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

Hint: 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.