Question

When $0.1$ mole of ${CH}_3{NH}_2$ (ionisation constant, $K_b=5\times {10}^{-4}$) is mixed with $0.08$ mole $HCl$ and the volume is made up to $1$ litre, find the $\left[H^{+}\right]$ of resulting solution:

• A. $8\times {10}^{-2}$
• B. $2\times {10}^{-11}$
• C. $1.23\times {10}^{-4}$
• D. $8\times {10}^{-11}$

Answer 1

The correct option is D $8\times {10}^{-11}$

${CH}_3{NH}_2+HCl⟶{CH}_3{NH}_3^{+}{Cl}^{-}$

$0.1mole$ $0.08mole$

$0.02mole$ $0$ $0.08mole$

$C{H}_{3}N{H}_2$ and $C{H}_{3}N{H}_3^{+}C{l}^{-}$ will form basic buffer.
So$pOH=p{K}_b+\log \frac{\left[Salt\right]}{\left[Base\right]}=-\log 5\times {10}^{-4}+\log \frac{0.08}{0.02}=3.903$

As we know that $pH+pOH=14$
$pH=10.0967so\left[H^{+}\right]=antilog(-10.0967)=8\times {10}^{-11}$