The volume of 0.02 M aqueous HBr required to neutralize 10.0 mL of 0.01 M aqueous $\mathrm{Ba}(\mathrm{OH})_{2}$ is (Assume complete neutralization)

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Equilibrium

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)

Answer: (C)

For neutralization, use the equivalence relation:

N_{1} V_{1} = N_{2} V_{2}

, where normality

N = M \times n

-factor. For HBr (monoprotic acid), n-factor = 1, so

N_{1} = 0.02

N. For

Ba (OH)_{2}

(diprotic base), n-factor = 2, so

N_{2} = 0.01 \times 2 = 0.02

N. Applying the equation:

0.02 \times V_{1} = 0.02 \times 10

, giving

V_{1} = 10

mL.

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Single correct option