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I have the following equation

$$K_a = \dfrac{c\alpha^2}{1-\alpha},$$

with units $c= \pu{mol/cm^3}$ and $\alpha$ is the same. I'm not sure how to get the correct units for $K_a$. What I have come up with so far is

$$K_a= \pu{mol^2/cm^6}.$$

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    $\begingroup$ This looks like Ostwald’s dilution law written for simple binary electrolyte $\ce{AB <=> A- + B+}$, so $[K_\mathrm{a}] = [C] = \pu{M}\, (\pu{mol L-1}, \pu{mol m-3})$. For more complex dissociation you need to reconstruct the formula, so there is no unified unit for $K_\mathrm{a}$. $\endgroup$ – andselisk Nov 9 '17 at 12:59
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    $\begingroup$ Ideally, you would write the equilibrium constant using normalised unitless concentrations so $K_\mathrm a$ has no unit at all. (If it had, you couldn’t logarithmatise it.) $\endgroup$ – Jan Nov 9 '17 at 13:02
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Andselisk correctly identified the law of dilution and the name Ostwald is often connected with it.
$$K_\text{dissociation} = \frac{\alpha^2}{1-\alpha}\cdot c$$ However, the degree of dissociation is $\alpha$ and has "no" unit, i.e. dimensionless quantity. Therefore the unit for the equilibrium constant is that of a concentration, in SI that would be $\pu{mol m-3}$.

It is derived from the Law of mass action; the derivation can be found on Wikipedia and various other sources.


In principle there is no fixed unit for the (generalised) equilibrium constant, as it is simply defined as a product of the involved quantities according to the IUPAC gold book.

Equilibrium Constant
Quantity characterizing the equilibrium of a chemical reaction and defined by an expression of the type $$K_x = \Pi_B x_B^{\nu_B},$$ where $\nu_B$ is the stoichiometric number of a reactant (negative) or product (positive) for the reaction and $x$ stands for a quantity which can be the equilibrium value either of pressure, fugacity, amount concentration, amount fraction, molality, relative activity or reciprocal absolute activity defining the pressure based, fugacity based, concentration based, amount fraction based, molality based, relative activity based or standard equilibrium constant (then denoted $K^\circ$ ), respectively.

The standard equilibrium constant is always dimensionless, as it is defined differently (IUPAC gold book).

Standard Equilibrium Constant $K$, $K^\circ$
(Synonym: thermodynamic equilibrium constant)
Quantity defined by $$K^\circ = \exp\left\{-\frac{\Delta_rG^\circ}{\mathcal{R}T}\right\}$$ where $\Delta_rG^\circ$ is the standard reaction Gibbs energy, $\mathcal{R}$ the gas constant and $T$ the thermodynamic temperature. Some chemists prefer the name thermodynamic equilibrium constant and the symbol $K$.

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  • $\begingroup$ Good answer. (+1) One problem with the 'equilibrium constant': unlike the standard equilibrium constant it is not constant at fixed temperatures, and will vary with composition. Only in ideal approximations will the quantity $K_x$ be truly constant at equilibrium. In this way the naming 'equilibrium constant' may be a bit misleading. doi: 10.1351/pac199466030533 This is more of a comment on IUPAC's recommendation than your post. $\endgroup$ – Linear Christmas Dec 4 '17 at 16:56
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By definition, the equilibrium constant has no units, as we're supposed to be using active masses instead of the molarity/ concentrations of the respective substances.

Active mass is a unit-less ratio of the mass reacting to the total mass present. Ideally, the equilibrium constant therefore doesn't have a unit. However, due to physical limitations, the exact active mass is difficult to calculate. Therefore, we use the concept of concentration instead, which is basically same as the active mass.
However, concentrations have units of $\pu{mol/cm3}$. This results in equilibrium constants having various different units depending on the equivalents of reactants in the balanced equation.

In your equation, the units that you calculated are perfectly right, but as far as I know, the $\alpha$ variable is a unit-less ratio, called the degree of dissociation, which is mathematically equal to: $$\frac{\text{amount of substance dissociated}}{\text{total amount of substance present}}$$ in case of Ostwald's dilution law. The actual units would be, as andselisk said, just $\text{mol/cm$^3$}$.

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    $\begingroup$ There is absolutely nothing wrong with expressing the same thing in different words. Some people might prefer your version. Therefore I have edited your version and undeleted it. (I can because I am a moderator. :D) One point though: Please note that the proper term for "(number of) moles" is amount of substance. The former would be the same as referring to the mass as "(number of) kilograms". $\endgroup$ – Martin - マーチン Nov 9 '17 at 14:32

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