Question

Molar conductance of a $0.5M-N{H}_{4}OH$ solution is $2.5\times 10^{-3}S{m}^{2}mo{l}^{-1}$. Then the $pH$ of the solution is
[Given: Molar conductances of $Ba(OH{)}_2,BaC{l}_2$ and $N{H}_{4}Cl$ at infinite dilutions are $525\times 10^{-4},280\times 10^{-4}$ and
$130\times 10^{-4}S{m}^{2}mo{l}^{-1}$, respectively $]$

Answer 1

Answer: 12.695

From Kohlrausch's Law,

${\Lambda }_{m\left(N{H}_{4}OH\right)}^{\circ }={\Lambda }_{m\left(N{H}_{4}Cl\right)}^{-}-\frac{1}{2}{\Lambda }_{m\left(BaC{l}_2\right)}^{\circ }+\frac{1}{2}{\Lambda }_{mBa(OH{)}_2}^{\circ }$

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

$=2.585\times 10^{-2}S{m}^{2}mo{l}^{-1}$

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

$=0.099$

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

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

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