What is the concentration of free `Cd^(2+)` in `0.005M CdC1_(2)? K_(1)` for chloride complexation of `Cd^(2+)` is `100, K_(2)` need not be considered.

Explanation

To find the concentration of free \( \text{Cd}^{2+} \) ions in a \( 0.005 \, M \) \( \text{CdCl}_2 \) solution, we can follow these steps:

### Step 1: Determine Initial Concentrations
When \( \text{CdCl}_2 \) dissociates, it produces \( \text{Cd}^{2+} \) and \( \text{Cl}^- \) ions. The dissociation can be represented as:
\[ \text{CdCl}_2 \rightarrow \text{Cd}^{2+} + 2 \text{Cl}^- \]
Given that the concentration of \( \text{CdCl}_2 \) is \( 0.005 \, M \):
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