Question

The values of ${\Lambda }_{eq}^0$ for $N{H}_{4}Cl,NaOH$ and $NaCl$ are $149.74,248.1$ and $126.4oh{m}^{-1}c{m}^{2}e{q}^{-1}$ respectively. The value of ${\Lambda }_{eq}^0\left(N{H}_{4}OH\right)$ is :

• A. 371.44
• B. 271.44
• C. 171.44
• D. Cannot be calculated from the data given

Answer 1

The correct option is B 271.44

Kohlrausch law:
At infinite dilution, when all the interionic effects disappear and dissociation of electrolyte is complete, each ion makes a definite contribution towards equivalent conductance of the electrolyte irrespective of the nature of the ion with which it is associated.
Kohlrausch law of independent migration of ions:
${\Lambda }_{eq}^0={\lambda }_{e{q}^{+}}^0+{\lambda }_{e{q}^{-}}^0$

${\lambda }_{e{q}^{+}}^0\to \text{Limiting equivalent conductivities of cation}$
${\lambda }_{e{q}^{-}}^0\to \text{Limiting equivalent conductivities of anion}$

${\Lambda }_{N{H}_{4}Cl}^0={\lambda }_{eq}^0\left(N{H}_4^{+}\right)+{\lambda }_{eq}^0\left(C{l}^{-}\right)$

${\Lambda }_{eq}^{0}NaOH={\lambda }_{eq}^0\left(N{a}^{+}\right)+{\lambda }_{eq}^0\left(O{H}^{-}\right)$

${\Lambda }_{eq}^{0}NaCl={\lambda }_{eq}^0\left(N{a}^{+}\right)+{\lambda }_{eq}^0\left(C{l}^{-}\right)$
Thus,
${\Lambda }_{eq}^0\left(N{H}_{4}OH\right)={\lambda }_{eq}^0\left(N{H}_4^{+}\right)+{\lambda }_{eq}^0\left(C{l}^{-}\right)+{\lambda }_{eq}^0\left(N{a}^{+}\right)+{\lambda }_{eq}^0\left(O{H}^{-}\right)-\left[{\lambda }_{eq}^0\right(N{a}^{+})+{\lambda }_{eq}^0(C{l}^{-}\left)\right]$
$\therefore$
${\Lambda }_{eq}^0\left(N{H}_{4}OH\right)={\lambda }_{eq}^0\left(N{H}_{4}Cl\right)+{\lambda }_{eq}^0\left(NaOH\right)-{\lambda }_{eq}^0\left(NaCl\right)$

${\Lambda }_{eq}^0\left(N{H}_{4}OH\right)=149.74+248.1-126.4$
${\Lambda }_{eq}^0\left(N{H}_{4}OH\right)=271.44oh{m}^{-1}c{m}^{2}e{q}^{-1}$