`M(OH)_(X)` has `K_(SP) 4xx10^(-12)` and solubility `10^(-4) M`. The value of `x` is:

Options

A. `1`
B. `2`
C. `3`
D. `4`

Explanation

To solve the problem, we need to determine the value of \( x \) in the compound \( M(OH)_x \) given its solubility and the solubility product constant \( K_{sp} \).

### Step-by-Step Solution:

1. **Write the Dissociation Equation:**
The dissociation of \( M(OH)_x \) in water can be represented as:
\[ M(OH)_x \rightleftharpoons M^{+} + x OH^{-} \]

2. **Define Solubility:**
Let the solubility of \( M(OH)_x \) be \( S \). According to the problem, \( S = 10^{-4} \, M \).

3. **Determine Ion Concentrations:**
From the dissociation equation:
- The concentration of \( M^{+} \) ions is \( S \).
- The concentration of \( OH^{-} \) ions is \( xS \).

4. **Write the Expression for \( K_{sp} \):**
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [M^{+}][OH^{-}]^x \]
Substituting the concentrations:
\[ K_{sp} = S \cdot (xS)^x \]

5. **Substitute Known Values:**
We know \( K_{sp} = 4 \times 10^{-12} \) and \( S = 10^{-4} \):
\[ 4 \times 10^{-12} = (10^{-4}) \cdot (x \cdot 10^{-4})^x \]

6. **Simplify the Equation:**
This can be rewritten as:
\[ 4 \times 10^{-12} = 10^{-4} \cdot (x^{x} \cdot 10^{-4x}) \]
Which simplifies to:
\[ 4 \times 10^{-12} = x^{x} \cdot 10^{-4 - 4x} \]

7. **Rearranging the Equation:**
Rearranging gives:
\[ 4 = x^{x} \cdot 10^{4 + 4x - 12} \]
Simplifying further:
\[ 4 = x^{x} \cdot 10^{4x - 8} \]

8. **Expressing in Terms of Powers of 10:**
To make it easier to solve, we can express \( 4 \) as \( 2^2 \):
\[ 2^2 = x^{x} \cdot 10^{4x - 8} \]

9. **Finding \( x \):**
To find \( x \), we can try different integer values. If we assume \( x = 2 \):
\[ 2^2 = 2^2 \cdot 10^{4(2) - 8} \]
This simplifies to:
\[ 4 = 4 \cdot 10^{0} \]
Which holds true.

### Conclusion:
Thus, the value of \( x \) is:
\[ \boxed{2} \]