JEE Main 2023 — Some Basic Concepts Of Chemistry Question with Solution

The reaction between $\mathrm{HBr}$ and $\mathrm{Ba}(\mathrm{OH}{)}_2$ is a neutralization reaction where one $\mathrm{Ba}(\mathrm{OH}{)}_2$ reacts with two $\mathrm{HBr}$ to form $\mathrm{BaB}{\mathrm{r}}_2$ and $2{\mathrm{H}}_2\mathrm{O}$:

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

We can use the stoichiometry of the reaction to find the volume of $\mathrm{HBr}$ required.

The number of moles of $\mathrm{Ba}(\mathrm{OH}{)}_2$ is given by its molarity times the volume in liters:

$n(\mathrm{Ba}(\mathrm{OH}{)}_2)=0.01\mathrm{M}\times \frac{10.0\mathrm{mL}}{1000mL/L}=0.0001\mathrm{moles}$

From the balanced equation, we can see that the reaction requires $2$ moles of $\mathrm{HBr}$ for each mole of $\mathrm{Ba}(\mathrm{OH}{)}_2$:

$n(\mathrm{HBr})=2\times n(\mathrm{Ba}(\mathrm{OH}{)}_2)=2\times 0.0001\mathrm{moles}=0.0002\mathrm{moles}$

Then, we find the volume of the $\mathrm{HBr}$ solution by dividing the number of moles by the molarity:

$V(\mathrm{HBr})=\frac{n(\mathrm{HBr})}{0.02\mathrm{M}}=\frac{0.0002\mathrm{moles}}{0.02mol/L}=0.01\mathrm{L}=10\mathrm{mL}$

Therefore, the volume of the $0.02\mathrm{M}$ $\mathrm{HBr}$ solution required to neutralize the $\mathrm{Ba}(\mathrm{OH}{)}_2$ solution is $10.0\mathrm{mL}$.