The equilibrium constant of acetic acid in an aqueous solution of concentration $c$ is given by

Question

$\begin{array}{1 1}K=\large\frac{c\Lambda_c^2}{\Lambda^{\infty}-\Lambda_c}\\K=\large\frac{c\Lambda_c^2}{\Lambda^{\infty}(\Lambda^{\infty}-\Lambda_c)}\\K=\large\frac{c\Lambda_c^2}{\Lambda^{\infty}+\Lambda_c}\\K=\large\frac{c\Lambda_c^2}{\Lambda^{\infty}(\Lambda^{\infty}+\Lambda_c)}\end{array} $

Answer 1

Answer : $K=\large\frac{c\Lambda_c^2}{\Lambda^{\infty}(\Lambda^{\infty}-\Lambda_c)}$

We have

$CH_3COOH \quad \leftrightharpoons \quad CH_3COO^-+H^+$

$K=\large\frac{[CH_3COO][H^+]}{[CH_3COOH]}=\frac{c\alpha^2}{(1-\alpha)}$

Since $\alpha=\Lambda_c/\Lambda^{\infty}$ we get

$K=\large\frac{c(\Lambda_c/\Lambda^{\infty})^2}{1-(\Lambda_c/\Lambda_{\infty})}$

$\Rightarrow \large\frac{c\Lambda_c^2}{\Lambda^{\infty}(\Lambda^{\infty}-\Lambda_c)}$