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]

• A. \(K_a Λ^o_m\)
• B. \(K_a \frac{Λ^o_m}{2}\)
• C. \(2 K_a Λ^o_m\)
• D. \(\frac{1}{(K_a Λ^o_m)}\)

Answer 1

The Correct Option is A

Given that plotting \(\frac{1}{Λ_{m}}\) against \(cΛ_{m}\) results in a straight line, we need to find the ratio \(\frac{P}{S}\) using the information provided:

• \(Λ_{m}\) = molar conductivity.
• \(Λ^{o}_{m}\) = limiting molar conductivity. 
• \(c\) = molar concentration.
• \(K_{a}\) = dissociation constant of HX.

For a monobasic weak acid HX, the relation between molar conductivity and concentration is given by:
\(Λ_{m} = Λ^{o}_{m} - K_{a}cΛ_{m}^{2}\)

Rearranging the equation:
\(\frac{1}{Λ_{m}} = \frac{1}{Λ^{o}_{m}} + K_{a}c\)
This equation is in the form \(y = mx + c\), where:

• \(y = \frac{1}{Λ_{m}}\)
• \(x = cΛ_{m}\)
• \(m = K_{a}\)
• \(c = \frac{1}{Λ^{o}_{m}}\)

So, the line has a y-intercept \(P = \frac{1}{Λ^{o}_{m}}\) and a slope \(S = K_{a}\).

Thus, the ratio \(\frac{P}{S} = \frac{\frac{1}{Λ^{o}_{m}}}{K_{a}} = \frac{1}{K_{a}Λ^{o}_{m}}\).

However, we seek \(K_{a}Λ^{o}_{m}\), which simplifies our ratio to \(\frac{1}{\frac{1}{K_{a}Λ^{o}_{m}}} = K_{a}Λ^{o}_{m}\).

Therefore, the correct answer is \(K_{a}Λ^{o}_{m}\).

Answer 2

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=Ka​1​+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.