Chemistry - Electrochemistry Question with Solution | TestHub

Molar conductivity ${\Lambda }_{\mathrm{M}}$ can be calculated using the formula:

${\Lambda }_{\mathrm{M}}=\frac{\mathrm{K}\times 1000}{\mathrm{C}}$.

The conductivity of 0.001 M HA is given as $\mathrm{K}=5\times 10^{-5}\mathrm{S}{\mathrm{cm}}^{-1}$

and $\mathrm{C}=0.001\mathrm{M}$,

thus, ${\Lambda }_{\mathrm{M}}=\frac{5\times 10^{-5}\times 1000}{0.001}=50\mathrm{S}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$.

Now, the total standard molar conductivity of HA will be:

${\Lambda }_{\mathrm{HA}}^{\circ }={\Lambda }_{\mathrm{HCl}}^{\circ }+{\Lambda }_{\mathrm{NaA}}^{\circ }-{\Lambda }_{\mathrm{NaCl}}^{\circ }$.

After substituting values, we get:

${\Lambda }_{\mathrm{HA}}^{\circ }=425.9+100.5-126.4=400\mathrm{S}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$.

Hence, the degree of dissociation $(\alpha )$

of HA will be:

The degree of dissociation of HA is 0.125.

Option A is correct.