During the detection of acidic radical present in a salt, a student gets a pale yellow precipitate soluble with…

Step 1: Focus on the reagents used

The test describes adding silver nitrate ($\ce{AgNO3}$) to an acidified sodium carbonate extract. This is the classic Silver Nitrate Test for halides.

Step 2: Note the observation

The reaction produces a pale yellow precipitate that is soluble with difficulty (sparingly soluble) in $\ce{NH4OH}$.

Step 3: Compare expected results for common halides with $\ce{AgNO3}$

1. Chloride ($\ce{Cl-}$): Yields a curdy white precipitate ($\ce{AgCl}$) which is readily soluble in dilute $\ce{NH4OH}$.
2. Bromide ($\ce{Br-}$): Yields a pale yellow precipitate ($\ce{AgBr}$) which is sparingly (with difficulty) soluble in $\ce{NH4OH}$.
3. Iodide ($\ce{I-}$): Yields a bright yellow precipitate ($\ce{AgI}$) which is completely insoluble in $\ce{NH4OH}$.

Step 4: Conclusion

The properties perfectly match the behavior of the bromide completely ($\ce{Br-}$).

Key Points to Remember:

• $\ce{AgCl}$: White ppt, fully soluble in dilute ammonia.
• $\ce{AgBr}$: Pale yellow ppt, partially/sparingly soluble in ammonia.
• $\ce{AgI}$: Bright yellow ppt, totally insoluble in ammonia.