`M(OH)_x` has a `K_(sp)` or `4xx10^(-9)` and its is solubility is `10^(-3)` M. The value of x is

Options

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

Answer

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

### Step-by-Step Solution:

1. **Understand the Dissociation of the Compound**:
The compound \( M(OH)_x \) dissociates in solution as follows:
\[ M(OH)_x \rightleftharpoons M^{+} + x OH^{-} \]

2. **Identify the Solubility**:
The solubility of the compound is given as \( 10^{-3} \, \text{M} \). This means that at equilibrium:
- The concentration of \( M^{+} \) ions is \( 10^{-3} \, \text{M} \).
- The concentration of \( OH^{-} \) ions is \( x \times 10^{-3} \, \text{M} \) (since there are \( x \) hydroxide ions produced for each formula unit of \( M(OH)_x \)).

3. **Write the Expression for \( K_{sp} \)**:
The solubility product \( K_{sp} \) is defined as:
\[ K_{sp} = [M^{+}][OH^{-}]^x \]
Substituting the concentrations:
\[ K_{sp} = (10^{-3})(x \times 10^{-3})^x \]

4. **Substitute the Given \( K_{sp} \)**:
We know that \( K_{sp} = 4 \times 10^{-9} \). Therefore, we can write:
\[ 4 \times 10^{-9} = (10^{-3})(x \times 10^{-3})^x \]

5. **Simplify the Equation**:
This can be rewritten as:
\[ 4 \times 10^{-9} = 10^{-3} \cdot (x^{x} \cdot 10^{-3x}) \]
Simplifying further gives:
\[ 4 \times 10^{-9} = x^{x} \cdot 10^{-3 - 3x} \]

6. **Rearranging the Equation**:
We can rearrange this to find:
\[ 4 \times 10^{-9} \cdot 10^{3 + 3x} = x^{x} \]
This simplifies to:
\[ 4 \cdot 10^{3 + 3x - 9} = x^{x} \]
\[ 4 \cdot 10^{3x - 6} = x^{x} \]

7. **Finding the Value of \( x \)**:
To solve for \( x \), we can compare the powers of 10. Let’s express \( 4 \) as \( 2^2 \):
\[ 2^2 \cdot 10^{3x - 6} = x^{x} \]
We can assume \( x = 2 \) and check:
\[ 2^2 \cdot 10^{3(2) - 6} = 2^{2} \]
This simplifies to:
\[ 4 \cdot 10^{0} = 4 \]
Thus, \( x = 2 \) satisfies the equation.

### Conclusion:
The value of \( x \) is \( 2 \).