For $0.01\ \text{M}$ acetic acid, the degree of dissociation $\alpha = 0.0414$ (obtained from $\Lambda_m/\Lambda_m^{0}$). Calculate the dissociation constant, $K_a$, of acetic acid at this concentration.

For 0.01M acetic acid, the degree of dissociation α = 0.0414 (obtained from Lambdam/Lambdam⁰). Calculate the d

Example · Example 23

0.01\ \text{M}

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For a weak monoprotic acid $\text{HA} \rightleftharpoons \text{H}^{+} + \text{A}^{-}$ dissociating to degree $\alpha$ from initial concentration $C$ , the equilibrium concentrations are $[\text{H}^{+}] = [\text{A}^{-}] = C\alpha$ and $[\text{HA}] = C(1-\alpha)$ , giving the dissociation constant $K_a = \dfrac{[\text{H}^{+}][\text{A}^{-}]}{[\text{HA}]} = \dfrac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \dfrac{C\alpha^{2}}{1-\alpha}$ (Ostwald's dilution law). Substituting $C=0.01\ \text{M}$ and $\alpha=0.0414$ : numerator $= 0.01\times(0.0414)^{2} = 0.01\times0.001714 = 1.714\times10^{-5}$ ; denominator $=1-0.0414=0.9586$ ; so $K_a = \dfrac{1.714\times10^{-5}}{0.9586} \approx 1.79\times10^{-5}$ . This value is very close to t …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set SEM43 marks
$Q. i) What is fuel cell? ii) At 298 K the molar conductances of KCl and LiCl at infinite dilution are Λ°M(KCl) = 149.9 and Λ°M(LiCl) = 115.0 ohm^-1 cm^2 mol^-1. Calculate the value of [Λ°M(KNO3) - Λ°M(LiNO3)].$
›Reveal solutionSolution
A fuel cell continuously turns a fuel + oxidant into electricity; and by Kohlrausch's law of independent ionic migration the common nitrate ion cancels, so Lambda(KNO3) - Lambda(LiNO3) equals Lambda(KCl) - Lambda(LiCl) = 34.9 ohm^-1 cm^2 mol^-1.

Fuel cell A fuel cell is a galvanic (voltaic) cell in which the chemical energy released by the combustion-type reaction of a fuel (such as hydrogen, methane or methanol) with an oxidant (oxygen) is converted directly and continuously into electrical energy, as long as the reactants are fed in from outside. In the H2-O2 fuel cell, at the anode H2 + 2OH- -> 2H2O + 2e- and at the cathode O2 + 2H2O + 4e- -> 4OH-, giving an overall reaction 2H2 + O2 -> 2H2O. It is efficient and pollution-free.

Using Kohlrausch's law of independent migration of ions At infinite dilution each ion contributes a fixed molar ionic conductance: Lambda(KNO3) = lambda(K+) + lambda(NO3-) Lambda(LiNO3) = lambda(Li+) + lambda(NO3-) Subtracting, the common lambda(NO3-) cancels: Lambda(KNO3) - Lambda(LiNO3) = lambda(K+) - lambda(Li+) The same difference lambda(K+) - lambda(Li+) is obtained from the chlorides (the common lambda(Cl-) cancels there too): lambda(K+) - lambda(Li+) = Lambda(KCl) - Lambda(LiCl) = 149.9 - 115.0 = 34.9 Therefore Lambda(KNO3) - Lambda(LiNO3) = 34.9 ohm^-1 cm^2 mol^-1. …
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CBSE 2023Set ANNUAL3 marks
$Q. (i) State the Kohlrausch's law of independent migration of ions.$
(ii) The specific conductance or conductivity of a 0.01 M acetic acid at 298 K is 1.65x10-4 S cm-1. Calculate molar conductivity of the solution and degree of dissociation of CH3COOH. Given that, lambda-degree(H+) = 349.1 and lambda-degree(CH3COO-) = 40.9 S cm2 mol-1. (1+2) OR
(i) Write two functions of salt bridge.
(ii) Write Nernst equation of the following galvanic cell and calculate emf of the cell at 298 K temperature: Cu(s) | Cu2+(0.130M) || Ag+(1.0x10-4M) | Ag(s). Given that E-degree(Cu2+/Cu) = +0.34V, E-degree(Ag+/Ag) = +0.80V. (1+2)
›Reveal solutionSolution
Kohlrausch's law lets ionic contributions to conductivity be added independently; applying it to the given data gives a degree of dissociation of about 4.2% for this dilute acetic acid solution.

Kohlrausch's law of independent migration of ions states that at infinite dilution, each ion migrates independently of its co-ion, and the limiting molar conductivity of the electrolyte is the sum of the limiting ionic conductivities of the cation and anion: $\Lambda_m^{\circ} = u_+\lambda_+^{\circ} + u_-\lambda_-^{\circ}$

Molar conductivity from specific conductance: $\Lambda_m = \kappa \times \frac{1000}{C} = 1.65\times10^{-4} \times \frac{1000}{0.01} = 16.5\ \text{S cm}^2\text{mol}^{-1}$ Limiting molar conductivity via Kohlrausch's law: …
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Q.For $0.01\ \text{M}$ acetic acid, the degree of dissociation $\alpha = 0.0414$ (obtained from $\Lambda_m/\Lambda_m^{0}$ ). Calculate the dissociation constant, $K_a$ , of acetic acid at this concentration.

Concept understanding — Kohlrausch's Law

Q.For 0.01 M0.01\ \text{M}0.01 M acetic acid, the degree of dissociation α=0.0414\alpha = 0.0414α=0.0414 (obtained from Λm/Λm0\Lambda_m/\Lambda_m^{0}Λm​/Λm0​). Calculate the dissociation constant, KaK_aKa​, of acetic acid at this concentration.

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Q.For $0.01\ \text{M}$ acetic acid, the degree of dissociation $\alpha = 0.0414$ (obtained from $\Lambda_m/\Lambda_m^{0}$ ). Calculate the dissociation constant, $K_a$ , of acetic acid at this concentration.

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