`K_(sp)` of `M(OH)_(x)`, is `27 xx 10^(-12)` and its solubility in water is `10^(-3)` mol `litre^(-1)` . Find the value of X

Answer

To find the value of \( x \) in the solubility product expression for \( M(OH)_x \), we can follow these steps:

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

### Step 2: Define solubility
Let the solubility of \( M(OH)_x \) in water be \( s \) mol/L. Given that \( s = 10^{-3} \) mol/L, we can express the concentrations of the ions at equilibrium:
- Concentration of \( M^{x+} \) = \( s = 10^{-3} \) mol/L
- Concentration of \( OH^- \) = \( x \cdot s = x \cdot 10^{-3} \) mol/L

### Step 3: Write the expression for \( K_{sp} \)
The solubility product \( K_{sp} \) can be expressed as:
\[ K_{sp} = [M^{x+}][OH^-]^x \]
Substituting the concentrations from Step 2:
\[ K_{sp} = (s)(x \cdot s)^x = s \cdot (x \cdot s)^x \]
This simplifies to:
\[ K_{sp} = s \cdot x^x \cdot s^x = x^x \cdot s^{x+1} \]

### Step 4: Substitute known values
We know \( K_{sp} = 27 \times 10^{-12} \) and \( s = 10^{-3} \):
\[ 27 \times 10^{-12} = x^x \cdot (10^{-3})^{x+1} \]
This can be rewritten as:
\[ 27 \times 10^{-12} = x^x \cdot 10^{-3(x+1)} \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ x^x = 27 \times 10^{-12} \cdot 10^{3(x+1)} \]
\[ x^x = 27 \times 10^{-12 + 3x + 3} \]
\[ x^x = 27 \times 10^{3x - 9} \]

### Step 6: Simplifying further
Since \( 27 = 3^3 \), we can write:
\[ x^x = 3^3 \cdot 10^{3x - 9} \]

### Step 7: Testing for integer values of \( x \)
We can test integer values for \( x \) to find a solution. Let's try \( x = 3 \):
\[ 3^3 = 27 \quad \text{and} \quad 10^{3(3) - 9} = 10^{0} = 1 \]
Thus:
\[ 3^3 \cdot 1 = 27 \]
This satisfies the equation.

### Conclusion
The value of \( x \) is \( 3 \).

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