A 20.00 ml sample of `Ba (OH)_2` solution is titrated with 0.245 M HCI. If 27.15 ml of HCI is required, then the molarity of the `Ba (OH)_2` solution will be :

Options

A. 0.166 M
B. 0.180 M
C. 0.333 M
D. 0.666 M

Answer

To find the molarity of the `Ba(OH)₂` solution, we can follow these steps:

### Step 1: Write the balanced chemical equation for the reaction.
The reaction between barium hydroxide and hydrochloric acid can be written as:
\[ \text{Ba(OH)}_2 + 2 \text{HCl} \rightarrow \text{BaCl}_2 + 2 \text{H}_2\text{O} \]

### Step 2: Determine the n-factor for both reactants.
- The n-factor for `Ba(OH)₂` is 2 because it can release 2 hydroxide ions (OH⁻).
- The n-factor for HCl is 1 because it can release 1 hydrogen ion (H⁺).

### Step 3: Calculate the equivalents of HCl used in the titration.
The number of equivalents of HCl can be calculated using the formula:
\[ \text{Equivalents} = \text{Molarity} \times \text{Volume (L)} \]
Given:
- Molarity of HCl = 0.245 M
- Volume of HCl = 27.15 mL = 0.02715 L

Calculating the equivalents of HCl:
\[ \text{Equivalents of HCl} = 0.245 \, \text{mol/L} \times 0.02715 \, \text{L} = 0.00666 \, \text{equivalents} \]

### Step 4: Use the stoichiometry of the reaction to find the equivalents of `Ba(OH)₂`.
From the balanced equation, 1 mole of `Ba(OH)₂` reacts with 2 moles of HCl. Therefore, the equivalents of `Ba(OH)₂` will be half of the equivalents of HCl:
\[ \text{Equivalents of Ba(OH)}_2 = \frac{0.00666}{2} = 0.00333 \, \text{equivalents} \]

### Step 5: Calculate the molarity of `Ba(OH)₂`.
Using the formula for equivalents:
\[ \text{Equivalents} = \text{Molarity} \times \text{Volume (L)} \]
We can rearrange this to find the molarity of `Ba(OH)₂`:
\[ \text{Molarity of Ba(OH)}_2 = \frac{\text{Equivalents}}{\text{Volume (L)}} \]
Given:
- Volume of `Ba(OH)₂` = 20.00 mL = 0.02000 L

Calculating the molarity:
\[ \text{Molarity of Ba(OH)}_2 = \frac{0.00333}{0.02000} = 0.1665 \, \text{M} \]

### Final Answer:
The molarity of the `Ba(OH)₂` solution is approximately **0.1665 M**.

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