How to Determine the Position of Equilibrium for an Acid–Base Reaction

In the previous post, we learned about the pKa and summarized the factors that affect the acidity and pKa value. The pKa values are mostly used in order the predict the position of the equilibrium of an acid-base reaction, as well as for choosing an acid or a base to protonate or deprotonate certain species.

For example. The reaction between acetic acid and sodium hydroxide is a simple acid-base reaction that you have probably used in your general chemistry course for the acid-base titration experiment:

 

 

So, we know that this reaction works, but how do we explain this in terms of pKa values?

Here is how it works. First, you need to understand and remember this important concept:

Any Acid-Base reaction favors the formation of a weaker acid and a weaker base.

The reason for this is that, like any chemical reaction or process, the acid-base reactions go towards a lower energy state. A strong acid or a base means that they have a lot of energy and are very reactive, while weaker acids and bases have lower energy.

Therefore, to predict the position of the equilibrium, follow these steps:

1) Identify the acid and the conjugate acid of the base:

 

 

In this case, it is the acetic acid and water, respectively. Remember, acid is the proton donor (CH3COOH), the base is the proton acceptor (OH), and the conjugate acid is the protonated form of the base (H2O).

2) Look up the pKa values of the acid and the conjugate acid:

 

pKa(CH3COOH) = 4.75,  pKa(H2O) = 15.7

 

3) Write the reaction equation pointing to the equilibrium position with a longer arrow:

 

 

Water is a 1011 times weaker acid. Therefore, the equilibrium is shifted to the right. For such large differences in acidity, the reaction is not even considered equilibrium, and very often acid-base reactions are shown with a single arrow.

 

 

Another example: What side do you think the position of equilibrium is for the following reaction?

 

 

Let’s follow the steps we just discussed.

1) Identify the acid and the conjugate acid. The acid is the alkyne, and the conjugate acid is ammonia (NH3).

2) Look up their pKa values:

pKa(CH3CCH) = 25,   pKa(NH3) = 38

 

3) Indicate the position of equilibrium with a larger arrow:

 

 

The position of equilibrium indicates that the reverse reaction does not happen to a significant extent, and if we were to have the products as starting material, an opposite direction arrow would be needed:

 

 

You can also predict the acid-base reaction without having the pKa values. Remember, in the beginning, we said that the position of equilibrium is always on the side of a weaker acid and a weaker base. Therefore, you can compare the base and the conjugate base instead of comparing the acid and the conjugate acid.

If the base is more stable (check the factors affecting the stability of the conjugate base) than the conjugate base, the reaction favors the reactants:

 

 

If, on the other hand, the conjugate base is more stable, then the equilibrium favors the products:

 

 

In all these examples, we had a reaction for which we needed to determine whether it is a feasible acid-base reaction or not.

However, sometimes you may be given one component and asked to choose a proper acid or a base to protonate or deprotonate it. This is also done based on the pKa values. Detailed coverage in the next post.

Practice

1.

Using the pKa values in the Table, determine the position of equilibrium for each of the following acid-base reactions: Is the acid strong enough to protonate the base or, which is the same to say, is the base strong enough to deptrotonate the acid in each reaction? Or, the question can also be formulated-will each of the following reactions proceed to a significant extent?

Answer

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2 thoughts on “How to Determine the Position of Equilibrium for an Acid–Base Reaction”

  1. Hello, I am a bit confused about the yes and no. For example, in problem B do you mean no as in no to the left and the correct answer is to the right ?

    Reply
    • Hi there,

      The larger arrow shows the direction of the equilibrium, and the term “yes” means the acid is strong enough to protonate the base, or we can also say that the base is strong enough to deprotonate the given species (the acid). In other words, the acid-base pair is a good match to react (equilibrium will shift to the right) The “No,” on the other hand, means the acid and the base are not strong enough to react as shown, which means the equilibrium will be shifted to the left – the reverse reaction is favored. Remember, the equilibrium shifts in favor of forming a weaker acid and a weaker base, and, in terms of pKa, this means forming an acid/conjugate acid with a greater pKa value.

      Reply

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