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A colourless solid A dissolves in water....

A colourless solid A dissolves in water. The aqueous solution gives a white precipitate B when NaOH or `NH_4 OH` is added. B dissolves in excess of NaOH but not in excess of `NH_4OH`. BaCl2 solution added to a solution of A gives a white precipitate C which is insoluble in dilute HCI. A may be 

A

`Al_(2)(SO_(4))_(3)`

B

`ZnSO_4`

C

`SnSO_4`

D

All of the above

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the information given step by step to identify the possible identity of the colorless solid A. ### Step 1: Identify the properties of solid A - A is a colorless solid that dissolves in water. - When dissolved, it gives a white precipitate B upon the addition of NaOH or NH₄OH. **Hint:** Look for common salts that are soluble in water and can react with bases to form precipitates. ### Step 2: Determine the nature of precipitate B - The precipitate B is formed when NaOH or NH₄OH is added. - This precipitate B dissolves in excess NaOH but not in excess NH₄OH. **Hint:** Consider the hydroxides that can form precipitates and their solubility in different conditions. ### Step 3: Analyze the behavior of precipitate B - Since B dissolves in excess NaOH, it suggests that B is likely a metal hydroxide that can form soluble aluminate, zincate, or stannate ions. - The fact that B does not dissolve in excess NH₄OH indicates that it is not an ammonium complex. **Hint:** Think about metal hydroxides that are amphoteric (soluble in strong bases) but not soluble in ammonium hydroxide. ### Step 4: Precipitate C formation - When BaCl₂ is added to the solution of A, it gives a white precipitate C that is insoluble in dilute HCl. - This indicates that C is likely barium sulfate (BaSO₄), which is known to be insoluble in dilute acids. **Hint:** Consider the sulfate salts that could be formed from the reaction of A with BaCl₂. ### Step 5: Identify possible candidates for A 1. **Al₂(SO₄)₃ (Aluminum sulfate)**: - Forms Al(OH)₃ (B) when reacted with NaOH, which is a white precipitate. - Al(OH)₃ dissolves in excess NaOH to form NaAlO₂. - When reacted with BaCl₂, it forms BaSO₄ (C), which is insoluble in dilute HCl. 2. **ZnSO₄ (Zinc sulfate)**: - Forms Zn(OH)₂ (B) when reacted with NaOH, which is a white precipitate. - Zn(OH)₂ dissolves in excess NaOH to form Na₂ZnO₂. - When reacted with BaCl₂, it also forms BaSO₄ (C), which is insoluble in dilute HCl. 3. **SnSO₄ (Tin(II) sulfate)**: - Forms Sn(OH)₂ (B) when reacted with NaOH, which is a white precipitate. - Sn(OH)₂ dissolves in excess NaOH to form Na₂SnO₂. - When reacted with BaCl₂, it forms BaSO₄ (C), which is insoluble in dilute HCl. ### Conclusion Since all three compounds (Al₂(SO₄)₃, ZnSO₄, and SnSO₄) fit the criteria outlined in the problem, the answer to the question is that A may be any of the following: Al₂(SO₄)₃, ZnSO₄, or SnSO₄. **Final Answer:** All of the above (Al₂(SO₄)₃, ZnSO₄, SnSO₄) can be the colorless solid A.
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