To a solution of colourless sodium salt, a solution of lead nitrate was added to have a white precipitate which dissolves in warm water and reprecipitates on cooling. Which of the following acid radical is present in the salt?
- CI-
- SO2-4
- S2-
- NO-3
The Correct Option is A
Solution and Explanation
The presence of Cl- (chloride) is indicated by the observations.
Here's a breakdown:
- Reaction with Lead Nitrate: The addition of lead nitrate (Pb(NO3)2) to a chloride salt will produce a white precipitate of lead(II) chloride (PbCl2): $Pb^{2+}(aq) + 2Cl^{-}(aq) \rightarrow PbCl_2(s)$
- Solubility in Warm Water: Lead(II) chloride (PbCl2) is unique in that it is soluble in hot water. The elevated temperature increases its solubility, causing the precipitate to dissolve: $PbCl_2(s) \xrightarrow{warm \, water} Pb^{2+}(aq) + 2Cl^{-}(aq)$
- Reprecipitation on Cooling: As the warm water cools, the solubility of lead(II) chloride decreases, causing it to reprecipitate out of solution as a white solid.
- Other Anions:
- Lead sulfate (PbSO4) is also a white precipitate, but it is largely insoluble in water, even when heated.
- Lead sulfide (PbS) is a black precipitate.
- Lead nitrate (Pb(NO3)2) is soluble, so nitrate (NO3-) would not form a precipitate.
Therefore, the given observations strongly suggest the presence of chloride (Cl-) ions in the original sodium salt.
Correct Answer: Cl-
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