On addition of aqueous sodium hydroxide to an aqueous solution of salt R, a white precipitate is obtained, soluble in an excess of the hydroxide. A white ppt is also obtained when R is treated with aqueous ammonia , but it is insoluble in excess. Which of the following could be R?

A: Al2(SO4)3
B: CaSO4
c: MgSO4
D: MnSO4
E: ZnSO4

Ans: A
Someone please explain?

To determine which of the given salts could be R, we need to analyze the given information:

1. Adding aqueous sodium hydroxide to the solution of salt R produces a white precipitate that is soluble in excess sodium hydroxide.
2. Treating salt R with aqueous ammonia also produces a white precipitate, but it is insoluble in excess ammonia.

Let's evaluate each option:

A: Al2(SO4)3 (aluminum sulfate)
- When aluminum sulfate reacts with sodium hydroxide, it forms a white precipitate of aluminum hydroxide, which is soluble in excess sodium hydroxide.
- When aluminum sulfate reacts with ammonia, it also forms a white precipitate of aluminum hydroxide, which is insoluble in excess ammonia.

B: CaSO4 (calcium sulfate)
- Calcium sulfate does not react with sodium hydroxide to form a white precipitate.
- Calcium sulfate also does not react with ammonia to form a white precipitate.

C: MgSO4 (magnesium sulfate)
- Magnesium sulfate does not react with sodium hydroxide to form a white precipitate.
- Magnesium sulfate also does not react with ammonia to form a white precipitate.

D: MnSO4 (manganese sulfate)
- Manganese sulfate does not react with sodium hydroxide to form a white precipitate.
- Manganese sulfate also does not react with ammonia to form a white precipitate.

E: ZnSO4 (zinc sulfate)
- Zinc sulfate does not react with sodium hydroxide to form a white precipitate.
- Zinc sulfate also does not react with ammonia to form a white precipitate.

Based on this analysis, only option A, Al2(SO4)3 (aluminum sulfate), matches the given observations. It reacts with both sodium hydroxide and ammonia to form a white precipitate, but only the precipitate formed with sodium hydroxide is soluble in excess. Therefore, A is the correct answer.

To identify the substance (R) responsible for the observed observations, we need to consider the reactions taking place when R is treated with sodium hydroxide and aqueous ammonia.

1. Reaction with sodium hydroxide:
Upon adding aqueous sodium hydroxide (NaOH) to the solution of salt R, a white precipitate is formed. However, the precipitate is soluble in an excess of sodium hydroxide. This indicates that the precipitate is a metal hydroxide that is only sparingly soluble in water but readily dissolves when excess hydroxide is added.

2. Reaction with aqueous ammonia:
When R is treated with aqueous ammonia (NH3), it forms a white precipitate. However, this precipitate remains insoluble even in excess ammonia. This suggests that the precipitate is a metal hydroxide that is not soluble in an excess of ammonia.

Now, let's examine the possible options: Al2(SO4)3, CaSO4, MgSO4, MnSO4, and ZnSO4.

- Al2(SO4)3: The reaction with NaOH results in the formation of white precipitate Al(OH)3, which is soluble in excess NaOH. The reaction with ammonia forms white precipitate Al(OH)3, which is insoluble in excess ammonia. This matches the observations, indicating that aluminum sulfate (Al2(SO4)3) could be R.

- CaSO4, MgSO4, MnSO4, and ZnSO4: These salts do not match the observations. While they may result in the formation of white precipitates with NaOH, these precipitates are either insoluble in excess NaOH or are not formed at all.

Therefore, based on the given reactions and observations, the substance responsible for the observed behavior is likely Al2(SO4)3. Therefore, option A is the correct answer.