The conductivity of a weak acid HA of concentration
Question:

The conductivity of a weak acid HA of concentration $0.001 \mathrm{~mol} \mathrm{~L}^{-1}$ is $2.0 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}$. If $\Lambda_{\mathrm{m}}^{\circ}(\mathrm{HA})=190 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$, the ionization constant $\left(\mathrm{K}_{\mathrm{a}}\right)$ of HA is equal to_______$\times 10^{-6}$

(Round off to the Nearest Integer)

Solution:

$\Lambda_{\mathrm{m}}=1000 \times \frac{\kappa}{\mathrm{M}}$

$=1000 \times \frac{2 \times 10^{-5}}{0.001}=20 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$

$\Rightarrow \alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\infty}}=\frac{20}{190}=\left(\frac{2}{19}\right)$

$\mathrm{HA} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-}$

$0.001(1-\alpha) 0.001 \alpha 0.001 \alpha$

$\Rightarrow \mathrm{k}_{\mathrm{a}}=0.001\left(\frac{\alpha^{2}}{1-\alpha}\right)=\frac{0.001 \times\left(\frac{2}{19}\right)^{2}}{1-\left(\frac{2}{19}\right)}$

$=12.3 \times 10^{-6}$

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