Question

What is the ionization constant of crotonic acid if the conductivity of a $0.001M$ crotonic acid solution is $3.83\times {10}^{-5}{\Omega }^{-1}c{m}^{-1}$?

Limiting molar conductivities at ${25}^{o}C$:
$HCl=426{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$
$NaCl=126{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$
$NaC\text{(sodium crotonate)}=83{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$.
(Crotonate ion is represented by $C^{-}$)

• A. ${10}^{-3}$
• B. $1.11\times {10}^{-5}$
• C. $1.11\times {10}^{-4}$
• D. $0.01$

Answer 1

The correct option is B

$1.11\times {10}^{-5}$

The molar conductivity of the dissociated form of crotonic acid is

${\Lambda }_m^0\left(HC\right)={\Lambda }_m^0\left(HCl\right)+{\Lambda }_m^0\left(NaC\right)-{\Lambda }_m^0\left(NaCl\right)$

${\Lambda }_m^0\left(HC\right)=(426+83-126){\Omega }^{-1}c{m}^{2}mo{l}^{-1}$

${\Lambda }_m^0\left(HC\right)=383{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$

$\text{Specific conductance,}\kappa =3.83\times {10}^{-5}{\Omega }^{-1}c{m}^{-1}$

The molar conductivity of $HC$ at $0.001M$ is,

${\Lambda }_m\left(HC\right)=\frac{\kappa }{C}\times 1000$

${\Lambda }_m\left(HC\right)=\frac{3.83\times {10}^{-5}{\Omega }^{-1}c{m}^{-1}}{0.001}\times 1000$

${\Lambda }_m\left(HC\right)=38.3{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$

The degree of dissociation,
$\alpha =\frac{{\Lambda }_m\left(HC\right)}{{\Lambda }_m^0\left(HC\right)}......eqn\left(1\right)$

Dissociation constant of weak electrolyte (Crotonic acid),

$K_a=\frac{C{\alpha }^2}{1-\alpha }........eqn\left(2\right)$

From equation (1) and (2),

$K_a=\frac{C\times (\frac{{\Lambda }_m}{{\Lambda }_m^0}{)}^2}{1-\frac{{\Lambda }_m}{{\Lambda }_m^0}}$

$K_a=\frac{C\times {\Lambda }_m^2}{{\Lambda }_m^0({\Lambda }_m^0-{\Lambda }_m)}$

$K_a=\frac{{10}^{-3}\times {38.3}^2}{383(383-38.3)}$

$K_a=1.1\times {10}^{-5}$