6-2.Equilibrium-II (Ionic Equilibrium)Medium

$0.001$ $M$ ઍનિલિન દ્રાવણની $pH$ કેટલી હશે ? એનિલિનનો આયનીકરણ અચળાંક કોષ્ટક માંથી લઈ શકાય છે. દ્રાવણમાં ઍનિલિનનો આયનીકરણ અંશ ગણો. એનિલિનના સંયુગ્મ ઍસિડનો આયનીકરણ અચળાંક પણ ગણો.

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, $CO \left( NH _{2}\right)_{2}$ $1.3 \times 10^{-14}$

Solution

$K_{b}=4.27 \times 10^{-10}$

$c=0.001 \,M\, pH$ $=?$

$a=?$

$k_{b}=c \alpha^{2}$

$4.27 \times 10^{-10}=0.001 \times \alpha^{2}$

$4270 \times 10^{-10}=\alpha^{2}$

$65.34 \times 10^{-5}=\alpha=6.53 \times 10^{-4}$

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 $2.34 \times 10^{-5}$

$25\,^o C$ તાપમાને બેઇઝ $BOH $નો વિયોજન અચળાંક $1.0 \times {10^{ - 12}}$ છે. તો બેઇઝના $0.01\, M$ જલીય દ્રાવણમાં હાઇડ્રોક્સિલ આયનની સાંદ્રતા .......... થશે.

Medium

Difficult

Medium

Medium

Difficult

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, $CO \left( NH _{2}\right)_{2}$	$1.3 \times 10^{-14}$