The volume of \( 0.02 \, \mathrm{M} \) aqueous \( \mathrm{HBr} \) required to neutralize \( 10.0 \, \mathrm{mL} \) of \( 0.01 \, \mathrm{M} \) aqueous \( \mathrm{Ba(OH)_2} \) is: (Assume complete neutralization)
\( 2.5 \, \mathrm{mL} \)
\( 7.5 \, \mathrm{mL} \)
\( 5.0 \, \mathrm{mL} \)
\( 10.0 \, \mathrm{mL} \)