Question

The volume of 0.02 M aqueous HBr required to neutralize 10.0 mL of 0.01 M aqueous Ba(OH)₂ is ____ . (Assume complete neutralization)

• A. 2.5 mL
• B. 5.0 mL
• C. 7.5 mL
• D. 10.0 mL

Answer 1

The Correct Option is D

Neutralization Calculation Problem 

For neutralization, the milliequivalents of acid must equal the milliequivalents of base.

Milliequivalents = Molarity × n-factor × Volume

For HBr, n-factor = 1. For Ba(OH)2, n-factor = 2.

Let V be the volume of HBr required.

m.eq. of HBr = m.eq. of Ba(OH)2

\( 0.02 \times 1 \times V = 0.01 \times 2 \times 10 \)

\( 0.02V = 0.2 \)

\( V = \frac{0.2}{0.02} = 10 \text{ mL} \)

Conclusion:

The volume of HBr required is 10 mL.

Answer 2

\[ \text{m.e.q of HBr} = \text{m.e.q of Ba(OH)}_2 \] \[ M_1 \times n_1 \times V_1 = M_2 \times n_2 \times V_2 \] \[ 0.02 \times 1 \times V_1 = 0.01 \times 2 \times 10 \] \[ V_1 = \frac{0.01 \times 10}{0.02} = 10 \, \text{mL} \] % Correct Answer Correct Answer:} (2) 10.0 mL