The volume of 0 . 02M aqueous HBr required to neutralize 10 . 0mL...

The balanced chemical equation for the reaction between $HBr$ and $Ba ( OH{)}_{2}$ is:

$2HBr + Ba ( OH{)}_{2} \to {BaBr}_{2} + 2{H}_{2}O$

From this equation, we can see that $2$ moles of $HBr$ react with $1$ mole of $Ba ( OH{)}_{2}$.

Equal equivalents will react.

Number of equivalents = Normality $\times$Volume

So,

${N}_{1}{V}_{1} = {N}_{2}{V}_{2}0 . 02 \times {V}_{1} = 0 . 02 \times 10{V}_{1} = 10ml$