QJEEM23Ch8738: JEE Main 2023 Chemistry - Class 11 JEEM23Ch8738 Solution: The volume of aqueous required to neutralize of aqueous is…

The reaction between $\mathrm{HBr}$ and $\mathrm{Ba(OH)_2}$ is a neutralization reaction where one $\mathrm{Ba(OH)_2}$ reacts with two $\mathrm{HBr}$ to form $\mathrm{BaBr_2}$ and $2\mathrm{H_2O}$ :

$\mathrm{Ba(OH)_2} + 2\mathrm{HBr} \rightarrow \mathrm{BaBr_2} + 2\mathrm{H_2O}$

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

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

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

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

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

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

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

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