Question

If $S{O_2}C{l_2}$ (sulphuryl oxide) reacts with water to give ${H_2}S{O_4}$ and $HCl$ , what volume $0.2M$ $Ba{(OH)_2}$ of is needed to completely neutralize $25mL$ of $0.2M$ $S{O_2}C{l_2}$ of solution.
A. $25mL$
B. $50mL$
C. $100mL$
D. $200mL$

Answer 1

Neutralization is a process where acid and base react with each other to form salt and water.
Acid $+$ base $\to$ salt $+$ water
To solve this problem we can use given formula,
Molarity $\times$ Volume $=$ number of moles

Complete step by step answer:
The reaction is,
$S{O_2}C{l_2} + 2{H_2}O \to {H_2}S{O_4} + 2HCl$
Millimoles of ${H_2}S{O_4}$ produced= $5$
Millimoles of $HCl$ produced= $10$
The neutralisation reaction will be
$S{O_2}C{l_2} + Ba{(OH)_2} \to BaS{O_4} + BaC{l_2} + 2{H_2}O$
Millimoles of $Ba{\left( {OH} \right)_2}$ required $=$ $5(for{H_2}S{O_4}) + \dfrac{{10}}{2}(forHCl) = 10$
$M \times V = 10 \\\ 0.2 \times V = 10 \\\ V = 50 \\\$
So, the correct answer is Option B.

Note: To neutralise any solution when number of moles of ${H^ + }$ becomes equal to number of moles of $O{H^ - }$ , That is moles ${H^ + }$ $=$ moles of $O{H^ - }$
To calculate we can use formulas and by putting the given values we can get the expected entity.