molar solubility of `Cd(OH)_(2)(Ksp=2.5xx10^(-14))` in 0.1 M KOH solution is :-

Options

A. `2.5xx10^(-7)`M
B. `5xx10^(-8)`M
C. `5xx10^(-7)`M
D. `2.5xx10^(-12)`M

Explanation

To find the molar solubility of \( Cd(OH)_2 \) in a 0.1 M KOH solution, we will follow these steps:

### Step 1: Write the dissociation equation for \( Cd(OH)_2 \)
The dissociation of cadmium hydroxide in water can be represented as:
\[ Cd(OH)_2 (s) \rightleftharpoons Cd^{2+} (aq) + 2OH^{-} (aq) \]

### Step 2: Define the solubility (S)
Let the molar solubility of \( Cd(OH)_2 \) in the solution be \( S \). At equilibrium, the concentrations will be:
- \( [Cd^{2+}] = S \)
- \( [OH^{-}] = 2S \)

### Step 3: Consider the common ion effect
Since we are adding \( KOH \) to the solution, which provides \( OH^{-} \) ions, we need to account for this common ion effect. In a 0.1 M KOH solution, the concentration of \( OH^{-} \) ions is:
\[ C = 0.1 \, M \]

Thus, at equilibrium, the concentration of \( OH^{-} \) will be:
\[ [OH^{-}] = 2S + C \approx C \quad (\text{since } 2S \text{ is very small compared to } C) \]
Therefore, we can approximate:
\[ [OH^{-}] \approx 0.1 \, M \]

### Step 4: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for \( Cd(OH)_2 \) is given by:
\[ K_{sp} = [Cd^{2+}][OH^{-}]^2 \]
Substituting the equilibrium concentrations, we have:
\[ K_{sp} = S \cdot (0.1)^2 \]

### Step 5: Substitute the value of \( K_{sp} \)
Given that \( K_{sp} = 2.5 \times 10^{-14} \), we can substitute this into the equation:
\[ 2.5 \times 10^{-14} = S \cdot (0.1)^2 \]
\[ 2.5 \times 10^{-14} = S \cdot 0.01 \]

### Step 6: Solve for S
Now, we can solve for \( S \):
\[ S = \frac{2.5 \times 10^{-14}}{0.01} \]
\[ S = 2.5 \times 10^{-12} \, M \]

### Conclusion
The molar solubility of \( Cd(OH)_2 \) in a 0.1 M KOH solution is:
\[ \boxed{2.5 \times 10^{-12} \, M} \]

---