Chemistry · Electrochemistry · NEET
α is the fraction of acid molecules that actually break into ions. If α = 0.05, it means only 5 out of every 100 acid molecules split into H+ and the acid ion; the other 95 stay whole. A weak acid always has a small α. A strong acid has α close to 1 because almost all of it breaks apart. This is why weak acids conduct electricity poorly.
Use α = Λm / Λ°m. Here Λm (molar conductivity) is the value measured at your given concentration, and Λ°m (limiting molar conductivity) is the value when the acid is fully broken apart (at infinite dilution). Divide the measured value by the full value and you get the fraction that dissociated. Example: Λm = 20, Λ°m = 400, so α = 20/400 = 0.05.
Only the ions that break away carry current. At infinite dilution (Λ°m) the acid is 100% broken, so it conducts the maximum. At normal concentration only a fraction α is broken, so it conducts only that fraction of the maximum. So Λm = α × Λ°m. Rearranging gives α = Λm / Λ°m. This shortcut works for weak electrolytes, not strong ones.
You cannot measure Λ°m for a weak acid directly, because it never fully dissociates even at high dilution. So you use Kohlrausch's law: add the limiting ion conductivities. For acetic acid, Λ°m = λ°(H+) + λ°(CH3COO−). NEET usually gives you these λ° values, or gives you Λ°m of NaCl, HCl and CH3COONa to combine. This is a very common trap step.
Use Ostwald's dilution law: Ka = cα² / (1 − α), where c is the concentration in mol/L. When α is small (weak acid), 1 − α is almost 1, so you can simplify to Ka ≈ cα². Always check: if α is small (like 0.05), use the shortcut; if α is larger, keep the (1 − α) term to be safe.
Use Ka ≈ cα² only when α is small (roughly α < 0.1), because then 1 − α ≈ 1 and the error is tiny. If α is not small (say 0.2 or more), use the full formula Ka = cα² / (1 − α) so you do not lose marks. In most NEET weak-acid questions α is small, so the shortcut is fine, but read the numbers first.
Put c in mol/L (mol dm⁻³). α has no unit because it is a fraction. So Ka comes out in mol/L (mol dm⁻³) too, for a monobasic acid. Do not convert Λm units into c; Λm is only used to find α. Keeping c in mol/L is what makes the answer match the NEET options.
The molar conductivity of 0.007 M acetic acid is 20 S cm² mol⁻¹. What is the dissociation constant of acetic acid? [λ°(H+) = 350 S cm² mol⁻¹, λ°(CH3COO−) = 50 S cm² mol⁻¹]
If the molar conductivity (Λm) of a 0.050 mol L⁻¹ solution of a monobasic weak acid is 90 S cm² mol⁻¹, its degree of dissociation will be [λ°+ = 349.6 S cm² mol⁻¹, λ°− = 50.4 S cm² mol⁻¹]
Try the real previous-year questions from this chapter — each with the answer and a full solution.
First α = Λm / Λ°m, then Ka = cα² / (1 − α). If α is small, Ka ≈ cα². Here c is in mol/L and Λ°m comes from Kohlrausch's law.
Λm is the molar conductivity measured at your given concentration. Λ°m (limiting molar conductivity) is the value when the acid is fully dissociated at infinite dilution. For weak acids Λ°m is found by adding limiting ion conductivities, not by measuring.
A weak acid holds its H+ tightly, so only a small fraction of molecules break into ions. That small fraction is α, and it makes weak acids conduct much less than strong acids at the same concentration.
No. α = Λm / Λ°m uses only the two conductivity values. You need concentration only for the next step, when you calculate Ka using Ka = cα².
Yes. Ka = cα² / (1 − α) is Ostwald's dilution law for a weak acid. Combining it with α = Λm / Λ°m lets you get Ka straight from conductivity data, which is exactly what NEET asks.
