Question

$0.1mol$ of $C{H}_{3}N{H}_2$ $(K_b=5\times {10}^{-4})$ is mixed with $0.08mol$ of $HCl$ and diluted to $1L$. Which statement(s) is/are correct?
(take $log2=0.3)$

• A. The concentration of $H^{+}$ ions is $8\times {10}^{-11}M$
• B. $HCl$ is left over in the solution because the weak base does not completely neutralise it
• C. This forms a basic buffer
• D. The pOH of solution is 5.9

Answer 1

Answer: (A), (C)

The correct options are
A The concentration of $H^{+}$ ions is $8\times {10}^{-11}M$
C This forms a basic buffer

$C{H}_{3}N{H}_2+HCl\to C{H}_{3}N{H}_3^{+}C{l}^{-}$
Initial 0.1 0.08
moles
Final (0.1-0.08) 0 0.08
All of the $HCl$ gets neutralised since there is a larger quantity of the base present. The low $K_b$ value of the base does not affect this.
Since weak base $C{H}_{3}N{H}_2$ is left along with its salt of strong base, it forms a basic buffer solution so using Henderson Hasselbalch equation:
$pOH=p{K}_b+log\frac{\left[salt\right]}{\left[base\right]}$
or
$\left[O{H}^{-}\right]=K_b\frac{\left[Base\right]}{\left[Salt\right]}=5\times {10}^{-4}\times \frac{0.02}{0.08}=1.25\times {10}^{-4}M$
and
$\left[H^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{{10}^{-14}}{1.25\times {10}^{-4}}=8\times {10}^{-11}M$
so, $pH=-log(8\times {10}^{-11})$
$=-log\left(2^3\right)+11=-3\times 0.3+11$
$pH=10.1$
$pOH=14-10.1=3.9$