The solubility product of `Ni(OH)_(2)` is `2.0xx10^(-15)`. The molar solubility of `Ni(OH)_(2)` in `0.1 M NaOH` solution is

Options

A. (a)`1.0xx10^(-5)`
B. (b)`2.0xx10^(-13)M`
C. (c )`4.44xx10^(-8)M`
D. (d)`2.0xx10^(-12)M`

Explanation

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

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

### Step 2: Write the Expression for the Solubility Product (Ksp)
The solubility product constant (Ksp) expression for Ni(OH)₂ is given by:
\[ K_{sp} = [\text{Ni}^{2+}][\text{OH}^-]^2 \]

Given that \( K_{sp} = 2.0 \times 10^{-15} \).

### Step 3: Determine the Concentration of OH⁻ in the Solution
In a 0.1 M NaOH solution, the concentration of hydroxide ions is:
\[ [\text{OH}^-] = 0.1 \, M \]

### Step 4: Let the Molar Solubility of Ni(OH)₂ be 's'
Let the molar solubility of Ni(OH)₂ in the 0.1 M NaOH solution be \( s \). Therefore:
- The concentration of Ni²⁺ ions will be \( s \).
- The concentration of OH⁻ ions will be \( 0.1 + 2s \). However, since \( s \) will be very small compared to 0.1 M, we can approximate this as:
\[ [\text{OH}^-] \approx 0.1 \, M \]

### Step 5: Substitute into the Ksp Expression
Now substituting the values into the Ksp expression:
\[ K_{sp} = [\text{Ni}^{2+}][\text{OH}^-]^2 \]
\[ 2.0 \times 10^{-15} = (s)(0.1)^2 \]

### Step 6: Solve for 's'
Now we can solve for \( s \):
\[ 2.0 \times 10^{-15} = s \times 0.01 \]
\[ s = \frac{2.0 \times 10^{-15}}{0.01} \]
\[ s = 2.0 \times 10^{-13} \, M \]

### Final Answer
The molar solubility of Ni(OH)₂ in a 0.1 M NaOH solution is:
\[ s = 2.0 \times 10^{-13} \, M \]

---