Animated Solution for Chemistry - Ionic Equilibrium: An acid HA ionises as $HA\rightleftharpoons H^{+}+A^{-}$. The pH of 1.0 M solution is 5. Its dissociation constant would be

Imagine you are standing in a laboratory, holding a beaker filled with a $1.0\text{ M}$ solution of an unknown acid, which we will simply call $\text{HA}$. You dip a pH meter into the clear liquid, and the digital display settles on a crisp, clean number: $5.0$.

What does this number tell us? If this were a strong acid like hydrochloric acid ($\text{HCl}$), a $1.0\text{ M}$ solution would have a pH of $0$. The fact that the pH is $5$ screams that this acid is incredibly weak. It is barely dissociating into its constituent ions. Our mission is to quantify exactly how weak it is by calculating its acid dissociation constant, $K_a$.

The pH scale is a logarithmic measure of the hydrogen ion concentration in a solution. The mathematical definition is:

By rearranging this equation, we can unlock the exact concentration of hydrogen ions floating in our beaker:

Substituting our measured pH of $5$, we find:

This is a tiny number—only $0.00001\text{ moles per liter}$ of hydrogen ions are present. The vast majority of the acid molecules are staying intact as $\text{HA}$.

To find the dissociation constant, we need to know the concentrations of all species at equilibrium. We use an ICE (Initial, Change, Equilibrium) table to map out the reaction:

Initial State: Before any dissociation happens, we have $1.0\text{ M}$ of $\text{HA}$, and $0\text{ M}$ of both ${\text{H}}^{+}$ and ${\text{A}}^{-}$.

Change: Let's say $x$ moles per liter of $\text{HA}$ dissociate. This will produce $x$ moles per liter of ${\text{H}}^{+}$ and $x$ moles per liter of ${\text{A}}^{-}$.

Equilibrium State: The final concentrations will be $(1.0-x)$ for $\text{HA}$, and $x$ for both ${\text{H}}^{+}$ and ${\text{A}}^{-}$.

But wait! We already know $x$. The equilibrium concentration of ${\text{H}}^{+}$ is exactly what we calculated from the pH. Therefore, $x=10^{-5}\text{ M}$.

This gives us our equilibrium concentrations:

The acid dissociation constant $K_a$ is defined by the equilibrium expression:

Let's plug in our values:

Now, we face a crucial moment in physical chemistry: The Approximation. Look at the denominator: $1.0-10^{-5}$. This is $1.0-0.00001$, which equals $0.99999$. In the grand scheme of significant figures and experimental error, $0.99999$ is practically indistinguishable from $1.0$.

Because the degree of dissociation is so incredibly small (only $0.001\%$ of the acid dissociates), we can safely ignore the $10^{-5}$ subtracted from the initial concentration. This simplifies our math beautifully:

With the denominator out of the way, the calculation is trivial:

And there we have it. The dissociation constant of our mystery acid is $10^{-10}$. This elegantly demonstrates how a simple pH reading, combined with the logical structure of an ICE table and a smart mathematical approximation, allows us to peer into the microscopic equilibrium of a chemical system.