An aqueous solution contains \(\mathrm{A} \mathrm{A}^{3+} \& \mathrm{Zn}^{2+}\) both. To this solution \(\mathrm{NH}_{4} \mathrm{OH}\) is added in excess. (1) only Al(OH) will be precipitated (2) only \(\mathrm{Zn}(\mathrm{OH})_{2}\) will be precipitated (3) both will be precipitated (4) no precipitate will appear

A precipitation reaction occurs when ions in a solution combine to form an insoluble compound, known as a precipitate. In this exercise, we're looking at how ions \( \text{A}^{3+} \) and \( \text{Zn}^{2+} \) interact with \( \text{NH}_4\text{OH} \) (ammonium hydroxide).
An important factor in determining whether a precipitation reaction occurs is the solubility of the resulting compound in water.
When \( \text{NH}_4\text{OH} \) is added to a solution containing \(\text{A}^{3+} \) and \( \text{Zn}^{2+} \), hydroxide ions (OH\textsuperscript{-}) are introduced into the mixture.
These hydroxide ions can potentially combine with \( \text{A}^{3+} \) and \( \text{Zn}^{2+} \) to form \( \text{A}(\text{OH})_3 \) and \( \text{Zn}(\text{OH})_2 \) respectively. These reactions can be represented as follows: \[ \text{A}^{3+} + 3\text{OH}^{-} \rightarrow \text{A}(\text{OH})_3 \] \[ \text{Zn}^{2+} + 2\text{OH}^{-} \rightarrow \text{Zn}(\text{OH})_2 \] Initially, both of these reactions will produce precipitates of aluminum hydroxide (Al(OH)_3) and zinc hydroxide (Zn(OH)_2). However, we need to consider solubility rules and further reactions to understand what ultimately precipitates.

Solubility rules help us predict the solubility of ionic compounds in water. These rules are key in determining which substances will form precipitates and which will remain dissolved.
For this exercise, we need to know how aluminum and zinc hydroxides behave in water.
According to solubility rules:

• Most hydroxides (OH\textsuperscript{-}) are insoluble in water, but there are exceptions. For instance, group 1 hydroxides are soluble.
• Aluminum hydroxide, \( \text{A}(\text{OH})_3 \), is insoluble in water and will form a precipitate.
• Zinc hydroxide, \( \text{Zn}(\text{OH})_2 \), is also insoluble in water but has some solubility in excess ammonium hydroxide.

When dealing with zinc hydroxide, another important reaction takes place in the presence of excess \( \text{NH}_4\text{OH} \): \[ \text{Zn}(\text{OH})_2 + 4\text{NH}_4\text{OH} \rightarrow [\text{Zn}(\text{NH}_3)_4]^{2+} \] This reaction forms a soluble complex ion, \( [\text{Zn}(\text{NH}_3)_4]^{2+} \), meaning \( \text{Zn}(\text{OH})_2 \) dissolves in the presence of excess ammonium hydroxide. Hence, while aluminum hydroxide will precipitate, zinc hydroxide will dissolve under these conditions.

Complex formation occurs when simple ions combine with other molecules or ions to form a larger, more complex ion. In this exercise, complex formation plays a crucial role in determining the solubility of zinc hydroxide.
When \( \text{Zn}(\text{OH})_2 \) reacts with an excess of \( \text{NH}_4\text{OH} \), it forms a complex ion called \( [\text{Zn}(\text{NH}_3)_4]^{2+} \). This reaction can be written as: \[ \text{Zn}(\text{OH})_2 + 4\text{NH}_4(\text{OH}) \rightarrow [\text{Zn}(\text{NH}_3)_4]^{2+} + 2\text{H}_2\text{O} \] The formation of this complex ion is important because it changes the solubility properties of zinc hydroxide.
In the presence of excess ammonium hydroxide, zinc hydroxide that initially forms as a precipitate dissolves to form the soluble complex ion. Consequently, no solid zinc hydroxide remains in the solution.
In summary, only \( \text{A}(\text{OH})_3 \) will be left as an insoluble precipitate, because the zinc hydroxide will have dissolved due to complex formation. Understanding these chemical interactions and solubility rules allows us to predict the outcomes of such reactions accurately.