25 mL of a solution of barium hydroxide on titration with 0.1 molar solution of hydrochloric acid gave a titre value of 35 mL. The molarity of barium hydroxide solution was

Options

A. `0.07`
B. `0.14`
C. `0.28`
D. `0.35`

Explanation

To find the molarity of the barium hydroxide (Ba(OH)₂) solution, we will follow these steps:

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

### Step 2: Identify the known values
- Volume of barium hydroxide solution (V₁) = 25 mL = 0.025 L
- Volume of hydrochloric acid solution (V₂) = 35 mL = 0.035 L
- Molarity of hydrochloric acid (M₂) = 0.1 M
- Number of moles of barium hydroxide (N₁) = 1 (from the balanced equation)
- Number of moles of hydrochloric acid (N₂) = 2 (from the balanced equation)

### Step 3: Use the molarity equation
The relationship between the molarity, volume, and number of moles is given by:
\[ M_1 V_1 N_1 = M_2 V_2 N_2 \]
Where:
- \( M_1 \) = Molarity of barium hydroxide
- \( V_1 \) = Volume of barium hydroxide solution
- \( N_1 \) = Number of moles of barium hydroxide
- \( M_2 \) = Molarity of hydrochloric acid
- \( V_2 \) = Volume of hydrochloric acid solution
- \( N_2 \) = Number of moles of hydrochloric acid

### Step 4: Rearrange the equation to solve for \( M_1 \)
Rearranging the equation gives:
\[ M_1 = \frac{M_2 V_2 N_2}{V_1 N_1} \]

### Step 5: Substitute the known values
Substituting the known values into the equation:
\[ M_1 = \frac{(0.1 \, \text{mol/L}) \times (0.035 \, \text{L}) \times 2}{0.025 \, \text{L} \times 1} \]

### Step 6: Calculate \( M_1 \)
Calculating the above expression:
\[ M_1 = \frac{0.1 \times 0.035 \times 2}{0.025} \]
\[ M_1 = \frac{0.007}{0.025} \]
\[ M_1 = 0.28 \, \text{mol/L} \]

### Step 7: Final calculation
Now, simplifying further:
\[ M_1 = 0.28 \, \text{mol/L} \]

### Conclusion
The molarity of the barium hydroxide solution is \( 0.28 \, \text{mol/L} \).