What happens when aluminium and zinc salts react with an excess of `NaOH`

Options

A. White precipitate of formed
B. White precipitate of both Zn and Al first formed redissolve in excess of `NaOH`
C. White precipitate of Al redissolves but of Zn does not
D. White precipitate of Zn redissolves and that of Al does not

Explanation

When aluminum and zinc salts react with an excess of sodium hydroxide (NaOH), the following reactions occur:

### Step 1: Reaction of Aluminum Salts with NaOH
1. **Initial Reaction**: When aluminum salts (like AlCl₃) react with NaOH, they initially form aluminum hydroxide (Al(OH)₃), which is a white precipitate.
\[ \text{Al}^{3+} + 3 \text{NaOH} \rightarrow \text{Al(OH)}_3 \downarrow + 3 \text{Na}^+ \]

2. **Excess NaOH**: When more NaOH is added (in excess), the aluminum hydroxide dissolves to form sodium aluminate (NaAlO₂), which is soluble in water.
\[ \text{Al(OH)}_3 + \text{OH}^- \rightarrow \text{NaAlO}_2 + \text{H}_2 \uparrow \]

### Step 2: Reaction of Zinc Salts with NaOH
1. **Initial Reaction**: When zinc salts (like ZnSO₄) react with NaOH, they also form zinc hydroxide (Zn(OH)₂), which is a white precipitate.
\[ \text{Zn}^{2+} + 2 \text{NaOH} \rightarrow \text{Zn(OH)}_2 \downarrow + 2 \text{Na}^+ \]

2. **Excess NaOH**: With the addition of excess NaOH, zinc hydroxide dissolves to form sodium zincate (Na₂ZnO₂), which is also soluble.
\[ \text{Zn(OH)}_2 + 2 \text{OH}^- \rightarrow \text{Na}_2\text{ZnO}_2 + \text{H}_2 \uparrow \]

### Conclusion
In summary, both aluminum and zinc salts react with excess NaOH to form soluble compounds after initially forming white precipitates. The correct option based on the reactions is that both aluminum and zinc initially form white precipitates, which then dissolve upon the addition of excess NaOH.