Calculate the molar solubility of `Ni(OH)_2` in 0.10 M NaOH. The solubility product of `Ni(OH)_2` is `2.0xx10^(-15)`.

Explanation

To calculate the molar solubility of `Ni(OH)₂` in a 0.10 M NaOH solution, we can follow these steps:

### Step 1: Write the dissociation equation for `Ni(OH)₂`
When `Ni(OH)₂` dissolves in water, it dissociates into its ions:
\[ \text{Ni(OH)}_2 (s) \rightleftharpoons \text{Ni}^{2+} (aq) + 2 \text{OH}^- (aq) \]

### Step 2: Define the solubility
Let the molar solubility of `Ni(OH)₂` be \( S \) mol/L. Therefore, at equilibrium:
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