Acid-Base Ex I. Why do Acid-Base Reactions Proceed Towards the Formation of Weaker Acids and Weaker Bases? – A Thermodynamic Point of View

III. Calculating the Values of K_eq and ∆G

I. Introduction

In Acid-Base VI, I mentioned that as a rule of thumb, stronger acids and bases react with each other to form weaker acids and bases. I believe that even high school students have unknowingly told about this when they studied acid-base, but have probably never realised that there is this unspoken rule!

So, why do acid-base reactions proceed towards the formation of weaker acids and weaker bases?

From the point of view of acid-base theory, we know that strong acids give away their protons easily and strong bases catch protons easily. This means that the conjugate bases of strong acids and conjugate acids of strong bases are both more stable and therefore, weak. It is because of the stability of weaker species that acid-base reactions favour the formation of weak acids and bases. You can review the concept of strong and weak acid-base and their stability in Acid-Base II.

In this post, we would like to have a look at their stability and what drives this process forward from the point of view of thermodynamics.

In general chemistry, we have learnt that whether a reaction proceeds or not depends on the Gibbs energy change that accompanies that reaction. A negative Gibbs energy change means that the reaction proceeds spontaneously, whilst a positive Gibbs energy change means that the reaction does not proceed spontaneously. You can read more or refresh your memory about Gibbs energy and reaction spontaneity from this post on my other website, ChemisTronomiX.

Therefore, a quick answer to the main topic of this post is that because the formation of weaker acids and weaker bases gives a negative Gibbs energy change. The aim of this post is to prove this to you, using real examples and numbers.

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II. Several Examples

Let’s start off with an easy example. Have look at the reaction of HCl with NH₃, along with their respective pK_a and pK_aH values at 25 ^oC (298 K).

Fig01 HCl and NH3 neutralisation with pKa Figure 1. Reaction of HCl with NH₃.

So we have 2 acids: HCl and NH₄⁺, and 2 bases: NH₃ and Cl^–. Let’s compare their strength relative to each other using their respective pK_a values:

HCl (pK_a = –7) is a stronger acid compared to NH₄⁺ (pK_a = 9.2)
NH₃ (pK_aH = 9.2) is a stronger base compared to Cl^– (pK_aH = –7)

Therefore, this reaction proceeds from strong acid-base to become weaker acid-base. We know from high school chemistry that this reaction, a neutralisation reaction, proceeds spontaneously.

Let’s get more organic-y now, shall we? Let’s have a look at the reaction between the α-H of diethyl malonate and ethoxide anion to form an enolate ion. The formation of enolate catalysed by a base is very commonly observed in aldol reaction, Claisen condensation, Dieckmann condensation, etc.

Fig02 diethyl malonate and ethoxide reaction with pKa
Figure 2. Reaction of diethyl malonate with ethoxide anion.

Again, we have 2 acids: diethyl malonate and ethanol, and 2 bases: ethoxide anion and the enolate ion. Let’s compare their strength relative to each other using their respective pK_a values:

Diethyl malonate (pK_a = 13) is a stronger acid compared to ethanol (pK_a = 16)
Ethoxide anion (pK_aH = 16) is a stronger base compared to the enolate ion (pK_aH = 13)

Therefore, the formation of enolate ion also proceeds from strong acid-base to become weaker acid-base.

Both of the examples above are talking about forming weaker acid-base from stronger acid-base. What about the reverse? What about the possibility of forming stronger acid-base from weaker acid-base? Well, in short, it’s not possible. Let’s have a look at an example.

When you dissolve NaCl in water, you get Na⁺ and Cl^–. Do you see Cl^– reacting with water to give HCl and OH^–? Like what’s shown in Figure 4 below.

Fig03 H2O and Cl- reaction with pKa Figure 3. Reaction of Cl^– with water.

So is the above reaction possible? Let’s see.

Here, we have 2 acids: water and HCl, and 2 bases: Cl^– and OH^–. Let’s compare their strength relative to each other using their respective pK_a values:

Water (pK_a = 15.7) is a weaker acid compared to HCl (pK_a = –7)
Cl^– (pK_aH = –7) is a weaker base compared to OH^– (pK_aH = 15.7)

Therefore, this reaction proceeds from weaker acid-base to become stronger acid-base. Sure enough, this does not happen – you don’t get NaOH and HCl when you dissolve table salt in your soup, do you?

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III. Calculating the Values of K_eq and ∆G

Now, let’s see the reason for this behaviour, from thermodynamic point of view.

As we have seen in Acid-Base V, we can predict the equilibrium constant K_eq of this neutralisation reaction using the formula below. If you’re unsure or forget how we got this, head over to Acid-Base V.

$K_{\text{eq}} = 10^{\text{(p}K_{\text{aH}}\text{ of base)}-\text{(p}K_{\text{a}}\text{ of acid)}}$

If we know the equilibrium constant, we can find out the Gibbs energy of the neutralisation reaction. The relationship between K_eq and ∆G is expressed in the following equation:

$\Delta G = -RT \text{ ln }K_{\text{eq}}$

And here’s the results for our three examples, compiled in a single table. You can read the explation below the table.

No.	Acid (pK_a)	Base (pK_aH)	K_eq	∆G (kJ mol^–1)
1.	HCl (–7)	NH₃ (9.2)	1.58 × 10¹⁶	–92.41
2.	Diethyl malonate (13)	Ethoxide anion (16)	1.00 × 10³	–17.11
3.	H₂O (15.7)	Cl^– (–7)	2.00 × 10^–²³	+129.49

Table 1. K_eq and ∆G values for all three reactions.

I think you know what this means already, but let’s just talk about it anyway.

In the first example:

Fig04 HCl and NH3 neutralisation with Keq and Gibbs energy Figure 4. Reaction of HCl with NH₃ and its K_eq and ∆G values.

For the neutralisation reaction between HCl and NH₃, K_eq of 1.58 × 10¹⁶ means that this equilibrium greatly favours the product formation and, for all intents and purposes, goes to completion. This is why this reaction is usually expressed with a one-way arrow instead of an equilibrium arrow. This K_eq yields ∆G at 298 K as –92.41 kJ mol^–1. Such a large negative value shows that this reaction is a highly exergonic reaction and therefore proceeds spontaneously at room temperature.

In the second example:

Fig05 diethyl malonate and ethoxide reaction with Keq and Gibbs energy
Figure 5. Reaction of diethyl malonate with ethoxide anion and its K_eq and ∆G values.

For the enolate formation between diethyl malonate and ethoxide anion, K_eq of 1.00 × 10³ means that this equilibrium favours the enolate formation. This K_eq yields ∆G at 298 K as –17.11 kJ mol^–1, meaning that this exergonic reaction proceeds spontaneously at room temperature.

In the third example:

Fig06 H2O and Cl- reaction with Keq and Gibbs energy Figure 6. Reaction of Cl^– with water and its K_eq and ∆G values.

For the reaction between Cl^– and H₂O, K_eq of 2.00 × 10^–²³ means that this equilibrium does not favour the product formation and, for all intents and purposes, does not take place. This K_eq yields ∆G at 298 K as +129.49 kJ mol^–1. Such a large positive value shows that this reaction is a highly endergonic reaction and therefore proceeds nonspontaneously at room temperature.

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IV. Conclusion

Overall we have seen and proved that acid-base reactions proceed spontaneously from stronger acid-base to form weaker acid-base. The main reason is that thermodynamically, this is an exergonic reaction and is therefore, spontaneous.

The reverse reaction, in which weaker acid-base react to form stronger acid-base, is therefore endergonic and nonspontaneous.

One thing you should always remember is that a nonspontaneous reaction does not mean that a reaction won’t happen, it just means it needs to be forced by a constant supply of energy to happen. If you’re confused, have a look at this post on my other website, ChemisTronomiX

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Page last updated: 15v17
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