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A metal chloride original solution (i.e....

A metal chloride original solution (i.e. `O.S`) on mixing with `K_(2)CrO_(4)` solution give a yellow precipitate soluble in aqueous sodium hydroxide.The metal may be:

A

mercury

B

iron

C

silver

D

lead

Text Solution

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To solve the problem, we need to identify the metal chloride that forms a yellow precipitate when mixed with potassium chromate (K₂CrO₄) and is soluble in aqueous sodium hydroxide (NaOH). ### Step-by-Step Solution: 1. **Understanding the Reaction**: - When a metal chloride solution is mixed with K₂CrO₄, a yellow precipitate is formed. This indicates that the metal ion reacts with the chromate ion (CrO₄²⁻) to form a metal chromate. 2. **Identifying the Yellow Precipitate**: - The yellow precipitate formed in this reaction is likely to be lead(II) chromate (PbCrO₄). This is a well-known yellow compound that precipitates when lead(II) ions (Pb²⁺) react with chromate ions. 3. **Solubility in Aqueous Sodium Hydroxide**: - The problem states that the yellow precipitate is soluble in aqueous sodium hydroxide. Lead(II) chromate (PbCrO₄) can dissolve in NaOH to form a soluble complex, specifically lead(II) hydroxide complex [Pb(OH)₄]⁻. 4. **Writing the Chemical Equations**: - The first reaction can be represented as: \[ \text{Pb}^{2+} + \text{CrO}_4^{2-} \rightarrow \text{PbCrO}_4 \, (\text{yellow precipitate}) \] - The second reaction, where the yellow precipitate dissolves in NaOH, can be represented as: \[ \text{PbCrO}_4 + 4 \text{OH}^- \rightarrow [\text{Pb(OH)}_4]^{-} \] 5. **Conclusion**: - Based on the above reactions, we conclude that the metal in question is lead (Pb). ### Final Answer: The metal may be **lead (Pb)**.
To solve the problem, we need to identify the metal chloride that forms a yellow precipitate when mixed with potassium chromate (K₂CrO₄) and is soluble in aqueous sodium hydroxide (NaOH). ### Step-by-Step Solution: 1. **Understanding the Reaction**: - When a metal chloride solution is mixed with K₂CrO₄, a yellow precipitate is formed. This indicates that the metal ion reacts with the chromate ion (CrO₄²⁻) to form a metal chromate. 2. **Identifying the Yellow Precipitate**: ...
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