The `K_(sp)` of `Ca(OH)_(2)` is `4.42 xx 10^(-5)` at `25^(@)C`. A 500 ml of saturated solution of `Ca(OH)_(2)` is mixed with an equal volume of `0.4M NaOH`. How much `Ca(OH)_(2)` in mg is precipitated ?

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the solubility of Ca(OH)₂ from Ksp
The solubility product (Ksp) expression for calcium hydroxide, Ca(OH)₂, is given by:
\[ K_{sp} = [Ca^{2+}][OH^-]^2 \]
Let the solubility of Ca(OH)₂ be \( S \) mol/L. Therefore, the concentration of \( Ca^{2+} \) ions will be \( S \) and the concentration of \( OH^- \) ions will be \( 2S \).

Substituting into the Ksp expression:
\[ K_{sp} = S(2S)^2 = 4S^3 \]
Given \( K_{sp} = 4.42 \times 10^{-5} \):
\[ 4S^3 = 4.42 \times 10^{-5} \]
\[ S^3 = \frac{4.42 \times 10^{-5}}{4} = 1.105 \times 10^{-5} \]
Taking the cube root:
\[ S = (1.105 \times 10^{-5})^{1/3} \approx 0.0223 \, \text{mol/L} \]

### Step 2: Calculate the moles of OH⁻ from NaOH
The concentration of NaOH is 0.4 M and the volume is 500 mL (0.5 L):
\[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 0.4 \, \text{mol/L} \times 0.5 \, \text{L} = 0.2 \, \text{mol} \]
Since NaOH dissociates completely, the moles of \( OH^- \) ions will also be 0.2 mol.

### Step 3: Calculate the new concentration of OH⁻ ions
In the saturated solution of Ca(OH)₂, the concentration of \( OH^- \) ions is \( 2S = 2 \times 0.0223 \, \text{mol/L} = 0.0446 \, \text{mol/L} \).

After mixing, the total volume becomes 1 L (500 mL + 500 mL). The total concentration of \( OH^- \) ions after mixing is:
\[ \text{Total OH}^- = 0.2 \, \text{mol} + 0.0446 \, \text{mol} = 0.2446 \, \text{mol} \]
Thus, the new concentration of \( OH^- \) in the total volume (1 L):
\[ [OH^-] = \frac{0.2446 \, \text{mol}}{1 \, \text{L}} = 0.2446 \, \text{mol/L} \]

### Step 4: Set up the new Ksp expression
Using the new concentration of \( OH^- \):
\[ K_{sp} = [Ca^{2+}][OH^-]^2 \]
Let \( x \) be the new solubility of Ca(OH)₂:
\[ K_{sp} = x(0.2446)^2 \]
Setting this equal to \( 4.42 \times 10^{-5} \):
\[ 4.42 \times 10^{-5} = x(0.2446)^2 \]
Calculating \( (0.2446)^2 \):
\[ (0.2446)^2 \approx 0.0598 \]
Now substituting back:
\[ 4.42 \times 10^{-5} = x \cdot 0.0598 \]
Solving for \( x \):
\[ x = \frac{4.42 \times 10^{-5}}{0.0598} \approx 7.39 \times 10^{-4} \, \text{mol/L} \]

### Step 5: Calculate the moles of Ca(OH)₂ that remain dissolved
The total volume is 1 L, so the moles of Ca(OH)₂ that remain dissolved:
\[ \text{Moles of Ca(OH)₂} = 7.39 \times 10^{-4} \, \text{mol} \]

### Step 6: Calculate the moles of Ca(OH)₂ that precipitated
Initially, the moles of Ca(OH)₂ in the saturated solution (before mixing) were:
\[ \text{Moles of Ca(OH)₂} = 0.0223 \, \text{mol/L} \times 0.5 \, \text{L} = 0.01115 \, \text{mol} \]
The moles that precipitated:
\[ \text{Moles precipitated} = 0.01115 \, \text{mol} - 7.39 \times 10^{-4} \, \text{mol} \approx 0.01041 \, \text{mol} \]

### Step 7: Convert moles of Ca(OH)₂ to mass in mg
The molar mass of Ca(OH)₂ is approximately 74 g/mol:
\[ \text{Mass} = \text{Moles} \times \text{Molar mass} = 0.01041 \, \text{mol} \times 74 \, \text{g/mol} \approx 0.771 \, \text{g} = 771 \, \text{mg} \]

### Final Answer
The amount of Ca(OH)₂ that precipitated is approximately **771 mg**.

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