The molar solubility of $Ni(OH)_{2}$ in 0.10 M NaOH when the ionic product of $Ni(OH)_{2}$ is $2.0 imes10^{-15}$ is:

Question

The molar solubility of $Ni(OH)_{2}$ in 0.10 M NaOH when the ionic product of $Ni(OH)_{2}$ is $2.0	imes10^{-15}$ is:

Accepted Answer

To determine the molar solubility of $ Ni(OH)_2 $ in a 0.10 M NaOH solution, we need to consider the solubility product constant ($ K_{sp} $) and the common ion effect.

### Step-by-Step Explanation:

1. **Dissociation of $ Ni(OH)_2 $:**
   $$
   Ni(OH)_2 (s) \rightleftharpoons Ni^{2+} (aq) + 2OH^- (aq)
   $$

2. **Expression for $ K_{sp} $:**
   The solubility product ($ K_{sp} $) for $ Ni(OH)_2 $ is given by:
   $$
   K_{sp} = [Ni^{2+}][OH^-]^2
   $$
   Given $ K_{sp} = 2.0 \times 10^{-15} $.

3. **Common Ion Effect:**
   In a 0.10 M NaOH solution, the concentration of $ OH^- $ ions is already 0.10 M due to the dissociation of NaOH:
   $$
   NaOH \rightarrow Na^+ + OH^-
   $$

4. **Calculating Molar Solubility:**
   Let the molar solubility of $ Ni(OH)_2 $ be $ s $. Then, the concentration of $ Ni^{2+} $ ions in the solution will be $ s $, and the concentration of $ OH^- $ ions will be $ 0.10 + 2s $. However, since $ s $ is very small compared to 0.10 M, we can approximate the concentration of $ OH^- $ as 0.10 M.

Substituting these into the $ K_{sp} $ expression:
   $$
   K_{sp} = [Ni^{2+}][OH^-]^2 = s \times (0.10)^2
   $$
   $$
   2.0 \times 10^{-15} = s \times (0.10)^2
   $$
   $$
   2.0 \times 10^{-15} = s \times 0.01
   $$
   $$
   s = \frac{2.0 \times 10^{-15}}{0.01} = 2.0 \times 10^{-13} \, \text{M}
   $$

5. **Conclusion:**
   The molar solubility of $ Ni(OH)_2 $ in 0.10 M NaOH is $ 2.0 \times 10^{-13} \, \text{M} $, which corresponds to Option 3.

### Explanation of Incorrect Options:

- **Option 1: $ 2 \times 10^{-15} \, \text{M} $**
  - This value is equal to the $ K_{sp} $ itself, which does not account for the common ion effect from the 0.10 M NaOH.

- **Option 2: $ 2 \times 10^{-14} \, \text{M} $**
  - This value is ten times smaller than the correct solubility calculated considering the common ion effect.

- **Option 4: $ 2 \times 10^{-12} \, \text{M} $**
  - This value is ten times larger than the correct solubility, which would imply a much higher solubility than possible under the given conditions.

Thus, Option 3 is the correct answer, as it accurately reflects the molar solubility of $ Ni(OH)_2 $ in the presence of a common ion from NaOH.