The solubility of magnesium hydroxide `[Mg(OH)_2]` at `300^@C` is `8.35xx10^(-3) g L^(-1)` . Find out its `K_(sp)` . [Mol.wt. of `Mg(OH)_2` =58]

Explanation

To find the solubility product constant \( K_{sp} \) of magnesium hydroxide \( Mg(OH)_2 \), we can follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of magnesium hydroxide in water can be represented as:
\[ Mg(OH)_2 (s) \rightleftharpoons Mg^{2+} (aq) + 2OH^{-} (aq) \]

### Step 2: Determine the Solubility
Given that the solubility of \( Mg(OH)_2 \) at \( 300^\circ C \) is \( 8.35 \times 10^{-3} \, g \, L^{-1} \), we need to convert this solubility into moles per liter (molarity).

### Step 3: Convert Grams to Moles
Using the molar mass of \( Mg(OH)_2 \), which is given as 58 g/mol, we can calculate the molarity:
\[ \text{Molarity} = \frac{\text{Solubility (g/L)}}{\text{Molar Mass (g/mol)}} \]
\[ \text{Molarity} = \frac{8.35 \times 10^{-3} \, g \, L^{-1}}{58 \, g \, mol^{-1}} = 1.44 \times 10^{-4} \, mol \, L^{-1} \]

### Step 4: Calculate Ion Concentrations
From the dissociation equation, we can see that:
- The concentration of \( Mg^{2+} \) ions will be equal to the molarity of \( Mg(OH)_2 \):
\[ [Mg^{2+}] = 1.44 \times 10^{-4} \, mol \, L^{-1} \]
- The concentration of \( OH^{-} \) ions will be twice that of \( Mg(OH)_2 \) because 2 moles of \( OH^{-} \) are produced for every mole of \( Mg(OH)_2 \):
\[ [OH^{-}] = 2 \times [Mg(OH)_2] = 2 \times 1.44 \times 10^{-4} = 2.88 \times 10^{-4} \, mol \, L^{-1} \]

### Step 5: Write the Expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) is given by the expression:
\[ K_{sp} = [Mg^{2+}][OH^{-}]^2 \]

### Step 6: Substitute the Values
Now, substituting the concentrations into the \( K_{sp} \) expression:
\[ K_{sp} = (1.44 \times 10^{-4}) \times (2.88 \times 10^{-4})^2 \]
\[ K_{sp} = (1.44 \times 10^{-4}) \times (8.2944 \times 10^{-8}) \]
\[ K_{sp} = 1.194 \times 10^{-11} \]

### Final Answer
Thus, the solubility product constant \( K_{sp} \) for magnesium hydroxide at \( 300^\circ C \) is:
\[ K_{sp} = 1.194 \times 10^{-11} \]

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