Question

Molar conductance of a $0.5\, M - NH _{4} OH$ solution is $2.5 \times 10^{-3}\, S\, m ^{2}\, mol ^{-1}$. Then the $pH$ of the solution is
[Given: Molar conductances of $Ba ( OH )_{2},\, BaCl _{2}$ and $NH _{4} Cl$ at infinite dilutions are $525 \times 10^{-4},\, 280 \times 10^{-4}$ and
$130 \times 10^{-4}\, S\, m ^{2}\, mol ^{-1}$, respectively $]$

Answer 1

From Kohlrausch's Law,

$\Lambda_{ m \left( NH _{4} OH \right)}^{\circ}=\Lambda_{ m \left( NH _{4} Cl \right)}^{-}-\frac{1}{2} \Lambda_{ m \left( BaCl _{2}\right)}^{\circ}+\frac{1}{2} \Lambda_{ m Ba ( OH )_{2}}^{\circ}$

$=130 \times 10^{-4}-\frac{1}{2} \times 280 \times 10^{-4}+\frac{1}{2} \times 525 \times 10^{-4}$

$=2.585 \times 10^{-2}\, S\, m ^{2}\, mol ^{-1}$

Now, $\alpha=\frac{\Lambda_{ m }}{\Lambda_{ m }^{\circ}} =\frac{2.5 \times 10^{-3}}{2.525 \times 10^{-2}}$

$=0.099$

Now, $\left[ OH ^{-}\right]=\alpha.\, c =0.099 \times 0.5=0.0495\, M$

$pOH =-\log [ OH ]=-\log (0.0495)=1.305$

and $pH$ of solution $=14-1.305=12.695$