Will a precipitate exist at equilbriun of 1/2 litre of a `4xx10^(-2)` M solution of NaOH and 1/2 litre of a solution of `2xx10^(-3)` solution of `AlCl_3` are mixed and diluted to `10^3` litres with water at room temperature ? `K_(sp)` of `Al(OH)_3=5xx10^(-33)`

Explanation

To determine whether a precipitate will exist at equilibrium when mixing a solution of NaOH with AlCl3, we can follow these steps:

### Step 1: Determine the concentrations after mixing
We start with:
- Volume of NaOH solution = 0.5 L
- Concentration of NaOH = \(4 \times 10^{-2} \, \text{M}\)
- Volume of AlCl3 solution = 0.5 L
- Concentration of AlCl3 = \(2 \times 10^{-3} \, \text{M}\)

When mixed, the total volume becomes:
\[ V_{\text{total}} = 0.5 \, \text{L} + 0.5 \, \text{L} = 1.0 \, \text{L} \]
Then, we dilute this mixture to \(10^3 \, \text{L}\).

### Step 2: Calculate the concentrations in the final volume
1. **Concentration of NaOH after dilution:**
\[ [\text{NaOH}] = \frac{0.5 \, \text{L} \times 4 \times 10^{-2} \, \text{M}}{10^3 \, \text{L}} = \frac{2 \times 10^{-2}}{10^3} = 2 \times 10^{-5} \, \text{M} \]

2. **Concentration of AlCl3 after dilution:**
\[ [\text{AlCl}_3] = \frac{0.5 \, \text{L} \times 2 \times 10^{-3} \, \text{M}}{10^3 \, \text{L}} = \frac{1 \times 10^{-3}}{10^3} = 1 \times 10^{-6} \, \text{M} \]

### Step 3: Calculate the concentrations of ions
- **Concentration of OH⁻ ions:**
Since NaOH dissociates completely:
\[ [\text{OH}^-] = [\text{NaOH}] = 2 \times 10^{-5} \, \text{M} \]

- **Concentration of Al³⁺ ions:**
AlCl3 dissociates to give:
\[ [\text{Al}^{3+}] = [\text{AlCl}_3] = 1 \times 10^{-6} \, \text{M} \]

### Step 4: Calculate the ionic product (IP)
The solubility product expression for Al(OH)₃ is:
\[ K_{sp} = [\text{Al}^{3+}][\text{OH}^-]^3 \]
Substituting the concentrations:
\[ IP = [\text{Al}^{3+}][\text{OH}^-]^3 = (1 \times 10^{-6}) \times (2 \times 10^{-5})^3 \]
Calculating \( (2 \times 10^{-5})^3 \):
\[ (2 \times 10^{-5})^3 = 8 \times 10^{-15} \]
Thus,
\[ IP = (1 \times 10^{-6}) \times (8 \times 10^{-15}) = 8 \times 10^{-21} \]

### Step 5: Compare IP with Ksp
Given \( K_{sp} \) of Al(OH)₃ is \( 5 \times 10^{-33} \).

Since:
\[ IP = 8 \times 10^{-21} \quad \text{and} \quad K_{sp} = 5 \times 10^{-33} \]
We find that:
\[ IP > K_{sp} \]

### Conclusion
Since the ionic product is greater than the solubility product, a precipitate of Al(OH)₃ will form.

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