Question

The reaction $Pb \left( NO _3\right)_2$ 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

• A. $PbCl _2$
• B. $PbCl _4$
• C. $\left[ PbCl _4\right]^{2-}$
• D. $\left[ PbCl _6\right]^{2-}$

Answer 1

The Correct Option is C

We are given the following reaction in aqueous solution:

\[ \text{Pb(NO}_3\text{)}_2 + \text{NaCl} \rightarrow \text{PbCl}_2 (\text{precipitate}) + 2 \text{NaNO}_3 \]

When the precipitate, \( \text{PbCl}_2 \), is formed and then dissolves upon the addition of HCl, this dissolution is due to the formation of a soluble complex ion.

Step 1: Precipitation of \( \text{PbCl}_2 \)

When \( \text{Pb(NO}_3\text{)}_2 \) and \( \text{NaCl} \) are mixed in water, a precipitation reaction occurs: \[ \text{Pb}^{2+} + 2 \text{Cl}^- \rightarrow \text{PbCl}_2 (s) \] Here, lead(II) chloride (\( \text{PbCl}_2 \)) forms as a white precipitate.

Step 2: Dissolution of the Precipitate

Upon adding excess HCl, the \( \text{PbCl}_2 \) precipitate dissolves, forming a soluble lead(II) chloride complex. This occurs because the chloride ions from HCl coordinate with lead ions to form a complex. The formation of the complex ion can be represented as: \[ \text{PbCl}_2 (s) + 2 \text{Cl}^- \rightleftharpoons [\text{PbCl}_4]^{2-} \] The \( [\text{PbCl}_4]^{2-} \) complex is soluble in water, which leads to the dissolution of the precipitate.

Step 3: Conclusion

The dissolution of \( \text{PbCl}_2 \) is due to the formation of the soluble \( [\text{PbCl}_4]^{2-} \) complex ion. Therefore, the correct answer is:

The correct option is C: \( [\text{PbCl}_4]^{2-} \)

Answer 2

When lead ions (Pb²⁺) react with chloride ions (Cl⁻), a white precipitate of lead(II) chloride (PbCl₂​) forms.
This precipitate is soluble in concentrated hydrochloric acid due to formation of tetrachloroplumbate (II) ion

So the correct answer option is (C)