So, we are essentially flooding the system with excess chloride ions.

What does this excess chloride do to our stubborn lead chloride precipitate? This is where the magic of coordination chemistry comes into play.

$$ 	ext{PbCl}_2	ext{(s)} + 2	ext{HCl(aq)} ightarrow 	ext{H}_2[	ext{PbCl}_4]	ext{(aq)} $$

Lead, being a heavy p-block element, has empty orbitals available and can expand its coordination sphere. The excess chloride ions from the acid act as strong ligands.

They attack the solid lead chloride lattice, coordinating with the central lead ion. This forms a new, highly stable, and most importantly, water-soluble coordination complex.

$$ 	ext{H}_2[	ext{PbCl}_4]	ext{(aq)} ightleftharpoons 2	ext{H}^+	ext{(aq)} + [	ext{PbCl}_4]^{2-}	ext{(aq)} $$

As this complex forms, the solid lattice breaks apart, and the precipitate completely dissolves back into a clear solution.

### The Final Conclusion

Therefore, the dissolution of the precipitate is entirely due to the formation of this specific complex anion, $[	ext{PbCl}_4]^{2-}$.

**Many insoluble metal halides can be forced to dissolve by adding an excess of the same halide ions, driving the formation of a soluble complex.**

This is a classic, high-yield concept in qualitative salt analysis that you will encounter frequently in your exams.

Animated Solution for Chemistry - Salt Analysis: The reaction $Pb(NO_{3})_{2}$ and $NaCl$ in water produces a precipitate that dissolves upon the addition of $HCl$ of appropriate concentration. The dissolution of the precipitate is due to the formation of

Select Answer:

Visualized Solution

$Initial Reactants$

$Precipitation Reaction$

$Balanced Equation$

$Addition of HCl$

$Complex Formation$

$Conclusion$

The Sigma Insight: Salt Analysis

The Initial Encounter

The White Precipitate

The Magic of Complexation

The Final Conclusion

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