Question

The $0.1\,mole$ of $C{H_3}N{H_2}\,\left( {{K_b}\, = \,5 \times {{10}^{ - 4}}} \right)$ is mixed with $0.08\,moles$of $HCl$ and diluted to $1\,L$. The $\left[ {{H^ + }} \right]$ in solution is :
(i) $8 \times {10^{ - 2}}\,M$
(ii) $8 \times {10^{ - 11}}\,M$
(iii) $1.6 \times {10^{ - 11}}\,M$
(iv) $8 \times {10^{ - 5}}\,M$

Answer 1

Hint:$C{H_3}N{H_2}$ when reacts with $HCl$forms $C{H_3}N{H_3}^ + C{l^ - }$. Since the solution contains methylamine and its hydrochloride salt it is actually a basic buffer. A basic buffer is the one which maintains the$pH$of the solution on slight dilution or upon addition of a slight amount of acid or base. Manipulating Henderson’s equation you can calculate $\left[ {O{H^ - }} \right]$, and hence find $\left[ {{H^ + }} \right]$ .

Complete step-by-step answer:$C{H_3}N{H_2}$ reacts with $HCl$ as $C{H_3}N{H_2} + HCl \to C{H_3}N{H_3}^ + C{l^ - }$. Since $0.1\,mole$ of $C{H_3}N{H_2}$ is mixed with $0.08\,moles$of $HCl$, where $HCl$ is the limiting reagent and hence after reaction $C{H_3}N{H_2}$ will be present in excess. Hence the reaction mixture will contain methylamine and its hydrochloric salt resulting in a basic buffer as $C{H_3}N{H_2}$ is a weak base and $C{H_3}N{H_3}^ + C{l^ - }$ is its salt with a strong acid.
The $pOH$ of a basic buffer solution can be estimated using Henderson Hasselbach equation which is given as:
$pOH\, = \,p{K_b}\, + \,{\log _{10}}\left( {\dfrac{{concentration\,of\,salt}}{{concentration\,of\,a\,base}}} \right)$.
$ \Rightarrow \, - {\log _{10}}\left( {\left[ {O{H^ - }} \right]} \right)\, = \, - {\log _{10}}\left( {{K_b}} \right) - {\log _{10}}\,\dfrac{{Concentration\,of\,base}}{{Concentration\,of\,salt}}$
$ \Rightarrow \,{\log _{10}}\left( {\left[ {O{H^ - }} \right]} \right)\, = \,{\log _{10}}\left( {{K_b}} \right) + {\log _{10}}\,\dfrac{{Concentration\,of\,base}}{{Concentration\,of\,salt}}$
$ \Rightarrow \,{\log _{10}}\left( {\left[ {O{H^ - }} \right]} \right)\, = \,{\log _{10}}\left( {{K_b}\, \times \,\dfrac{{Concentration\,of\,base}}{{Concentration\,of\,salt}}} \right)$
Taking antilog we get,
$\left[ {O{H^ - }} \right]\, = \,{K_b}\, \times \,\dfrac{{Concentration\,of\,base}}{{Concentration\,of\,salt}}..........\left( 1 \right)$.
Therefore, concentration of base $ = \,\left[ {C{H_3}N{H_2}} \right]$ and concentration of salt $ = \,\left[ {C{H_3}N{H_3}^ + C{l^ - }} \right]$.
Now, $0.1\,mole$ of $C{H_3}N{H_2}$ is mixed with $0.08\,moles$of $HCl$, where $HCl$ is the limiting reagent. Hence $0.08\,moles$of $HCl$reacts with $0.08\,moles$ of $C{H_3}N{H_2}$ to form $0.08\,moles$ of $C{H_3}N{H_3}^ + C{l^ - }$. Hence after reaction $\left( {1.00 - 0.08} \right)\,moles\, = \,0.02\,moles$of $C{H_3}N{H_2}$ will be left in the reaction mixture.
Therefore, $\left[ {C{H_3}N{H_2}} \right]\, = \,\dfrac{{0.02\,moles}}{{1\,L}}\, = \,0.02\,M$.
Also, \[\left[ {C{H_3}N{H_3}^ + C{l^ - }} \right]\, = \,\dfrac{{0.08\,moles}}{{1\,L}}\, = \,0.08\,M\].
${K_b}\, = \,5 \times {10^{ - 4}}$.
Putting all the values in equation $\left( 1 \right)$ we get,
$\left[ {O{H^ - }} \right]\, = \,5\, \times {10^{ - 4}} \times \,\dfrac{{0.02}}{{0.08}}\,\, = \,5 \times {10^{ - 4}} \times 0.25\, = \,1.25 \times {10^{ - 4}}\,M$.
Now, \[\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]\, = \,{K_w}\, = \,{10^{ - 14}}\].
\[\, \Rightarrow \,\left[ {{H^ + }} \right]\, = \,\dfrac{{{{10}^{ - 14}}}}{{\left[ {O{H^ - }} \right]}}\, = \,\dfrac{{{{10}^{ - 14}}}}{{1.25 \times {{10}^{ - 4}}}}\,M\, = \,0.8 \times {10^{ - 10}}\,M\, = \,8 \times {10^{ - 11}}\,M\]

Hence the correct answer is (ii) $8 \times {10^{ - 11}}\,M$.

Note:For this question you must have a basic idea about buffers and its types. You can also calculate the $pOH$ of the solution using Henderson Hasselbach equation, then find $pH$of the solution using $pH + pOH\, = \,14$. Ultimately you can calculate $\left[ {{H^ + }} \right]$ using $pH\, = \, - {\log _{10}}\left( {\left[ {{H^ + }} \right]} \right)$.