What is the $pH$ of $0.001 \,M$ aniline solution? The ionization constant of aniline can be taken from Table . Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Base | $K _{ b }$ |
Dimethylamine, $\left( CH _{3}\right)_{2} NH$ | $5.4 \times 10^{-4}$ |
Triethylamine, $\left( C _{2} H _{5}\right)_{3} N$ | $6.45 \times 10^{-5}$ |
Ammonia, $NH _{3}$ or $NH _{4} OH$ | $1.77 \times 10^{-5}$ |
Quinine, ( $A$ plant product) | $1.10 \times 10^{-6}$ |
Pyridine, $C _{5} H _{5} N$ | $1.77 \times 10^{-9}$ |
Aniline, $C _{6} H _{5} NH _{2}$ | $4.27 \times 10^{-10}$ |
Urea, $CO \left( NH _{2}\right)_{2}$ | $1.3 \times 10^{-14}$ |
$K_{b}=4.27 \times 10^{-10}$
$c=0.001 \,M\, pH$ $=?$
$a=?$
$k_{b}=c \alpha^{2}$
$4.27 \times 10^{-10}=0.001 \times \alpha^{2}$
$4270 \times 10^{-10}=\alpha^{2}$
$65.34 \times 10^{-5}=\alpha=6.53 \times 10^{-4}$
Then, [anion] $=c \alpha=.001 \times 65.34 \times 10^{-5}$
$=.065 \times 10^{-5}$
$ pOH =-\log \left(.065 \times 10^{-5}\right) $
$=6.187$
$pH =7.813$
Now,
$K_{a} \times K_{b}=K_{w}$
$\therefore 4.27 \times 10^{-10} \times K_{a}=K_{w}$
$ K_{a} =\frac{10^{-14}}{4.27 \times 10^{-10}}$
$=2.34 \times 10^{-5} $
Thus, the ionization constant of the conjugate acid of aniline is $2.34 \times 10^{-5}$
Which of the following base is weakest
Given
$(i)$ $\begin{gathered}
HCN\left( {aq} \right) + {H_2}O\left( l \right) \rightleftharpoons {H_3}{O^ + }\left( {aq} \right) + C{N^ - }\left( {aq} \right) \hfill \\
{K_a} = 6.2 \times {10^{ - 10}} \hfill \\
\end{gathered} $
$(ii)$ $\begin{gathered}
C{N^ - }\left( {aq} \right) + {H_2}O\left( l \right) \rightleftharpoons HCN\left( {aq} \right) + O{H^ - }\left( {aq} \right) \hfill \\
{K_b} = 1.6 \times {10^{ - 5}} \hfill \\
\end{gathered} $
These equilibria show the following order of the relative base strength
$pH$ of $0.1\,\, M$ $N{H_3}$ aqueous solution is $({K_b} = 1.8 \times {10^{ - 5}})$
The dissociation constant of a substituted benzoic acid at $25^{\circ} \mathrm{C}$ is $1.0 \times 10^{-4}$. The $\mathrm{pH}$ of a $0.01 \ \mathrm{M}$ solution of its sodium salt is
A weak acid $HA$ has a $K_a$ of $1.00 \times 10^{-5} $. If $0.100\,mol$ of this acid is dissolved in one litre of water the percentage of acid dissociated at equilibrium is closest to.....$\%$