# Question Video: Deriving the Relationship between 𝐾_𝑎 and Degree of Dissociation Chemistry

According to Oswald’s law of dilution, which of the following is the correct relationship between the dissociation constant (𝐾_𝑎) and the degree of dissociation (𝛼)? Assume that 1 − 𝛼 ≌ 1. [A] 𝐾_(𝑎) = √(𝛼/𝐶_𝑎) [B] 𝐾_(𝑎) = 𝛼 𝐶_𝑎 [C] 𝐾_(𝑎) = √(𝐶_(𝑎)/𝛼) [D] 𝐾_(𝑎) = 𝛼 𝐶 𝑎² [E] 𝐾_(𝑎) = 𝛼²𝐶_(𝑎)

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### Video Transcript

According to Oswald’s law of dilution, which of the following is the correct relationship between the dissociation constant 𝐾 𝑎 and the degree of dissociation 𝛼? Assume that one minus 𝛼 is approximately equal to one. (A) 𝐾 𝑎 equals the square root of 𝛼 divided by 𝐶 𝑎. (B) 𝐾 𝑎 equals 𝛼 multiplied by 𝐶 𝑎. (C) 𝐾 𝑎 equals the square root of 𝐶 𝑎 divided by 𝛼. (D) 𝐾 𝑎 equals 𝛼 multiplied by 𝐶 𝑎 squared. (E) 𝐾 𝑎 equals 𝛼 squared multiplied by 𝐶 𝑎.

Some molecules when placed into water dissociate into separate ions. Dissociation is the separation of a substance into individual components. Degree of dissociation is a measure of the proportion of the substance that actually dissociated. Mathematically, it’s given the symbol 𝛼, and it equates to the amount of dissociated substance divided by the total amount of substance that is dissociated and undissociated. Degree of dissociation is usually expressed as a decimal, and it can be expressed in terms of moles of dissociated substance divided by total moles of substance dissociated and undissociated. We can apply it to all homogeneous systems that have arrived in an equilibrium state.

If our starting molecule is that of a weak acid, shown here as HA, we’ll find that it only dissociates partly in water. We can see that the dissociation of HA is a reversible process, and eventually equilibrium will be established. For our weak acid, we can write out an expression for the acid dissociation constant denoted by 𝐾 𝑎. In our 𝐾 𝑎 expression, the product ion concentrations are seen multiplied together on the top half of the 𝐾 𝑎 fraction, whilst the reactant molecule concentration, in this case HA, is seen on the bottom half of our fraction.

The size of the 𝐾 𝑎 value gives us an indication of the strength of the acid. Smaller values of 𝐾 𝑎 mean weaker acids. Larger values of 𝐾 𝑎 mean stronger acids. Remember that for a weak acid, all three species seen in the equation here will be present at equilibrium. In a weak acid, there will always be some HA left in the solution as it’s not fully dissociated. The amount of H+ ions and A− ions in this scenario depends on the strength of the acid. In the context of our weak acid HA, the degree of dissociation or 𝛼 for HA can be viewed as the proportion of HA that actually dissociates.

We can see how the acid concentration changes from an initial concentration, denoted 𝐶 𝑎, as equilibrium is approached. 𝐶 𝑎, which represents the starting concentration of our weak acid, is seen in all the possible answers. So, for our weak acid, we’ll take the initial concentration before any dissociation occurs as being 𝐶 𝑎. For this scenario, the concentrations of H+ and A− would both be zero as no dissociation has taken place yet. For the weak acid, the change in concentration as equilibrium is attained would be negative 𝐶 𝑎 multiplied by 𝛼. Here, we are adjusting the original concentration of HA by the proportion of HA that dissociates or 𝛼 to find the new concentration at equilibrium.

The change in concentration is negative here as HA is being used up. It’s a reactant in this process. The changes in the concentrations of our product ions H+ and A− are both 𝐶 𝑎 times 𝛼, respectively. These changes in concentrations both take positive values as they’re being formed from zero. The changes of these product ion concentrations are identical to each other due to the stoichiometry of the reaction occurring. For each species, by adjusting the initial concentration by the change in concentration, we can get the equilibrium concentrations. The equilibrium concentration for HA becomes 𝐶 𝑎 subtract 𝐶 𝑎 multiplied by 𝛼. Since the initial concentrations of H+ and A− were both zero initially, the adjusted concentrations are both 𝐶 𝑎 multiplied by 𝛼.

Now let’s focus more closely on the equilibrium concentrations for HA, H+, and A−. We’ve got them in terms of 𝐶 𝑎 and 𝛼, but we can simplify them a little further. When factorized, the equilibrium concentration for HA simplifies to 𝐶 𝑎 multiplied by one minus 𝛼. Since we now have our equilibrium concentrations in terms of 𝐶 𝑎, the initial concentration of the acid, and 𝛼, the degree of dissociation, we’ll substitute these values into our 𝐾 𝑎 expression. We’ll substitute the concentration of HA into the bottom of the 𝐾 𝑎 expression. And we’ll substitute the concentrations of H+ and A− into the top half of the 𝐾 𝑎 expression.

We can now see more clearly the relationship between the acid dissociation constant 𝐾 𝑎, the initial concentration of the weak acid, and the degree of dissociation 𝛼. To simplify this expression a little further, we can cancel 𝐶 𝑎 on the top and bottom half of this fraction. This gives us 𝐾 𝑎 equals 𝛼 squared multiplied by 𝐶 𝑎 all divided by one minus 𝛼. But we’re also told in the question that one minus 𝛼 is approximately equal to one. This means that for a weak acid, the degree of dissociation or 𝛼 is a very small quantity indeed. If one minus 𝛼 is approximately equal to one, our 𝐾 𝑎 expression can be further simplified.

Our 𝐾 𝑎 expression therefore simplifies to 𝛼 squared multiplied by 𝐶 𝑎 divided by one. So 𝐾 𝑎 equals 𝛼 squared multiplied by 𝐶 𝑎, which agrees with answer (E). The correct answer is 𝐾 𝑎 equals 𝛼 squared multiplied by 𝐶 𝑎.