At 298 K, an aqueous solution of a weak acid HA of concentration 0.010 mol L−1 has a conductivity of 7.0 × 10−4 S cm−1. If the limiting molar conductivity of HA at 298 K is 350 S cm2 mol−1, what is the value of its dissociation constant, Ka? (Do not neglect α in comparison with 1.)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 5.0 × 10−4 mol L−1
For a weak electrolyte, the degree of dissociation is the ratio of the molar conductivity at the given concentration to the limiting molar conductivity, α = Λm/Λ°m. The dissociation constant then follows from Ka = cα2/(1 − α).
Molar conductivity (κ in S cm−1, c in mol L−1): \( \displaystyle \Lambda_m = \frac{\kappa \times 1000}{c} = \frac{7.0 \times 10^{-4}\ \mathrm{S\,cm^{-1}} \times 1000\ \mathrm{cm^3\,L^{-1}}}{0.010\ \mathrm{mol\,L^{-1}}} = 70\ \mathrm{S\,cm^2\,mol^{-1}} \)
Degree of dissociation: \( \displaystyle \alpha = \frac{\Lambda_m}{\Lambda^{\circ}_m} = \frac{70}{350} = 0.20 \)
Dissociation constant: \( \displaystyle K_a = \frac{c\alpha^2}{1-\alpha} = \frac{0.010 \times (0.20)^2}{1-0.20} = \frac{4.0 \times 10^{-4}}{0.80} = 5.0 \times 10^{-4}\ \mathrm{mol\,L^{-1}} \)
NCERT (2026-27 edition):
2.4.2 Variation of Conductivity and Molar Conductivity with Concentration – “…At any concentration c, if α is the degree of dissociation then it can be approximated to the ratio of molar conductivity Λm at the concentration c to limiting molar conductivity, Λ°m. Thus we have:”
\( \displaystyle \alpha = \frac{\Lambda_m}{\Lambda^{\circ}_m} \qquad (2.26) \)
“But we know that for a weak electrolyte like acetic acid …”
\( \displaystyle K_a = \frac{c\alpha^2}{(1-\alpha)} = \frac{c\Lambda_m^2}{\Lambda^{\circ 2}_m\left(1-\frac{\Lambda_m}{\Lambda^{\circ}_m}\right)} = \frac{c\Lambda_m^2}{\Lambda^{\circ}_m\left(\Lambda^{\circ}_m-\Lambda_m\right)} \qquad (2.27) \)