Question

The conductivities at infinite solution of $N{H}_{4}Cl$, $NaOH$ and $NaCl$ are $130,218,120oh{m}^{-1}c{m}^{2}e{q}^{-1}$. If equivalent conductance of ${\displaystyle \frac{N}{1000}}$ solution of $N{H}_{4}OH$ is $10oh{m}^{-1}c{m}^{2}e{q}^{-1}$, then degree of dissociation of $N{H}_{4}OH$ at this dilution is:
A.$0.005$
B.$0.043$
C.$0.01$
D.$0.02$

Answer 1

Hint: We can say equivalent conductivity as the conducting power of all the ions produced by dissolving one gram equivalent of an electrolyte present in solution. We can calculate the degree of dissociation using ${\wedge }_m$ and ${{\wedge }^{\circ }}_m\left(N{H}_{4}OH\right)$. We can calculate the ${{\wedge }^{\circ }}_m\left(N{H}_{4}OH\right)$ using the values of ${{\wedge }^{\circ }}_m\left(N{H}_{4}Cl\right)$, ${{\wedge }^{\circ }}_m\left(NaOH\right)$ and ${{\wedge }^{\circ }}_m\left(NaCl\right)$.

Complete step by step answer:
Given data contains,
The conductivity of $N{H}_{4}Cl$ is $130oh{m}^{-1}c{m}^{2}e{q}^{-1}$.
The conductivity of $NaOH$ is $218oh{m}^{-1}c{m}^{2}e{q}^{-1}$.
The conductivity of $NaCl$ is $210oh{m}^{-1}c{m}^{2}e{q}^{-1}$.
The reaction of ammonium chloride with sodium hydroxide results in the formation of ammonium hydroxide and sodium chloride. We can write the chemical equation as,
$N{H}_{4}OH+NaCl\to NaCl+N{H}_{4}OH$
At infinite dilution, the conductivity of $N{H}_{4}OH$ is,
${{\wedge }^{\circ }}_{N{H}_{4}OH}={{\wedge }^{\circ }}_{N{H_4}^{+}}+{{\wedge }^{\circ }}_{O{H}^{-}}$
Now,
${{\wedge }^{\circ }}_m\left(N{H}_{4}OH\right)={{\wedge }^{\circ }}_m\left(N{H}_{4}Cl\right)+{{\wedge }^{\circ }}_m\left(NaOH\right)-{{\wedge }^{\circ }}_m\left(NaCl\right)$
The equivalent conductance of $N{H}_{4}OH$ is calculated by adding the conductivities of $N{H}_{4}Cl$, $NaOH$ and subtracting the obtained conductivity with conductivity of $NaCl$.
Let us now substitute the values the conductivities of $N{H}_{4}Cl$, $NaOH$ and $NaCl$ to get ${{\wedge }^{\circ }}_m\left(N{H}_{4}OH\right)$.
${{\wedge }^{\circ }}_m\left(N{H}_{4}OH\right)={{\wedge }^{\circ }}_m\left(N{H}_{4}Cl\right)+{{\wedge }^{\circ }}_m\left(NaOH\right)-{{\wedge }^{\circ }}_m\left(NaCl\right)$
${{\wedge }^{\circ }}_m\left(N{H}_{4}OH\right)=130+218-120$
${{\wedge }^{\circ }}_m\left(N{H}_{4}OH\right)=228$
The value of ${{\wedge }^{\circ }}_m\left(N{H}_{4}OH\right)$ is $228$.
The equivalence conduction of solution of ${\displaystyle \frac{N}{100}}N{H}_{4}OH$ is $10oh{m}^{-1}c{m}^{2}e{q}^{-1}$.
Equivalent conductance $\left({\wedge }_m\right)$ is $10oh{m}^{-1}c{m}^{2}e{q}^{-1}$.
We can calculate the degree of dissociation with ${\wedge }_m$ and ${{\wedge }^{\circ }}_m$. The formula to calculate the degree of dissociation is given as,
$\alpha ={\displaystyle \frac{{\wedge }_m}{{{\wedge }^{\circ }}_m}}$
Here, $\alpha$ represents the degree of dissociation of $N{H}_{4}OH$.
Equivalent conductance $\left({\wedge }_m\right)$ is $10oh{m}^{-1}c{m}^{2}e{q}^{-1}$.
The value of ${{\wedge }^{\circ }}_m\left(N{H}_{4}OH\right)$ is $228$.
Using the values of ${\wedge }_m$and ${{\wedge }^{\circ }}_m$, we can calculate the degree of dissociation as,
$\alpha ={\displaystyle \frac{{\wedge }_m}{{{\wedge }^{\circ }}_m}}$
$\alpha ={\displaystyle \frac{10}{228}}$
$\alpha =0.043$
The degree of dissociation of $N{H}_{4}OH$ is $0.043$.
Therefore, the option (B) is correct.

Note: We can say that conductivity (or) specific conductivity is the capacity of solution to conduct electricity. We can represent the SI unit of conductivity as S/m. The unit that is used to express molar conductivity is $S{m}^{2}mo{l}^{-1}$. We can obtain the limiting molar conductivity of any electrolytes with the help of Kohlraush’s law.