Question

Hydrofluoric acid is a weak acid. At ${25}^{\circ }C$, the molar conductivity of $0.002MHF$ is $176.2oh{m}^{-1}c{m}^{2}mo{l}^{-1}$. If its ${\wedge }_m^{\infty }=405.2oh{m}^{-1}c{m}^{2}mo{l}^{-1}$, calculate its degree of dissociation and equilibrium constant at the given concentration.

Answer 1

$\alpha =\frac{{\Lambda }_c}{{\Lambda }_m^0}=176.2/405.2=0.435$

$K=\frac{\left[H^{+}\right]\left[F^{-}\right]}{\left[HF\right]}=\frac{C{\alpha }^2}{1-\alpha }=\frac{0.002\times {0.435}^2}{1-0.435}=6.70\times {10}^{-4}M$