AP EAMCET · Chemistry · Electrochemistry
. 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
- A \(1.67 \times 10^{-4}\)
- B \(1.67 \times 10^{-5}\)
- C \(1.67 \times 10^{-3}\)
- D \(1.67 \times 10^{-6}\)
Answer & Solution
Correct Answer
(B) \(1.67 \times 10^{-5}\)
Step-by-step Solution
Detailed explanation
\[ \begin{gathered} \text {Conductivity }=3.12 \times 10^{-4} \times 0.5 \mathrm{~S} \mathrm{~cm}^{-1} \\ =1.56 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1} \end{gathered} \] Molar conductivity \(=\frac{1.56 \times 10^{-4}}{0.01}=\mathrm{S} \mathrm{cm}^3 \mathrm{~mol}^{-1}\)…
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