Mastering Acid-Base Equilibrium: pKa & Keq Calculations

1. From Acid Strength (Ka) to a Logarithmic Scale (pKa)

You know that an acid's strength is measured by its acidity constant, $K_a$, which is the equilibrium constant for its dissociation in water.

$\text{HA}+{\text{H}}_2\text{O}\rightleftharpoons {\text{A}}^{-}+{\text{H}}_3{\text{O}}^{+}$
$K_a=\frac{[{\text{A}}^{-}][{\text{H}}_3{\text{O}}^{+}]}{[\text{HA}]}$

The range of $K_a$ values is enormous, spanning many orders of magnitude.

To make these numbers more manageable, chemists use the pKa scale, defined as:

$p{K}_a=-{\log }_{10}(K_a)$

Due to the negative sign, this creates an inverse relationship:

• A stronger acid has a larger $K_a$ and a lower pKa.
• A weaker acid has a smaller $K_a$ and a higher pKa.

This logarithmic scale is crucial; a difference of one pKa unit corresponds to a tenfold difference in acidity.

2. Predicting the Position of Equilibrium

An acid-base reaction is a competition for a proton. The equilibrium will always lie on the side of the more stable species—the weaker acid and the weaker base.

Consider a general reaction:
$\underset{\text{Acid 1}}{\mathrm{HA}}+\underset{\text{Base 2}}{{\text{B}}^{-}}\rightleftharpoons \underset{\text{Base 1}}{{\text{A}}^{-}}+\underset{\text{Acid 2}}{\mathrm{HB}}$

The Rule: The equilibrium favors the formation of the acid with the higher pKa value.

This is because a higher pKa signifies a weaker, more stable acid that is less inclined to donate its proton.

The following videos provide excellent demonstrations of this principle and how to apply it.

3. Calculating the Equilibrium Constant (Keq)

Simply knowing which side is favored is useful, but calculating the equilibrium constant, $K_{eq}$, tells us by how much. We can derive a simple and powerful formula to do this.

For our general reaction HA + B⁻ ⇌ A⁻ + HB, the equilibrium expression is:
$K_{eq}=\frac{[{\text{A}}^{-}][\text{HB}]}{[\text{HA}][{\text{B}}^{-}]}$

We can cleverly rearrange this by multiplying the numerator and denominator by $[{\text{H}}_3{\text{O}}^{+}]$:
$K_{eq}=\left(\frac{[{\text{A}}^{-}][{\text{H}}_3{\text{O}}^{+}]}{[\text{HA}]}\right)\times \left(\frac{[\text{HB}]}{[{\text{B}}^{-}][{\text{H}}_3{\text{O}}^{+}]}\right)$
You should recognize the first term as $K_{a(\text{HA})}$ and the second term as $1/K_{a(\text{textHB})}$. Therefore:
$K_{eq}=\frac{K_{a(\text{Reactant Acid})}}{K_{a(\text{Product Acid})}}$
Taking the base-10 logarithm of both sides and multiplying by -1 gives:
$-\log (K_{eq})=-\log (K_{a(\text{Reactant})})-(-\log (K_{a(\text{Product})}))$
$p{K}_{eq}=p{K}_{a(\text{Reactant Acid})}-p{K}_{a(\text{Product Acid})}$
Since $K_{eq}=10^{-p{K}_{eq}}$, we arrive at the final, practical formula:
$K_{eq}=10^{(p{K}_{a(\text{Product Acid})}-p{K}_{a(\text{Reactant Acid})})}$
Or, more succinctly:
$K_{eq}=10^{\Delta p{K}_a}\text{where}\Delta p{K}_a=p{K}_{a(\text{product acid})}-p{K}_{a(\text{reactant acid})}$

Let's watch this calculation in action.

4. The Thermodynamic Basis: Why pKa Predicts Equilibrium

Your interest is in the fundamental "why," so let's connect this to thermodynamics. The rule that "equilibrium favors the weaker acid" is not just a mnemonic; it is a direct consequence of the second law of thermodynamics.

A reaction is spontaneous (favored) if the change in Gibbs free energy, $\Delta G^{\circ }$, is negative. $\Delta G^{\circ }$ is related to the equilibrium constant by the cornerstone equation:
$\Delta G^{\circ }=-RT\ln (K_{eq})$
Where $R$ is the gas constant and $T$ is the temperature in Kelvin.

By substituting our expression for $K_{eq}$ in terms of $K_a$ values:
$\Delta G^{\circ }=-RT\ln \left(\frac{K_{a(\text{Reactant})}}{K_{a(\text{Product})}}\right)=-RT(\ln K_{a(\text{Reactant})}-\ln K_{a(\text{Product})})$
Since $\ln (x)=2.303{\log }_{10}(x)$ and $p{K}_a=-{\log }_{10}(K_a)$, we can rewrite the Gibbs free energy for the dissociation of any acid as $\Delta G^{\circ }=2.303RT\cdot p{K}_a$. Substituting this back into our equation for the overall reaction gives:
$\Delta G_{\text{reaction}}^{\circ }=\Delta G_{\text{Reactant Acid}}^{\circ }-\Delta G_{\text{Product Acid}}^{\circ }$
$\Delta G_{\text{reaction}}^{\circ }=2.303RT\cdot p{K}_{a(\text{Reactant})}-2.303RT\cdot p{K}_{a(\text{Product})}$
This simplifies to a profoundly elegant relationship:
$\Delta G^{\circ }=2.303RT(p{K}_{a(\text{Reactant Acid})}-p{K}_{a(\text{Product Acid})})$

This equation mathematically proves our rule. For a reaction to be spontaneous ($\Delta G^{\circ }<0$), the term $(p{K}_{a(\text{Reactant})}-p{K}_{a(\text{Product})})$ must be negative. This is only true if $p{K}_{a(\text{Reactant})}<p{K}_{a(\text{Product})}$ — meaning the reactant acid must be stronger (lower pKa) than the product acid.

The following paper provides exactly this derivation, which you should find satisfying as it formalizes the qualitative rules taught in most textbooks.

Conclusion

Today you've made a significant leap from qualitative reasoning to quantitative prediction in acid-base chemistry. You can now use a single, powerful number—pKa—to determine not just the direction but also the magnitude of an acid-base equilibrium.

Key Takeaways:

• Acid-base equilibria always favor the side with the weaker acid and weaker base.
• The weaker acid is the one with the higher pKa value.
• The equilibrium constant can be calculated directly from the pKa values of the reactant and product acids: $K_{eq}=10^{(p{K}_{a(\text{product})}-p{K}_{a(\text{reactant})})}$.
• This relationship is fundamentally grounded in thermodynamics, where the Gibbs free energy change of the reaction, $\Delta G^{\circ }$, is directly proportional to the difference in the pKa values.

Preview of the Next Lesson:
We have focused on equilibrium (a thermodynamic property). However, being a strong base is not the same as being a fast-acting one. In our next lesson, we will explore this distinction by learning to distinguish between nucleophilicity and basicity, delving into the kinetic aspects of reactions and the roles of polarizability and solvent effects.