This was a question which came in a objective type examination.

Which of the following have a $\lambda^\infty$ (molar conductivity at infinite dilution) larger than $\ce{KCl}$?

A. $\ce{CH3COOH}$
B. $\ce{HC}$
C. $\ce{NaOH}$
D. None of the above

I was not provided any experimental data to arrive at the answer.I know that $\ce{KCl}$ is a strong electrolyte and so my guess was that it must have greater molar conductivity than at least $\ce{CH_3COOH}$. But the answer key gives all A, B, and C. Is there a definite rule here?


4 Answers 4


I would agree with the answer key. The trick here is it's limiting molar conductivity, the molar conductivity at infinite dilution. For a neutral electrolyte compound, I was taught the notation $\Lambda ^0$, and $\lambda ^0$ was reserved for individual ions. (The $0$ superscript represents zero concentration, equivalent to $\infty$ dilution.)

Anyway, limiting molar conductivity is an interesting property because it does not represent a physically possibly scenario (a solute conducting electric charge when no solute is present). It can be extrapolated by the Kolhrausch Law, where it $\Lambda^0$ is the y-intercept of molar conductivity $\Lambda$ vs $\sqrt C$, concentration (note this relationship is only valid for strong electrolytes).

The degree of ionization of a weak electrolyte depends on the concentration, and tends to unity (100% dissociation) in the limit of concentration approaching zero. Limiting molar conductivity $\Lambda^0$ is defined (only) for this exact limiting scenario.

So when we consider limiting molar conductivity, we are doing so in a (abstract) condition where both strong and weak electrolyte fully dissociate. Thus for $\Lambda^0$ it makes no difference whether the electrolytes being compared are "strong" or "weak" at finite concentrations (for we are comparing them at "zero" concentration, i.e. infinite dilution). So we should have no problem accepting that a "weak" electrolyte might have a higher limiting molar conductivity value than a "strong" one, as in this case.

Why might $AcOH$ (organic chemist abbreviation), $HCl$, and $NaOH$ have higher $\Lambda^0$ values than $KCl$? Well, the $H^+$ ion is better at transporting its charge through an aqueous medium than $K^+$ because it can "water hop" via the Grotthuss mechanism. This makes acidic compounds very effective electrolyte conductors. $OH^-$ is also an extremely effective charge carrier by an analogous "deprotonation-chain" mechanism. Because they use water itself to "tunnel"* through the bulk solution, protons and hydroxide have extremely high ion mobility in aqueous solutions and thus provide abnormally high molar conductivity.

So in general acids and bases would be expected to out-conduct a pH-neutral salt ($KCl$), and in the limit of zero concentration/infinite dilution, even "weak" acids and bases will have higher limiting molar conductivity values.

*Not in the quantum sense.


The gist here is that in water $\ce{H+}$ can often be the charge carrier. The reason that $\ce{H+}$ can move so fast through water is that a particular $\ce{H+}$ ion doesn't have to move from one point to another in the solution. The $\ce{H+}$ can attach to one side of a water molecule and "kick off" a different $\ce{H+}$ ion on the other side of the water molecule. So the $\ce{H+}$ effectively moves through solution via a chain reaction of water molecules.

In infinite dilution solutions solution acetic acid and hydrochloric acid will be completely ionized to $\ce{H+}$ and the anion. The $\ce{H+}$ moves in the chain reaction as noted above as so it moves very fast compared to $\ce{K+}$ which must actually travel the distance as the specific ion.

In infinite dilution solutions solution NaOH will be completely ionized to $\ce{OH^-}$ and the $\ce{OH^-}$ groups moves via a chain reaction like $\ce{H+}$, but a bit slower. $\ce{OH-}$ will move faster than $\ce{Cl^-}$ however.

So the answer is A, B, and C.


A good measure is to see the molar conductivities of individual ions, you will observe that:

$\ce{H+}$ >> $\ce{OH-}$ > Rest

and in general, if you have a lot of charge you come out in front.

At infinite dilution, all ions are completely dissociated.
Hence, simply by the fact that they have $\ce{OH-}$, $\ce{H+}$, you can say with high level of probability that they have more conductivity than $\ce{KCl}$. Also, all are monovalent, another nice thing for us.

The law I used was that at infinite dilution degree of dissociation = 1 for all and sum of individual ions' conductivity is equal to the compound's molar conductivity at infinite dilution.


The key is WRONG. There is NO WAY acetic acid has a lambda° (don't know where you got ∞ from, but a rose by any other name will smell the same) that is greater than KCl's. At 0.001 m/L KCl's Λ is 146, HCl's is over 400, NaOH's is 245 and HOAc's is 49 (S cm²). For obvious (I hope!) reasons the $\ce{H+}$ and OH(-) ions' mobility in water is different from all other ions. The amount of energy "available" to an ion in solution depends mostly on the temperature of the solution, which means the kinetic energy of any two ions will be (on average) the same at the same temperature. Since K.E. = ½mv² we reach the approximation v = SqrRoot(2KE/m) for an ion of mass m. Obviously, as m increases, v decreases. So, you'd THINK that $\ce{K+}$ being larger than $\ce{Na+}$, that it would be slower, and this is correct - so far as it goes. It is more complicated, because in real solutions, ions are never naked. That is, they are always surrounded by solvation layers. This means that as charge of an ion increases, the ion will (in polar solvents) surround itself with more and more solvent molecules, and it will have to "carry" these along as it moves. This complicates the picture, since now it's not just the mass of the ion we have to consider, but also the mass of the solvation sheath - which depends on the electric field [which is greater for smaller spheres (with the same charge).] The good news is that your 'ions' are all mono-valent. So, its just their relative size that matters. The bad news is that $\ce{AcOH}$, and $\ce{HCl}$, and NaOH all dissociate to ions which can move really fast in water, so you're not just comparing Cl with, say, Br or F. For $\ce{AcOH}$ (acetic acid) we might assume that at infinite dilution it will be completely ionized, but the mass of the acetate anion still substantially slow down any charge movement. In other words, its like a team race, the speed can't be any faster than its slowest runner. Neither NaOH nor HCl have that problem, so we'd definitely expect them to move faster than KCl in aqueous solution. Chemistry has both theoretical logic, and pragmatic - it is great for people who both like to understand why and who like to get their hands dirty by DOING stuff. Unfortunately, some of the ways it is taught are all about "rules" and formulas.

  • $\begingroup$ "Neither NaOH nor HCl has that problem,..." How do that account for NaOH or HCl to have higher molar conductivity than KCl". $\endgroup$
    – Pink
    Commented Dec 22, 2016 at 21:03
  • $\begingroup$ The commentary is not necessary and, in fact, discouraged. $\endgroup$
    – jonsca
    Commented Dec 22, 2016 at 21:18
  • $\begingroup$ I don't think this is a place to post rants. $\endgroup$
    – Mrigank
    Commented Dec 23, 2016 at 2:42
  • $\begingroup$ Is K+ slower than Na+? Because of its higher charge density, Na+ will have a larger hydration sphere, and thus a larger effective radius. So Na+ will have a lower ion mobility - it will be slower. $\endgroup$ Commented Dec 23, 2016 at 3:24

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