During the detection of acidic radical present in a salt, a student gets a pale yellow precipitate soluble with difficulty in $\text{NH}_4\text{OH}$ solution when sodium carbonate extract was first acidified with dil. $\text{HNO}_3$ and then $\text{AgNO}_3$ solution was added. This indicates the presence of:
- $\text{Br}^-$
- $\text{CO}_3^{2-}$
- $\text{I}^-$
- $\text{Cl}^-$
The Correct Option is A
Approach Solution - 1
To identify the acidic radical present in a salt, a student performs the following steps:
- The sodium carbonate extract of the salt is acidified with dilute \(\text{HNO}_3\).
- \(\text{AgNO}_3\) solution is added to the acidified solution.
Upon adding \(\text{AgNO}_3\), a pale yellow precipitate is formed which is soluble with difficulty in \(\text{NH}_4\text{OH}\) (ammonium hydroxide) solution. This behavior suggests the presence of the bromide ion \(\text{Br}^−\). Here's the reasoning:
- Formation of Precipitate: When \(\text{AgNO}_3\) is added, it reacts with the halide ions to form silver halides. Each halide ion forms a distinct colored precipitate. For bromide ions, the reaction is:
- The \(\text{AgBr}\) formed is a pale yellow precipitate.
- Solubility in \(\text{NH}_4\text{OH}\): The solubility of silver halides in ammonium hydroxide can distinguish between them:
- \(\text{AgCl}\) is white and soluble in dilute \(\text{NH}_4\text{OH}\).
- \(\text{AgBr}\) is pale yellow and partially soluble with difficulty in dilute \(\text{NH}_4\text{OH}\).
- \(\text{AgI}\) is yellow and insoluble in dilute \(\text{NH}_4\text{OH}\).
Given the choices and explanations, the presence of bromide ion \(\text{Br}^−\) is confirmed by the formation of a pale yellow precipitate which is soluble with difficulty in \(\text{NH}_4\text{OH}\). Therefore, the correct answer is: \(\text{Br}^−\)
Approach Solution -2
The reactions with \( \text{Ag}^+ \) ions produce different colored precipitates depending on the halide present:
\[ \text{Ag}^+ + \text{I}^- \rightarrow \text{AgI} \quad (\text{Yellow ppt.}) \]
\[ \text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl} \quad (\text{White ppt.}) \]
\[ \text{Ag}^+ + \text{Br}^- \rightarrow \text{AgBr} \quad (\text{Pale yellow ppt.}) \]
Since a pale yellow precipitate was obtained, it indicates the presence of \( \text{Br}^- \) ions.
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