COMEDK · Chemistry · 18. Electrochemistry
The conductivity of 0.01 M solution of \(\mathrm{CH}_3 \mathrm{COOH}\) at 298 K is \(1.65 \times 10^{-4} \mathrm{Scm}^{-1}\)
What is the \(\mathrm{pK}_{\mathrm{a}}\) value of the acid if \(\lambda^0\left(\mathrm{H}^{+}\right)\)and \(\lambda^0\left(\mathrm{CH}_3 \mathrm{COO}\right)^{-1}\) are \(349.1 \mathrm{Scm}^2 \mathrm{~mol}^{-1}\) and \(40.9 \mathrm{Scm}^2 \mathrm{~mol}^{-1}\) respectively?
- A \(1.87\)
- B \(3.47\)
- C \(2.95\)
- D \(4.75\)
Answer & Solution
Correct Answer
(D) \(4.75\)
Step-by-step Solution
Detailed explanation
The molar conductivity \(\Lambda_m\) is given by the formula \(\Lambda_m = \dfrac{\kappa \times 1000}{C}\), where \(\kappa = 1.65 \times 10^{-4} \text{ S cm}^{-1}\) and \(C = 0.01 \text{ M}\).…
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