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

The balanced chemical equation for the reaction between $\mathrm{HBr}$ and $\mathrm{Ba}(\mathrm{OH}{)}_2$ is:

$2\mathrm{HBr}+\mathrm{Ba}(\mathrm{OH}{)}_2 \to {\mathrm{BaBr}}_2+2{\mathrm{H}}_2\mathrm{O}$

From this equation, we can see that $2$ moles of $\mathrm{HBr}$ react with $1$ mole of $\mathrm{Ba}(\mathrm{OH}{)}_2$.

Equal equivalents will react. 

Number of equivalents = Normality $\times$Volume

So,

$$\begin{gathered} {\mathrm{N}}_1{\mathrm{V}}_1={\mathrm{N}}_2{\mathrm{V}}_2 \\ 0.02\times {\mathrm{V}}_1=0.02\times 10 \\ {\mathrm{V}}_1=10 \mathrm{ml} \end{gathered}$$