Equilibrium & Acid Base

January 30, 2019
Teaching & LearningSarah Wegwerth

In many ways organic chemistry is very different from general chemistry. However, there are things that you learned in general chemistry that are essential for grasping organic chemistry concepts. In particular, we will be reviewing equilibrium and how it applies to acid-base reactions.

Often these concepts are reviewed during the first semester of organic chemistry and heavily used second semester. If you feel like you could use a quick review then keep reading.

The theme for this post: Lower energy is favorable.

NEW! Super Short Version :) 


At equilibrium the rate of the forward and reverse reaction is equal

A generic equilibrium reaction


1 – reactants are going to products AND products are going to reactants at the same rate

2 – the concentrations of reactants and products remains constant

This does NOT mean the concentration of reactants = concentration of products

Concentrations are used to calculate the equilibrium constant (Keq)

Remember our theme: lower energy is favorable. The side of lower energy is favored at equilibrium. By favored we mean there is a higher concentration.

This means…

Acid Base Chemistry

Acid base reactions are reversible and therefore equilibrium reactions.

With acid base reactions we focus on the extent to which theACID (reactant) is deprotonated to the conjugate acid (product). Therefore, as the equivalent to the equilibrium constant, we look at the Ka.

To eliminate the need for scientific notation we then convert the Ka to the pKa

Because of the -log function, now the smaller the pKa value the more the products are favored. This means the stronger the acid the smaller the pKa value.


In short,

·     The smaller the pKa the stronger the acid

·     Acid-base equilibriums favor the side with the acid with a larger pKa

TL;DR Version (The Original)


All reactions are reversible. (In some reactions the product is so heavily favored that we say the reaction is irreversible and that it goes to completion.) This means that the reaction can go from starting materials to product and vice versa. In fact, the mechanism for going from products to starting materials is just the reverse of the mechanism for the forward reaction. For example the mechanism for the protection of the ketone in Aldehyde and Ketone 14 is reversed to accomplish the deprotection in Aldehyde and Ketone 16.

Let’s consider this reversible nature of reactions further using a simple generic reaction

Both the forward and reverse reactions never stop occurring but at some point there is no longer an observable change in the concentration of reactants and products. When this has happened we say the reaction has reached a state of equilibrium.

The extent to which starting material is converted to product is governed by thermodynamics (changes in energy). Remember, it is favorable for a system to go from high energy to low energy. Therefore, the side that is lower in energy is favored at equilibrium.

Quantitatively, we use the equilibrium constant (Keq) to describe the reaction at equilibrium. Using the concentrations at equilibrium we can calculate Keq using the following equation:

Based on this equation we can conclude that at equilibrium:

If Keq > 1 then [C][D] > [A][B]

If Keq < 1 then [C][D] < [A][B]

If Keq = 1 then [C][D] = [A][B]

Let’s take this a step further, since the extent of conversion is a function of the relative energies we can restate the first two above statements as follows:

The larger the value of Keq, the lower in energy the products are relatively to starting materials.  

The smaller the value of Keq, the higher in energy the products are relatively to starting material

I’m from Minnesota, and one of my favorite winter activities is sledding. We can use the analogy of sledding and climbing back up the hill to visualize equilibrium and the associated energetics changes.

When Keq > 1 (the greater than symbol points to the right towards, or products) energetically the reaction is sliding down the hill to get to products.

When Keq < 1 (the less than symbol points to the left, or toward starting materials) energetically the reaction is climbing up the hill to get to products.  

No doubt about it, going down the hill is much more favorable!!!

Gif animation of sliding down a hill with the Keq>1 trailing behind.

No matter how you think about it, knowing which side of the equilibrium is favored informs you whether the starting materials or products are energetically favorable. When is this useful? A good example can be seen by looking at acid-base chemistry.

Acid and Base Chemistry

The strength of an acid and base are measured by equilibrium constants. Let’s explore this using a generic acid-base reaction:
From this we can define the equilibrium constant:

Generally, we can assume the concentration of water to be nearly constant since acid-base reactions are usually done in dilute aqueous solutions. As a result, we can use a new term, Ka in which the concentration of water has been removed from the expression by multiplying both sides of the above equation by [H2O]. The result is

Similar to equilibrium constants, Ka measures the extent to which products are formed. In this case, we are measuring the extent to which the acid dissociates. The stronger the acid the more it dissociates. Therefore, the larger the Ka value the stronger the acid. The value of Ka is usually a very large or a very small number that require use of scientific notation. Therefore, chemists prefer to use pKa values instead. Remember, pKa is related to Ka using the following equation:

If you want, you can review logarithmic math functions elsewhere, but the key thing to remember is that the smaller the pKa value the stronger the acid. This is reverse of the trend we saw for Ka because we looking at the -log. Also keep in mind that a change in 1 pKa unit corresponds to an order of magnitude change in Ka.

Now that we have a better idea of equilibrium and pKa let’s take a look at how we can use pKa values to predict which side of the equilibrium will be favored in acid-base reactions.

Let’s use Acid Base 4 as an example

Remember, the products of an acid base reaction are the conjugate acid and conjugate base. Therefore, we can compare pKa values of the acid and conjugate acid to determine which side of the equilibrium is favored. In this example the pKa on the left is approximately 4 and on the right is 15.7.  Because acids and bases are paired as conjugates, the stronger acid is on the same side as the stronger base, so the reaction that is energetically favored is the one where stronger/stronger → weaker/weaker.

Key things to remember for acid-base reactions:

  • The stronger the acid the lower the pKa value
  • Equilibrium favors the side of the weaker acid

Therefore, pKa values can be used to determine which side of the equilibrium will be favored.

It is helpful to  get a good feeling of the relative pKa of common acids and/or functional groups. Your instructor and your textbook will highlight the ones you need to know.

Remember: Acidity increases as the stability of  the conjugate base increases. To review how resonance and inductive effects can impact the stability of the conjugate base watch the following videos from the app Mechanisms:


Inductive Effects:

Practice determining which side of an acid-base equilibrium is favored can be found following this link. Read through the answers to learn about examples of common equilibria you will likely encounter during the second semester of organic chemistry.

Co- Author

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