The degree of ionization of $0.10 \mathrm{M}$ lactic acid is $4.0 \%$ The value of $K_c$ is

The degree of ionization of $0.10 \mathrm{M}$ lactic acid is $4.0 \%$
The value of $K_c$ is
  1. $1.66 \times 10^{-5}$
  2. $1.66 \times 10^{-4}$
  3. $1.66 \times 10^{-3}$
  4. $1.66 \times 10^{-2}$

Solution

$\%$ dissociation $=4 \%$ degree of dissociation $(\alpha)=\frac{4}{100}=0.04$ For lactic acid $ \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{COO}^{-}+\mathrm{H}^{+} $ Initial concentration $\begin{array}{cll}C \text { mol L } & 0 & 0 \\ \text { At. equilibrium } C(1-\alpha) & C \alpha & C \alpha\end{array}$ $\begin{aligned} \therefore \quad K_c & =\frac{C \alpha \cdot C \alpha}{C(1-\alpha)}=\frac{C \alpha^2}{(1-\alpha)} \\ & =\frac{0.1 \times 0.04 \times 0.04}{(1-0.04)} \\ & =\frac{1.6 \times 10^{-4}}{0.96}=1.66 \times 10^{-4}\end{aligned}$

Asked in: AP EAMCET 2013

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