The molar conductivity of $0.025 \mathrm{~mol} \mathrm{~L}^{-1}$ methanoic acid is $46.1 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$. Calculate its degree of dissociation and dissociation constant. Given $\lambda^{\circ}\left(\mathrm{H}^{+}\right)=349.6 \mathrm{~S} \mathrm{~cm}{ }^2$ $\mathrm{mol}^{-1}$ and $\lambda^0\left(\mathrm{HCOO}^{-}\right)=54.6 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$.

Question

The molar conductivity of $0.025 \mathrm{~mol} \mathrm{~L}^{-1}$ methanoic acid is $46.1 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$. Calculate its degree of dissociation and dissociation constant. Given $\lambda^{\circ}\left(\mathrm{H}^{+}\right)=349.6 \mathrm{~S} \mathrm{~cm}{ }^2$ $\mathrm{mol}^{-1}$ and $\lambda^0\left(\mathrm{HCOO}^{-}\right)=54.6 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$.,

MARKS App · Accepted Answer

$\begin{aligned} \Lambda_{\mathrm{m}}^{\mathrm{o}}(\mathrm{HCOOH}) &=\lambda^{\circ}\left(\mathrm{H}^{+}\right)+\lambda^{\circ}\left(\mathrm{HCOO}^{-}\right) \ &=349.6+54.6 \ &=404.2 \mathrm{~S} \mathrm{~cm}{ }^2 \mathrm{~mol}^{-1} \ \Lambda_{\mathrm{m}}^{\mathrm{C}} &=46.1 \mathrm{~S} \mathrm{~cm}{ }^2 \mathrm{~mol}^{-1} \ \therefore \quad \alpha &=\frac{\Lambda_{\mathrm{m}}^{\mathrm{C}}}{\Lambda_{\mathrm{m}}^{\mathrm{o}}}=\frac{46.1}{404.2}=0.114 \end{aligned}$ $\therefore \quad K_a=\frac{c \alpha \cdot c \alpha}{c(1-\alpha)}=\frac{c \alpha^2}{1-\alpha}$ $=\frac{0.025 \times(0.114)^2}{1-0.114}$ $=3.67 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$

Search any question & find its solution

Looking for more such questions to practice?