`0.001 mol` of solid `NaC1` was added to `1.0L` of `0.01M Hg (NO_(3))_(2)`. Calculate `[c1^(Theta)]` equilibrated with newly formed `HgC1^(o+). K_(1)` for `HgC1^(o+)` formation is `5.5 xx 10^(6)`, neglect the `K_(2)` equilibrium.

Explanation

To solve the problem, we need to calculate the concentration of \( \text{Cl}^- \) that is in equilibrium with the newly formed \( \text{HgCl}^+ \) after adding \( 0.001 \, \text{mol} \) of solid \( \text{NaCl} \) to \( 1.0 \, \text{L} \) of \( 0.01 \, \text{M} \, \text{Hg(NO}_3)_2 \). The formation constant \( K_1 \) for \( \text{HgCl}^+ \) is given as \( 5.5 \times 10^6 \).

### Step-by-Step Solution:

1. **Identify Initial Concentrations:**
- The initial concentration of \( \text{Hg}^{2+} \) from \( \text{Hg(NO}_3)_2 \) is \( 0.01 \, \text{M} \) in \( 1.0 \, \text{L} \), which means there are \( 0.01 \, \text{mol} \) of \( \text{Hg}^{2+} \).
- The solid \( \text{NaCl} \) provides \( 0.001 \, \text{mol} \) of \( \text{Cl}^- \).

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