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 \( \text{Cd(OH)}_2 \) in a 0.1 M KOH solution, we can follow these steps:

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

### Step 2: Define the molar solubility
Let the molar solubility of \( \text{Cd(OH)}_2 \) be \( S \). Therefore, at equilibrium:
- The concentration of \( \text{Cd}^{2+} \) will be \( S \).
- The concentration of \( \text{OH}^- \) will be \( 2S \).

### Step 3: Consider the effect of the added KOH
Since we are adding 0.1 M KOH, the concentration of \( \text{OH}^- \) from KOH will be \( C = 0.1 \, \text{M} \). Thus, the total concentration of \( \text{OH}^- \) in the solution will be:
\[ \text{[OH}^-] = 2S + C = 2S + 0.1 \]

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

### Step 5: Substitute the value of \( K_{sp} \)
We know that \( K_{sp} = 2.5 \times 10^{-14} \). Therefore, we can set up the equation:
\[ 2.5 \times 10^{-14} = S (2S + 0.1)^2 \]

### Step 6: Simplify the equation
Since \( 2S \) is expected to be very small compared to \( 0.1 \) (because \( S \) will be small), we can approximate:
\[ 2S + 0.1 \approx 0.1 \]
Thus, the equation simplifies to:
\[ 2.5 \times 10^{-14} \approx S (0.1)^2 \]
\[ 2.5 \times 10^{-14} = S \times 0.01 \]

### Step 7: Solve for \( S \)
Rearranging gives:
\[ S = \frac{2.5 \times 10^{-14}}{0.01} = 2.5 \times 10^{-12} \, \text{M} \]

### Final Answer
The molar solubility of \( \text{Cd(OH)}_2 \) in 0.1 M KOH solution is:
\[ S = 2.5 \times 10^{-12} \, \text{M} \]