Calculate the solubility of `PbI_(2)` in water at `25^@C` which is 90% dissociated `K_(sp) (PbI_(2))=1.39 xx 10^(-8)" at "25^@C`.

Explanation

To calculate the solubility of lead iodide (PbI₂) in water at 25°C, given that it is 90% dissociated and the solubility product constant (Ksp) is 1.39 × 10^(-8), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of lead iodide in water can be represented as:
\[ \text{PbI}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{I}^- (aq) \]

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