Find out the solubility of `Ni(OH)_2` in `0.1 M` NaOH Given that the ionic product of `Ni(OH)_2` is `2xx10^(-15)`.

Options

A. `2xx10^(-13)`M
B. `2xx10^(-8)` M
C. `1xx10^(-13)`M
D. `1xx10^8` M

Answer

To find the solubility of `Ni(OH)₂` in `0.1 M` NaOH, we can follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of nickel hydroxide can be represented as:
\[ \text{Ni(OH)}_2 (s) \rightleftharpoons \text{Ni}^{2+} (aq) + 2 \text{OH}^- (aq) \]

### Step 2: Define Solubility
Let the solubility of `Ni(OH)₂` in `0.1 M` NaOH be \( S \) mol/L. Therefore, when `Ni(OH)₂` dissolves:
- The concentration of `Ni²⁺` ions will be \( S \).
- The concentration of `OH⁻` ions will be \( 0.1 + 2S \) (since there are already 0.1 M from NaOH and 2 moles of OH⁻ produced per mole of `Ni(OH)₂`).

### Step 3: Write the Expression for Ionic Product
The ionic product \( K_{sp} \) for `Ni(OH)₂` is given by:
\[ K_{sp} = [\text{Ni}^{2+}][\text{OH}^-]^2 \]
Substituting the concentrations:
\[ K_{sp} = S \cdot (0.1 + 2S)^2 \]

### Step 4: Substitute the Given Value of \( K_{sp} \)
We know that \( K_{sp} = 2 \times 10^{-15} \):
\[ 2 \times 10^{-15} = S \cdot (0.1 + 2S)^2 \]

### Step 5: Assume \( S \) is Small Compared to 0.1
Since the solubility of `Ni(OH)₂` is expected to be very low compared to 0.1 M, we can simplify:
\[ 0.1 + 2S \approx 0.1 \]
Thus, we can rewrite the equation as:
\[ 2 \times 10^{-15} = S \cdot (0.1)^2 \]

### Step 6: Solve for \( S \)
Now, substituting \( 0.1^2 = 0.01 \):
\[ 2 \times 10^{-15} = S \cdot 0.01 \]
\[ S = \frac{2 \times 10^{-15}}{0.01} \]
\[ S = 2 \times 10^{-13} \]

### Conclusion
The solubility of `Ni(OH)₂` in `0.1 M` NaOH is:
\[ S = 2 \times 10^{-13} \, \text{mol/L} \]

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