`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 and \( K_{sp} \).

### Step-by-Step Solution:

1. **Understanding the Dissociation:**
The compound \( M(OH)_x \) dissociates in water as follows:
\[ M(OH)_x \rightleftharpoons M^x + xOH^- \]
Here, \( M \) is the metal cation and \( OH^- \) is the hydroxide ion.

2. **Setting Up the Solubility:**
Let the solubility of \( M(OH)_x \) be \( S = 10^{-3} \) M.
When it dissolves, it produces \( x \) moles of \( OH^- \) ions for every mole of \( M(OH)_x \) that dissolves. Therefore, the concentration of \( M^x \) will be \( S \) and the concentration of \( OH^- \) will be \( xS \).

3. **Expressing \( K_{sp} \):**
The solubility product constant \( K_{sp} \) is given by:
\[ K_{sp} = [M^x][OH^-]^x \]
Substituting the concentrations:
\[ K_{sp} = S \cdot (xS)^x \]
This simplifies to:
\[ K_{sp} = S \cdot x^x \cdot S^x = S^{x+1} \cdot x^x \]

4. **Substituting Known Values:**
Given \( K_{sp} = 4 \times 10^{-9} \) and \( S = 10^{-3} \):
\[ 4 \times 10^{-9} = (10^{-3})^{x+1} \cdot x^x \]

5. **Simplifying the Equation:**
\[ 4 \times 10^{-9} = 10^{-3(x+1)} \cdot x^x \]
To isolate \( x^x \), we can express \( 10^{-9} \) as \( 10^{-3(x+1)} \):
\[ 10^{-3(x+1)} = 4 \times 10^{-9} \]
This gives us:
\[ -3(x+1) = -9 \implies 3(x+1) = 9 \implies x + 1 = 3 \implies x = 2 \]

6. **Final Verification:**
We can verify by substituting \( x = 2 \) back into the equation:
\[ K_{sp} = (10^{-3})^{2+1} \cdot 2^2 = (10^{-3})^3 \cdot 4 = 10^{-9} \cdot 4 = 4 \times 10^{-9} \]
This confirms our solution is correct.

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