`500mL` of `2 xx 10^(-3)M A1C1_(3)` and `500mol` of `4 xx 10^(-2)M` solution of `NaOH` are mixed and solution is diluted to `10^(-2)L` with water at room temperature wil a precipiate exist? Given:
`K_(sp)`of `A1(OH)_(3) = 5 xx 10^(-33)`.

Explanation

To determine whether a precipitate will form when mixing the solutions of \( \text{AlCl}_3 \) and \( \text{NaOH} \), we need to follow these steps:

### Step 1: Calculate the concentrations of \( \text{Al}^{3+} \) and \( \text{OH}^- \)

1. **Find the concentration of \( \text{Al}^{3+} \)**:
- The initial concentration of \( \text{AlCl}_3 \) is \( 2 \times 10^{-3} \, M \) and the volume is \( 500 \, mL \).
- After mixing with \( 500 \, mL \) of \( \text{NaOH} \), the total volume becomes \( 1000 \, mL \) or \( 1 \, L \).
- Using the dilution formula \( M_1 V_1 = M_2 V_2 \):
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