The molar solubility of `Cd(OH)_(2)` is `1.84xx10^(-5)M` in water. The expected solubility of `Cd(OH)_(2)` in a buffer solution of `pH=10` is `2.49xx10^(-x)M`. The numerical value of x is.

Explanation

To solve the problem, we need to determine the numerical value of \( x \) in the expected solubility of \( Cd(OH)_2 \) in a buffer solution of \( pH = 10 \).

### Step-by-Step Solution:

1. **Dissociation of \( Cd(OH)_2 \)**:
The dissociation of \( Cd(OH)_2 \) in water can be represented as:
\[ Cd(OH)_2 (s) \rightleftharpoons Cd^{2+} (aq) + 2OH^{-} (aq) \]
Let the molar solubility of \( Cd(OH)_2 \) in pure water be \( S \). Therefore, the concentration of \( Cd^{2+} \) will be \( S \) and the concentration of \( OH^{-} \) will be \( 2S \).

2. **Expression for \( K_{sp} \)**:
The solubility product constant \( K_{sp} \) can be expressed as:
\[ K_{sp} = [Cd^{2+}][OH^{-}]^2 = S \cdot (2S)^2 = 4S^3 \]
Given that \( S = 1.84 \times 10^{-5} \, M \), we can calculate \( K_{sp} \):
\[ K_{sp} = 4 \cdot (1.84 \times 10^{-5})^3 \]

3. **Calculating \( K_{sp} \)**:
First, calculate \( (1.84 \times 10^{-5})^3 \):
\[ (1.84)^3 = 6.244 \quad \text{and} \quad (10^{-5})^3 = 10^{-15} \]
Thus,
\[ (1.84 \times 10^{-5})^3 = 6.244 \times 10^{-15} \]
Now, substituting back into the \( K_{sp} \) expression:
\[ K_{sp} = 4 \cdot 6.244 \times 10^{-15} = 2.4976 \times 10^{-14} \]

4. **Solubility in Buffer Solution**:
In a buffer solution with \( pH = 10 \), the \( pOH \) is:
\[ pOH = 14 - pH = 4 \]
Therefore, the concentration of \( OH^{-} \) in the buffer is:
\[ [OH^{-}] = 10^{-4} \, M \]

5. **Setting up the solubility equation**:
Let \( S' \) be the solubility of \( Cd(OH)_2 \) in the buffer. The equilibrium can be expressed as:
\[ K_{sp} = [Cd^{2+}][OH^{-}]^2 = S' \cdot (2S' + 10^{-4})^2 \]
Since \( S' \) is expected to be very small compared to \( 10^{-4} \), we can approximate:
\[ (2S' + 10^{-4}) \approx 10^{-4} \]
Therefore:
\[ K_{sp} = S' \cdot (10^{-4})^2 = S' \cdot 10^{-8} \]

6. **Solving for \( S' \)**:
Equating the two expressions for \( K_{sp} \):
\[ S' \cdot 10^{-8} = 2.4976 \times 10^{-14} \]
Solving for \( S' \):
\[ S' = \frac{2.4976 \times 10^{-14}}{10^{-8}} = 2.4976 \times 10^{-6} \, M \]

7. **Finding \( x \)**:
The problem states that the expected solubility \( S' = 2.49 \times 10^{-x} \, M \). From our calculation:
\[ 2.4976 \times 10^{-6} = 2.49 \times 10^{-x} \]
This implies:
\[ x = 6 \]

### Final Answer:
The numerical value of \( x \) is \( 6 \).