The number of moles of hydroxide `(OH^(-))` ion in 0.3 litre of 0.005 M solution of `Ba(OH)_(2)` is

Options

A. 0.005
B. 0.003
C. 0.0015
D. 0.0075

Answer

To find the number of moles of hydroxide ions (OH⁻) in a 0.3 liter solution of 0.005 M Ba(OH)₂, we can follow these steps:

### Step 1: Determine the number of moles of Ba(OH)₂
The formula for calculating the number of moles is:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in liters)} \]
Given:
- Molarity of Ba(OH)₂ = 0.005 M
- Volume of solution = 0.3 L

Calculating the number of moles of Ba(OH)₂:
\[ \text{Number of moles of Ba(OH)₂} = 0.005 \, \text{mol/L} \times 0.3 \, \text{L} = 0.0015 \, \text{mol} \]

### Step 2: Determine the number of moles of OH⁻ ions produced
When Ba(OH)₂ dissociates in solution, it produces one barium ion (Ba²⁺) and two hydroxide ions (OH⁻):
\[ \text{Ba(OH)₂} \rightarrow \text{Ba}^{2+} + 2 \text{OH}^- \]
This means that for every mole of Ba(OH)₂, two moles of OH⁻ are produced.

Calculating the number of moles of OH⁻:
\[ \text{Number of moles of OH}^- = 2 \times \text{Number of moles of Ba(OH)₂} = 2 \times 0.0015 \, \text{mol} = 0.003 \, \text{mol} \]

### Conclusion
The number of moles of hydroxide ions (OH⁻) in the solution is:
\[ \text{Number of moles of OH}^- = 0.003 \, \text{mol} \]

### Final Answer
The number of moles of hydroxide ion in 0.3 liters of 0.005 M solution of Ba(OH)₂ is **0.003 moles**.

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