Question

\[S{O_2}C{l_2}\](sulphuryl chloride) reacts with water to give a mixture of
\[{H_2}S{O_4}\] and \[HCl\]. What volume of \[0.2\,M\,Ba{\left( {OH} \right)_2}\] is needed to
completely neutralize \[25\,mL\] of \[0.2\,M\,S{O_2}C{l_2}\] solution.
(A) \[25\,mL\]
(B) \[50\,mL\]
(C) \[100\,mL\]
(D) \[200\,mL\]

Answer 1

Hint: As we know that \[{H_2}S{O_4}\] and \[HCl\] both are strong acids, whereas \[\,Ba{\left( {OH} \right)_2}\] is a strong base so this is a strong acid strong base titration. If we know the number of moles of \[{H_2}S{O_4}\] and \[HCl\] then we will easily find out the number of moles of \[\,Ba{\left( {OH} \right)_2}\] and hence we can easily calculate the volume of \[0.2\,M\,Ba{\left( {OH} \right)_2}\].

Complete step by step answer:
The reaction of sulfuryl chloride reacts with water occurs as
\[S{O_2}C{l_2} + 2{H_2}O \to {H_2}S{O_4} + 2HCl\]
Now we will calculate the number of moles of \[25\,mL\] of \[0.2\,M\,S{O_2}C{l_2}\] by the
formula
\[concentration(C) = \dfrac{{number\,of\,moles(n)}}{{volume\,(V)}} - - - - (i)\]
Rearranging the above formula, we get as
\[n = V\,\times\,C\]
Putting the values \[C = 0.2\,M,V = 25mL\]

\[
\Rightarrow n = 25\,mL\,\times\,0.2\,\,M\\
\Rightarrow n = 5\,mmol = 5\,\times{10^{ - 3}}mole
\]
$\Rightarrow\,S{O_2}C{l_2}\:\:+\:\:2{H_2}O\:\to\:{H_2}S{O_4}\:\:+\:\:2HCl$

$5\times{10^{ -3}}moles$	0	0
0	$5\times\,{10^{ -3}}moles$	$10\times{10^{-3}}moles$

Therefore, the number of moles of \[\,Ba{\left( {OH} \right)_2}\]to neutralize \[{H_2}S{O_4}\]\[\,
= 5\times\,{10^{ - 3}}\]
And the number of moles of \[\,Ba{\left( {OH} \right)_2}\] to neutralize \[HCl\]\[\, = 5\times\,{10^{ - 3}}\]
So, the total number of moles of \[\,Ba{\left( {OH} \right)_2}\] to neutralize both acids
\[\,=\,10\times\,{10^{ - 3}}\]
From the equation\[(i)\]we can easily find the volume of \[\,Ba{\left( {OH} \right)_2}\] as
\[
\Rightarrow V = \dfrac{n}{C}\\
\Rightarrow V = \dfrac{{10\,\times\,{{10}^{ - 3}}mole}}{{0.2\,M}}\\
\Rightarrow V = 50\,\times\,{10^{ - 3}}L\\
\therefore V = 50\,mL
\]

Note: The volume of \[\,Ba{\left( {OH} \right)_2}\] is \[50\,mL\]. In acid base titration, the
number of moles of hydroxyl ions must be equal to the number of moles of hydrogen ions for
the complete neutralization.