. The conductivity of $0.01 \mathrm{M}$ aqueous acetic acid measured with a conductivity cell of cell…

. The conductivity of $0.01 \mathrm{M}$ aqueous acetic acid measured with a conductivity cell of cell constant of $0.5 \mathrm{~cm}^{-1}$ at $298 \mathrm{~K}$ is $3.12 \times 10^{-4} \mathrm{~S}$. If the limiting conductivities of $\mathrm{H}^{+}$and $\mathrm{CH}_3 \mathrm{COO}^{-}$ at the same temperature are 349 , and $41 \mathrm{~S} \mathrm{~cm}^2$ $\mathrm{mol}^{-1}$ respectively, the dissociation constant of acetic acid is
  1. $1.67 \times 10^{-4}$
  2. $1.67 \times 10^{-5}$
  3. $1.67 \times 10^{-3}$
  4. $1.67 \times 10^{-6}$

Solution

$ \begin{gathered} \text {Conductivity }=3.12 \times 10^{-4} \times 0.5 \mathrm{~S} \mathrm{~cm}^{-1} \\ =1.56 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1} \end{gathered} $ Molar conductivity $=\frac{1.56 \times 10^{-4}}{0.01}=\mathrm{S} \mathrm{cm}^3 \mathrm{~mol}^{-1}$ $=1.56 \times 10^{-2} \mathrm{Scm}^3 \mathrm{~mol}^{-1}=1.56 \mathrm{~S} \mathrm{~cm}^3 \mathrm{~mol}^{-1}$ Degree of dissociation $d=\frac{1.56}{390}=0.004=4 \times 10^{-3}$ Dissociation constant $=\frac{\left(4 \times 10^{-3}\right)^2}{1-\left(4 \times 10^{-3}\right)}$ $=\frac{16 \times 10^{-6}}{1-0.004}=\frac{16 \times 10^{-6}}{0.996}=16.0 \times 10^{-6}=1.60 \times 10^{-5}$

Asked in: AP EAMCET 2018 (23 Apr Shift 2)

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