Question

One mole $Ba{F}_2$ is treated with two moles of $H_{2}S{O}_4$. To make the resulting mixture neutral, $NaOH$ is added. $NaOH$ required in this process is :

• A. $4$ mol $NaOH$
• B. $2$ mol $NaOH$
• C. $3$ mol $NaOH$
• D. $1$ mol $NaOH$

Answer 1

The correct option is A $4$ mol $NaOH$

The reactions are as follows :

$\underset{1mol}{Ba{F}_2}+\underset{1mol}{H_{2}S{O}_4}⟶\underset{1mol}{BaS{O}_4}+\underset{1mol}{H_2{F}_2}$

Given: 1 mol 2 mol

Thus, unreacted $H_{2}S{O}_4=1mol$

$H_2{F}_{2}formed=1mol$

$\underset{1mol}{H_{2}S{O}_4}+\underset{2mol}{2NaOH}⟶N{a}_{2}S{O}_4+2H_{2}O$

$\underset{1mol}{H_2{F}_2}+\underset{2mol}{2NaOH}⟶N{a}_2{F}_2+2H_{2}O$

Thus, total $NaOH$ required by $1$ mol $H_2{F}_2$ (formed) and $1$ mol $H_{2}S{O}_4$ (unreacted) $=4$ mol.

Hence the correct option is A.