Question

The conductivity at infinite dilution of ${NH}_{4}Cl,NaOH$ and $NaCl$ are $130,218,$ $120{ohm}^{-1}{cm}^2{eq}^{-1}$. If equivalent conductance of $\frac{N}{100}$ solution of ${NH}_{4}OH$ IS $10$, then degree of dissociation of ${NH}_{4}OH$ at this dilution is:

• A. $0.005$
• B. $0.043$
• C. $0.015$
• D. $0.025$

Answer 1

Answer: (B)

$0.043$

$NaCl\rightleftharpoons {Na}^{+}+{Cl}^{-}$ ${\Lambda }_1^{\infty }eq=120oh{m}^{-1}{cm}^2{eq}^{-1}⟶\left(1\right)$

${NH}_{4}Cl\rightleftharpoons {NH}_4^{+}+{Cl}^{-}$ ${\Lambda }_2^{\infty }eq=130oh{m}^{-1}{cm}^2{eq}^{-1}⟶\left(2\right)$

$NaOH\rightleftharpoons {Na}^{+}+\stackrel{-}{O}H$ ${\Lambda }_3^{\infty }eq=218oh{m}^{-1}{cm}^2{eq}^{-1}⟶\left(3\right)$

as ${NH}_{4}OH\rightleftharpoons {NH}_4^{+}+\stackrel{-}{O}H$

by adding eqn (2) and (3) and substracting (1) from them we get.

${NH}_{4}Cl+NaOH-NaCl\rightleftharpoons {NH}_4^{+}+\stackrel{-}{O}H$

${\Lambda }^{8}eq={\Lambda }_2^{\infty }eq+{\Lambda }_3^{\infty }eq-{\Lambda }_1^{8}eq$

${\Lambda }^{\infty }eq=(218+130-120)oh{m}^{-1}{cm}^2{eq}^{-1}$

${\Lambda }^{\infty }eq=228oh{m}^{-1}{cm}^2{eq}^{-1}$

Given $\Lambda eq$ of $NaOH=10oh{m}^{-1}{cm}^2{eq}^{-1}$

$\alpha =\frac{\Lambda eq}{{\Lambda }^{\infty }eq}=\frac{1.0}{228}$

$\alpha =0.043=$ degree of dissociation.