Question:
Plotting \(\frac{1}{Λ_{m}}\) against \(cΛ_{m}\) for aqueous solutions of a monobasic weak acid (HX) resulted in a straight line with y-axis intercept of P and slope of S. The ratio \(\frac{P}{S}\) is
[\(Λ_{m}\) = molar conductivity,
\(Λ^{o}_{m}\)= limiting molar conductivity,
\(c\) = molar concentration,
\(K_{a}\) = dissociation constant of HX]
Plotting \(\frac{1}{Λ_{m}}\) against \(cΛ_{m}\) for aqueous solutions of a monobasic weak acid (HX) resulted in a straight line with y-axis intercept of P and slope of S. The ratio \(\frac{P}{S}\) is
[\(Λ_{m}\) = molar conductivity,
\(Λ^{o}_{m}\)= limiting molar conductivity,
\(c\) = molar concentration,
\(K_{a}\) = dissociation constant of HX]
[\(Λ_{m}\) = molar conductivity,
\(Λ^{o}_{m}\)= limiting molar conductivity,
\(c\) = molar concentration,
\(K_{a}\) = dissociation constant of HX]
Updated On: May 24, 2024
\(K_a Λ^o_m\)
\(K_a \frac{Λ^o_m}{2}\)
\(2 K_a Λ^o_m\)
\(\frac{1}{(K_a Λ^o_m)}\)
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The Correct Option is A
Solution and Explanation
Correct option is (A) Ka Λ°m.
When plotting 1/Λm against cΛm for aqueous solutions of a monobasic weak acid (HX), a linear relationship is observed, yielding a straight line with a y-axis intercept of P and a slope of S. The ratio P/S can be represented as:
0P/S=Ka1+Λm0
Where:
- Λm is the molar conductivity,
- Λm0 is the limiting molar conductivity,
- c is the molar concentration,
- Ka is the dissociation constant of HX.
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