A solution was prepared by dissolving 0.0005 mol of `Ba(OH)_(2)` in 100 mL of the solution. If the base is assumed to ionise completely, the pOH of the solution will be

Options

A. 1
B. 5
C. 2
D. 14

Answer

To solve the problem, we need to determine the pOH of a solution prepared by dissolving 0.0005 moles of Ba(OH)₂ in 100 mL of water. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the dissociation reaction
When barium hydroxide, Ba(OH)₂, dissolves in water, it dissociates completely into barium ions (Ba²⁺) and hydroxide ions (OH⁻):
\[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2 \text{OH}^- \]

### Step 2: Calculate the molarity of the Ba(OH)₂ solution
Molarity (M) is defined as the number of moles of solute per liter of solution. We have:
- Moles of Ba(OH)₂ = 0.0005 mol
- Volume of solution = 100 mL = 0.1 L

Using the formula for molarity:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
\[ \text{Molarity} = \frac{0.0005 \, \text{mol}}{0.1 \, \text{L}} = 0.005 \, \text{M} \]

### Step 3: Determine the concentration of hydroxide ions
Since Ba(OH)₂ dissociates to produce 2 moles of OH⁻ for every mole of Ba(OH)₂, the concentration of OH⁻ ions will be:
\[ \text{[OH}^-] = 2 \times \text{[Ba(OH)}_2] \]
\[ \text{[OH}^-] = 2 \times 0.005 \, \text{M} = 0.01 \, \text{M} \]

### Step 4: Calculate the pOH of the solution
The pOH of a solution can be calculated using the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Substituting the concentration of OH⁻:
\[ \text{pOH} = -\log(0.01) \]
Since \(0.01 = 10^{-2}\):
\[ \text{pOH} = -(-2) = 2 \]

### Final Answer
The pOH of the solution is **2**.