pH of `Ca(OH)_(2)` solution is 12. Millimoles of `Ca(OH)_(2)` present in 100mL of solution will be

Options

A. 1
B. 0.5
C. 0.05
D. 5

Explanation

To solve the problem of finding the millimoles of \( \text{Ca(OH)}_2 \) in a 100 mL solution with a pH of 12, we can follow these steps:

### Step 1: Determine the pOH
Since the pH of the solution is given as 12, we can calculate the pOH using the formula:
\[ \text{pOH} = 14 - \text{pH} \]
Substituting the given pH:
\[ \text{pOH} = 14 - 12 = 2 \]

### Step 2: Calculate the concentration of hydroxide ions \([OH^-]\)
The relationship between pOH and the concentration of hydroxide ions is given by:
\[ \text{pOH} = -\log[OH^-] \]
Substituting the pOH value:
\[ 2 = -\log[OH^-] \]
To find \([OH^-]\), we can rewrite the equation:
\[ [OH^-] = 10^{-\text{pOH}} = 10^{-2} \, \text{M} \]

### Step 3: Calculate the concentration of \( \text{Ca(OH)}_2 \)
Since \( \text{Ca(OH)}_2 \) dissociates into one \( \text{Ca}^{2+} \) ion and two \( \text{OH}^- \) ions, the concentration of \( \text{Ca(OH)}_2 \) can be found using the relationship:
\[ [\text{Ca(OH)}_2] = \frac{[OH^-]}{2} \]
Substituting the concentration of hydroxide ions:
\[ [\text{Ca(OH)}_2] = \frac{10^{-2}}{2} = 5 \times 10^{-3} \, \text{M} \]

### Step 4: Calculate the number of millimoles in 100 mL
To find the number of millimoles in 100 mL, we use the formula:
\[ \text{millimoles} = \text{concentration (M)} \times \text{volume (L)} \times 1000 \]
Converting 100 mL to liters:
\[ \text{volume} = 100 \, \text{mL} = 0.1 \, \text{L} \]
Now substituting the values:
\[ \text{millimoles} = 5 \times 10^{-3} \, \text{M} \times 0.1 \, \text{L} \times 1000 = 0.5 \, \text{mmol} \]

### Final Answer
The number of millimoles of \( \text{Ca(OH)}_2 \) present in 100 mL of solution is **0.5 mmol**.

---