Question

At ${18^ \circ }C$, the conductivities at infinite dilutions of $N{H_4}Cl,NaOH$and $NaCl$are$129.8,217.4$ and $108.9mho$ respectively. If the equivalent conductivity of$\dfrac{N}{{100}}$ solution of $N{H_4}OH$ is $9.93mho$, calculate the degree of dissociation of $N{H_4}OH$ at this dilution.

Answer 1

Kohlrausch law is used for the calculation of molar conductivities of weak electrolyte, for calculating degree of dissociation, for the calculation of dissociation constant of weak electrolyte. Using Kohlrausch law we will find the degree of dissociation of ammonium hydroxide.

Complete step by step answer:
The relationship between molar conductivity and degree of dissociation is given by the Kohlrausch law.
Kohlrausch law states that the migration of an ion at the infinite dilution depends on the nature of the solvent and on the potential gradient.
It is given by the equation as follows:
$\Lambda _{eq}^\infty = \Lambda _c^\infty + \Lambda _a^\infty$
Where, $\Lambda _{eq}^\infty =$ equivalent conductivity at infinite dilution
$\Lambda _c^\infty =$ conductivity of cation at infinite dilution
$\Lambda _a^\infty =$ conductivity of anion at infinite dilution.
So, now we will write the equations for $N{H_4}Cl,NaOH$and $NaCl$ as follows:
$\Lambda _{N{H_4}Cl}^\infty = \Lambda _{NH_4^ + }^\infty + \Lambda _{C{l^ - }}^\infty$ ….1
$\Lambda _{NaOH}^\infty = \Lambda _{N{a^ + }}^\infty + \Lambda _{O{H^ - }}^\infty$ ……2
$\Lambda _{NaCl}^\infty = \Lambda _{Na_{}^ + }^\infty + \Lambda _{C{l^ - }}^\infty$ …….3
The molar conductivities of $\Lambda _{N{H_4}Cl}^\infty = 129.8mho$,$\Lambda _{NaOH}^\infty = 217.4mho$,$\Lambda _{NaCl}^\infty = 108.9mho$
We will add equation 1 and equation 2 and we will subtract equation 3.
$= \Lambda _{NH_4^ + }^\infty + \Lambda _{C{l^ - }}^\infty + \Lambda _{N{a^ + }}^\infty + \Lambda _{O{H^ - }}^\infty - \Lambda _{Na_{}^ + }^\infty - \Lambda _{C{l^ - }}^\infty$
$= \Lambda _{NH_4^ + }^\infty + \Lambda _{O{H^ - }}^\infty$
Substituting the molar conductivities value we get,
$\Lambda _{NH_4^ + }^\infty + \Lambda _{O{H^ - }}^\infty = 129.8 + 217.4 - 108.9$
$\Lambda _{NH_4^ + }^\infty + \Lambda _{O{H^ - }}^\infty = 238.3mho$
This is the molar conductivity of $N{H_4}OH$ .
Now we will find the degree of dissociation of $N{H_4}OH$ using the formula: $\alpha = \dfrac{{{\Lambda _v}}}{{{\Lambda _\infty }}}$
Where, $\alpha =$ degree of dissociation, ${\Lambda _V} =$ equivalent conductivity, ${\Lambda _\infty } =$ molar conductivity at infinite dilution.
Given data: ${\Lambda _{N{H_4}OH}} = 238.3mho$,${\Lambda _V} = 9.93mho$
To find : $\alpha = ?$
Formula to be used: $\alpha = \dfrac{{{\Lambda _v}}}{{{\Lambda _\infty }}}$
So Solution:
$\alpha = \dfrac{{{\Lambda _v}}}{{{\Lambda _\infty }}}$
Substituting the value we get,
$\alpha = \dfrac{{9.93}}{{238.3}}$
$\alpha = 0.0416$.
The degree of dissociation of $N{H_4}OH$ is $0.0416$

Note: The conductivity of ions at the infinite dilution is constant and does not depend on the nature of co-ions. Mho is unit for conductance whereas ohm is the unit of resistance. If the concentration of an electrolyte is almost zero then at that point molar conductivity will be known as limiting molar conductivity.