Question

The dissociation constant of monobasic acid is$1.6\times {10}^{-5}$ and the molar conductivity at infinite dilution is $380\times {10}^{-4}{\mathrm{Sm}}^2{\mathrm{mol}}^{-1}$. The specific conductance of the $0.01\mathrm{M}$ acid solution is

Answer 1

Step 1: Given data

Dissociation constant$=$$1.6\times {10}^{-5}$

Molar conductivity at infinite dilution$=$$380\times {10}^{-4}{\mathrm{Sm}}^2{\mathrm{mol}}^{-1}$

Concentration of acid$=$$0.01\mathrm{M}$

Step 2: Calculating specific conductance

degree of dissociation($\mathrm{\alpha }$)$=\sqrt{\frac{\mathrm{dissociation}\mathrm{constant}}{\mathrm{concentration}\mathrm{of}\mathrm{acid}}}$

$\mathrm{\alpha }=\sqrt{\frac{1.6\times {10}^{-5}}{0.01\mathrm{M}}}\Rightarrow 0.04$

Also, $\mathrm{\alpha }$$=\frac{\mathrm{molar}\mathrm{conductivity}}{\mathrm{molar}\mathrm{conductivity}\mathrm{at}\mathrm{infinite}\mathrm{dilution}}$

$0.04=\frac{{\mathrm{\lambda }}_{\mathrm{m}}}{380\times {10}^{-4}}$

${\mathrm{\lambda }}_{\mathrm{m}}=0.00152{\mathrm{Sm}}^2{\mathrm{mol}}^{-1}\mathrm{or}15.2{\mathrm{Scm}}^2{\mathrm{mol}}^{-1}$

Also, ${\mathrm{\lambda }}_{\mathrm{m}}=\frac{1000\mathrm{k}}{\mathrm{c}}$

$\Rightarrow 15.2=\frac{1000\mathrm{k}}{0.01}$

$\mathrm{k}=1.52\times {10}^{-4}{\mathrm{Scm}}^{-1}\mathrm{or}1.52\times {10}^{-2}{\mathrm{Sm}}^{-1}$

Therefore, the specific conductance of the $0.01\mathrm{M}$ acid solution is $1.52\times {10}^{-2}{\mathrm{Sm}}^{-1}$