What is the difference between molar conductance and molar conductivity?

Question

I'm learning (or at least trying to learn) about electrochemistry, but a major obstacle to that, is that different books I refer use different terms for the same symbols. So in a last ditch attempt to clear stuff up, I've resorted to Chem.SE.

So here's what I intend to do; I'll list out everything I think I've understood, as well as pose a couple of questions regarding some of them. I'd really appreciate it if someone would take the time to go through what I've listed out, checking them for errors and then clearing those queries which I've got. So here I go

Symbols used:

Resistance ($R$) , Resistivity or specific resistance ($\rho$), Conductance ($C$), Conductivity or specific conductance ($\kappa$), Area of cross-section of the electrode ($A$), distance between the electrodes ($L$),

and the REALLY confusing bit, Molar conductance according to some books and Molar conductivity according to others and one book uses both terms, both represented by $\mathrm{Λ_{m}}$

Now, $$ R = \rho \frac{L}{A}\\ R = \frac1C\\ $$ Therefore $$\frac1\rho = \frac1R\cdot\frac{L}A = C\cdot\frac{L}A = \kappa$$

Now if I've got this right, then,

1. Conductance is the degree to which the solution conducts electricity.
2. Conductivity is the conductance per unit volume of the solution; it may also be considered as the concentration of ions per unit volume of solution.
3. Molar Conductivity is the conductance of the entire solution having 1 mole of electrolyte dissolved in it.

Q1. So what's Molar Conductance?

Q2. Is there a difference between Molar Conductivity and Molar Conductance?

Also, according to Ostwald's Dilution Law, greater the dilution, greater the dissociation of the electrolyte in solution.

Regarding dilution of an electrolyte solution, this is what I've understood

1. As dilution increases, Conductivity (ion concentration per unit volume) DECREASES.
2. As dilution increases, Molar conductivity (Conductance of 1 mole of electrolyte in the total solution) should INCREASE in accordance with Ostwald's Law

Q3. How does dilution affect Molar Conductance?

Q4. How is Conductance affected upon dilution?

I suppose if the above statements are proof-read and the queries answered, I might get fairly good idea about this....

Also if you feel there is are any additional points worth mentioning, by all means go ahead and put it in the answer.

And finally, if anyone could recommend a decent site that deals with the above-mentioned terms and concepts in a fairly lucid manner, it'd be appreciated.

Answer 1

I can understand your frustration. The use of terminology is often inconsistent and confused (much to my chagrin). I think you've got the general idea, the conductance ($G$) can be defined as follows:

$$G = \frac{1}{R}$$

i.e. the ease with which a current can flow. As you said, $$R = \rho \frac{l}{A}$$

one can now identify, $$G = \kappa\frac{A}{l}$$ where the conductivity $$\kappa = \frac{1}{\rho}$$

Molar "any quantity" always has the dimensions (it is helpful to think in terms of dimensions) $\text{"quantity" } \mathrm{mol^{-1}}$

so, it follows molar conductivity $$ \Lambda_m = \frac{\kappa}{c}$$ where $c$ is the molar concentration. It is useful to define molar conductivity because, as you already know, conductivity changes with concentration.

Now, to address the effect of change of concentration on molar conductivity, we need to consider the case of weak and strong electrolytes separately.

For a strong electrolyte, we can assume ~ 100% disassociation into constituent ions. A typical example is an $\ce{MX}$ salt like $\ce{KCl}$

$$\ce{MX} \rightleftharpoons \ce{M^+} + \ce{X^-}$$ The equilibrium constant for this reaction is $$ K = \frac{[\ce{M^+}][\ce{X^-}]}{[\ce{MX}]}$$ and thus, with decreasing molar concentration of the electrolyte, equilibrium shifts towards the disassociated ions.

At sufficiently low concentrations, the following relations are obeyed:

$$\Lambda _{m}=\Lambda _{m}^{0}-K{\sqrt {c}}$$

where $ \Lambda _{m}^{0}$ is known as the limiting molar conductivity, $K$ is an empirical constant and $c$ is the electrolyte concentration (Limiting here means "at the limit of the infinite dilution").

In effect, the observed conductivity of a strong electrolyte becomes directly proportional to concentration, at sufficiently low concentrations

As the concentration is increased however, the conductivity no longer rises in proportion.

Moreover, the conductivity of a solution of a salt is equal to the sum of conductivity contributions from the cation and anion.

$$ \Lambda_{m}^{0}= \nu_{+}\lambda_{+}^{0}+\nu _{-}\lambda _{-}^{0}$$ where: $ \nu _{+}$ and $\nu _{-}$ are the number of moles of cations and anions, respectively, which are created from the dissociation of $\pu{1 mol}$ of the dissolved electrolyte, and $\lambda _{+}^{0}$, $\lambda _{-}^{0}$ are the limiting molar conductivity of each individual ion.

The situation becomes slightly more complex for weak electrolytes, which never fully disassociate into their constituent ions. We no longer have a limit of dilution below which the relationship between conductivity and concentration becomes linear. We always have a mixture of ions and complete molecules in equilibrium. Hence, the solution becomes ever more fully dissociated at weaker concentrations.

For low concentrations of "well behaved" weak electrolytes, the degree of dissociation of the weak electrolyte becomes proportional to the inverse square root of the concentration.

A typical example would be a monoprotic weak acid like acetic acid (again, from your graph):

$$\ce{AB} \rightleftharpoons \ce{A^+} + \ce{B^-} $$

Let $\alpha$ be the fraction of dissociated electrolyte, then $ \alpha c_0$ is the concentration of each ionic species. And $(1 - \alpha)$, and $(1 - \alpha)c_0 $ gives the fraction, and concentration of undissociated electrolyte. The dissociation constant is:

$$K = \frac{\alpha^2 c_0}{1-\alpha}$$

for weak electrolytes, $\alpha$ is tiny, so the denominator is nearly equal to one so, $$ K \approxeq \alpha^2 c_0$$ and $$ \alpha = \sqrt{\frac{K}{c_0}}$$ (like I said earlier)

for conductivities, one can now write the following relation $$\frac{1}{\Lambda_m} = \frac{1}{\Lambda_m^0} + \frac{\Lambda_mc}{K (\Lambda_m^0)^2}$$ This fits the curve seen in your graph.

Caveat, all these arguments hold for dilute solutions. Things get out of hand at high concentrations, and one has to account for some additional phenomenon (for example, acetic acid will form hydrogen bonded dimers).

Anyway, long story short, conductivity increases with increasing dilution (though differently for strong and weak electrolytes). From the definitions I outlined at the very start, guessing how conductance changes is trivial.

Answer 2

Let's get the terms right, firstly, because that seems to generate some confusion.

Conductance=ability of a component to conduct electric current (it might also refer to heat and others, but we'll stick to the ones in your example for now).

It depends on: the material of your component (that is actually its conductivity), the length of the component (l) and the cross section of the component (A). It's unit measurement is S(from conductivity)*m(from length)/m2 (from area) which yields S/m.

Conductivity is the ability of a material to conduct electric current, regardless of its dimensions. It is also called specific conductance. And it's the reverse of resistivity, measured in S (siemens, mentioned also above).

If I have a cell/component of 1 sqm cross section and 1 m length, conductance and conductivity would be equal in value. If these are different values for the component Im computing, I adjust the conductivity of the material in general with the dimensions, and get the conductance of my particular component (such as a conductor, piece of metal, electrolyte environment etc)

So I would say about a metal that it has high conductivity and about a rod (or other specific piece) of the same metal that is has a (assumably) high conductance.

Now, moving to the questions:

Q1: Molar Conductance = conductance of all the ions produced by ionization of 1 g mole of an electrolyte when present in V mL of solution. Mathematical relation is miu=k * V ; k is the conductivity=specific conductance

Q2: Molar Conductivity=lambda= k/ molecular concentration which makes it equal to molar conductance per mole. So this is pretty much the difference between them, if we could name it as such.

Note that these refer to ionic conductivities, for electrolytes.

Q3: The variation of molar conductance with concentration can be explained on the basis of conducting ability of ions for weak and strong electrolytes. A strong electrolyte is a compound that is totally dissociated in water in its component ions while a weak one is an electrolyte where we have present the compound in itself, as well as component ions separately. I remind you that ionic conductivity/conductance both rely on the presence of ions in solution.

For weak electrolytes: as we increase the volume (thus the dilution), we encourage the equilibrium to shift towards dissociation. Therefore, we will have more ions in solution and a implicit increase in molar conductance. It basically means that 1 mole dissolved into 10 mL of water gives out less ions than 1 mole of the same, in the same conditions, in 100 mL of water.

For strong electrolytes, the observed effect is the same (molar cond. increases with V increasing), but the reasons are different. In concentrated solutions (less solvent) the strong opposingly charged ions tend to attract each other. Give them enough space (i.e increase V) to move freely of these attractions and you get an increased molar conductance.

Q4: Conductance, we said is conductivity* cross area/ length of component, and the reverse of resistance. So conductance=current intensity/tension; The intensity of the current in turn, is directly proportional with ion concentration (no of ions/volume) and ionic mobility (nevermind this for a bit). Therefore, if we have lower ions concentration (same ions/bigger volume), we will have a smaller current and hence, a decrease in conductance.

Hope these help with sth. If you have some follow-up questions, please feel free. In the meanwhile, I will think of some resources to send you as trustworthy source.

Cheers.