Calculate the `[OH^(-)]` of a solution after 100 mL of 0.1 M `MgCl_(2)` is added to 100 mL 0.2 M NaOH `K_(sp) ` of `Mg(OH)_(2)` is `1.2 xx10^(-11)` .

Explanation

To calculate the concentration of hydroxide ions \([OH^-]\) after mixing 100 mL of 0.1 M \(MgCl_2\) with 100 mL of 0.2 M \(NaOH\), we will follow these steps:

### Step 1: Determine the moles of \(MgCl_2\) and \(NaOH\)
- **Moles of \(MgCl_2\)**:
\[ \text{Moles of } MgCl_2 = \text{Concentration} \times \text{Volume} = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{mol} \]

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