`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 for the value of \( x \) in the compound \( M(OH)_x \) given the solubility product \( K_{sp} = 4 \times 10^{-12} \) and solubility \( S = 10^{-4} \, M \), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of the compound \( M(OH)_x \) in water can be represented as:
\[ M(OH)_x \rightleftharpoons M^{x+} + x OH^{-} \]

### Step 2: Define the solubility
Let the solubility of \( M(OH)_x \) be \( S = 10^{-4} \, M \). At equilibrium, the concentrations of the ions will be:
- Concentration of \( M^{x+} \) = \( S = 10^{-4} \, M \)
- Concentration of \( OH^{-} \) = \( xS = x \times 10^{-4} \, M \)

### Step 3: Write the expression for \( K_{sp} \)
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [M^{x+}][OH^{-}]^x \]
Substituting the concentrations from Step 2:
\[ K_{sp} = (10^{-4})(x \times 10^{-4})^x \]
This simplifies to:
\[ K_{sp} = (10^{-4})(x^x \times 10^{-4x}) = x^x \times 10^{-4(1+x)} \]

### Step 4: Set up the equation with the given \( K_{sp} \)
We know \( K_{sp} = 4 \times 10^{-12} \), so we can set up the equation:
\[ x^x \times 10^{-4(1+x)} = 4 \times 10^{-12} \]

### Step 5: Simplify the equation
From the equation, we can separate the powers of 10:
\[ x^x \times 10^{-4(1+x)} = 4 \times 10^{-12} \]
This implies:
\[ -4(1+x) = -12 \quad \text{(for the powers of 10)} \]
Solving for \( x \):
\[ 4(1+x) = 12 \\ 1+x = 3 \\ x = 2 \]

### Step 6: Verify the value of \( x \)
Now substituting \( x = 2 \) back into the equation for \( K_{sp} \):
\[ K_{sp} = (10^{-4})(2 \times 10^{-4})^2 = (10^{-4})(4 \times 10^{-8}) = 4 \times 10^{-12} \]
This confirms that our value of \( x \) is correct.

### Final Answer
The value of \( x \) is:
\[ \boxed{2} \]