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

$\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} $

1 Answer

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
$K=\large\frac{c(\Lambda_c/\Lambda^{\infty})^2}{1-(\Lambda_c/\Lambda_{\infty})}$
$\Rightarrow \large\frac{c\Lambda_c^2}{\Lambda^{\infty}(\Lambda^{\infty}-\Lambda_c)}$
answered Jul 18, 2014 by sreemathi.v
 

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