Calculate the number of equivalents in 10 litre of `0.5 Mba(OH)_(2)` solution (Ba = 137)

Options

A. 0.1
B. 10
C. 100
D. 1

Answer

To calculate the number of equivalents in a 10-litre solution of 0.5 M barium hydroxide (Ba(OH)₂), we can follow these steps:

### Step 1: Determine the Molarity and Volume
Given:
- Molarity (M) = 0.5 M
- Volume (V) = 10 L

### Step 2: Calculate the Normality
The normality (N) is related to molarity (M) by the formula:
\[ N = M \times n_f \]
where \( n_f \) is the number of equivalents per mole of solute.

For barium hydroxide, Ba(OH)₂:
- Barium (Ba) has a valency of +2.
- Each hydroxide ion (OH⁻) has a valency of -1, and there are 2 hydroxide ions in barium hydroxide.

Thus, the total number of equivalents for Ba(OH)₂ is:
\[ n_f = 2 \text{ (from Ba)} + 2 \text{ (from 2 OH)} = 2 + 2 = 4 \]

Now we can calculate the normality:
\[ N = 0.5 \, \text{M} \times 2 = 1 \, \text{N} \]

### Step 3: Calculate the Number of Equivalents
The number of equivalents (Eq) can be calculated using the formula:
\[ \text{Number of equivalents} = N \times V \]
Substituting the values we have:
\[ \text{Number of equivalents} = 1 \, \text{N} \times 10 \, \text{L} = 10 \, \text{Eq} \]

### Final Answer
The number of equivalents in 10 litres of 0.5 M Ba(OH)₂ solution is **10 equivalents**.

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