## Contents

I. Introduction

II. Acidity and Basicity

II.1. Acidity (Ka and pKa)

II.2. Basicity (Kb and pKb)

II.3. Seeing a Base Through Its Conjugate Acid: KaH and pKaH

II.3.1. Step i

II.3.2. Step ii

II.3.3. Summary and Other Details

III. Conclusion

IV. Try It Yourself!

## I. Introduction

In Acid-Base I, we have learnt what strong and weak acids/bases are. We also learnt the relationship between them and their respective conjugate pairs. So that’s good and all, but here we will move on to the more common way that an organic chemist would do when looking at a reaction.

So now that your memory is, hopefully, refreshed about what acid/base is and what conjugate pairs are, we shall start looking at the concept from an organic chemist’s point of view.

In this post I re-introduce the concept of acidity (Ka and pKa) and basicity (Kb and pKb). After that, we learn how to think about the basicity of a base using the Ka of its conjugate acid. This is something that is very useful and is used a lot in organic chemistry.

From my own experience, it was hard for me to think in terms of Ka for bases because in general chemistry, I was more used to using Kb to gauge basicity of a base. I’m guessing that many people also run into the same problem, so I will discuss this important topic in a thorough manner. Here, I introduce the concept of KaH and pKaH.

The aims of this post are:

• To refresh your memory about acidity (Ka and pKa) and basicity (Kb and pKb)
• To start thinking and seeing bases through their conjugate acids by introducing KaH and pKaH

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## II. Acidity and Basicity

This is another extension of Brønsted-Lowry acid/base theory. First, let’s address this question: how acidic is an acid and how basic is a base? Let’s look at it the way an organic chemist does.

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### II.1. Acidity (Ka and pKa)

Brønsted-Lowry theory states that an acid is a proton (H+) donor. This implies that the strength of an acid is determined by how readily it releases H+.

Strong acids release H+ more readily, whilst weaker acids, not so much. This implies that strong acids also release their conjugate bases readily. This means that the conjugate bases of strong acids are stable (that’s why it can be released readily). Ultimately, this translates to: strong acids give weak conjugate bases.

Weak acids find it hard to release their H+, and do not do it readily. Once a weak acid releases H+ and a conjugate base, they quickly get back together to form the acid again. This means that the conjugate bases of weak acids are unstable (so once the conjugate base is released, it quickly goes back to its acid). Ultimately, this translates to: weak acids give strong conjugate bases.

We can determine whether an acid is strong or weak through its Ka (acid dissociation constant) value. I believe that this has been covered in general chemistry, so I will not get into too much detail.

In solution, an acid can dissociate to release proton and a conjugate base in equilibrium. The K for this dissociation reaction is the Ka, and its value is different for every acid.

The value of Ka tend to be very large (10x) for strong acids and very small (10–x) for weak acids. Furthermore, there are a lot more weak acids compared to strong acids. Owing to all of these reasons, chemists like to simplify Ka by using its negative log, known as pKa.

$\text{p}K_{\text{a}} = - \log{K_{\text{a}}}$

Important note: because pKa is a negative log function, a large Ka translates to a small pKa and vice versa. And yes, pKas can have a negative value.

Let’s have a look at a famous weak acid: acetic acid (HOAc).

In water, HOAc dissociates to release proton (H+ which subsequently protonates water and becomes H3O+) and acetate ion (its conjugate base, AcO in equilibrium. The Ka value for acetic acid is known to be 1.8 × 10–5 (pKa HOAc= 4.74). This means that there are only a very small number of H+ and AcO ions at any given time in the equilibrium, whilst the majority of them exist as HOAc.

Now let’s compare it to a strong acid: HCl.

In water, HCl releases proton (H+, which also protonates water and becomes H3O+) and chloride ion (its conjugate base). It releases proton so easily that we can essentially say that HCl completely ionises. However, it does have a Ka value, which is about 107 (pKa HCl = –7), so we can look at the ionisation reaction through the perspective of an equilibrium reaction. Such a large Ka value (small pKa value) means that there’s a large number of H+ and Cl in the solution at any given time in the equilibrium, and only a very small number exists as HCl.

To summarise:

• Strong acids give weak conjugate bases
• Weak acids give strong conjugate bases
• The smaller the pKa value of an acid is, the stronger the acid is, the weaker its conjugate base is

We can come to another conclusion after we have a look at the next section: basicity.

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### II.2. Basicity (Kb and pKb)

Brønsted-Lowry theory states that a base is a proton (H+) acceptor. This implies that the strength of a base is determined by how readily it accepts H+.

Strong bases accept H+ more readily, whilst weaker bases, not so much. This implies that strong bases also release their conjugate acids readily. This means that the conjugate acids of strong bases are stable (that’s why it can be released readily). Ultimately, this translates to: strong bases give weak conjugate acids.

Weak bases find it hard to accept H+. Once a weak base accepts H+ and a conjugate acid, they quickly get back together to form the base again. This means that the conjugate acids of weak bases are unstable (so once the conjugate acid is released, it quickly goes back to its base). Ultimately, this translates to: weak bases give strong conjugate acids.

Similar to the acids, we can determine the strength of a base through its Kb (base dissociation constant) value. In solution, a base can accept a proton and becomes its conjugate acid in an equilibrium. The K for this dissociation reaction is the Kb, and its value is different for every base.

Also similar to the acids, the value of Kb tend to be very large (10x) for strong bases and very small (10–x) for weak bases. Furthermore, there are also a lot more weak bases compared to strong bases. Therefore, pKb can be used to simplify this.

$\text{p}K_{\text{b}} = - \log{K_{\text{b}}}$

Important note: because pKb is a negative log function, a large Kb translates to a small pKb and vice versa. And yes, pKbs can have a negative value.

Let’s have a look at a famous weak base: ammonia (NH3).

In water, ammonia readily accepts proton (H+) from H2O to form NH4+ and OH in equilibrium. The K for this dissociation reaction is known as Kb, and its value for ammonia is known to be 1.8 × 10–5 (pKb NH3 = 4.74). This means that there are only a very small number of NH4+ and OH ions at any given time in the equilibrium, whilst the majority of them exist as NH3 and H2O.

Note: the fact that the Kb value of NH3 is the same as the Ka value of HOAc is just a coincidence.

Now let’s compare it to a strong acid: NaNH2 (a much stronger base than NaOH).

In water, NaNH2 completely dissociates to form Na+ and NH2 ions. The NH2 ion accepts proton (H+) from H2O to form NH3 and OH in equilibrium. However, it does have a Kb value, which is about 1024 (pKb NH2 = –24). So we can look at the ionisation reaction through the perspective of an equilibrium reaction. Such a large Kb value means that there’s a large number of NH3 and OH in the solution at any given time in the equilibrium, and only an extremely small number exists as NH2.

To summarise:

• Strong bases give weak conjugate acids
• Weak bases give strong conjugate acids
• The smaller the pKb value of a base is, the stronger the base is, the weaker its conjugate acid is

Now, we can come up with another conclusion if we combine everything we’ve got so far at the next section.

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### II.3. Seeing a Base Through Its Conjugate Acid: KaH and pKaH

Organic chemists like myself don’t really use Kb or pKb that often. We like to see the strength of a base from the strength of its conjugate acid. Why? You shall see.

First, have a look at the reactions below.

I’m sure you’ll see the similarity between the two reactions right away. Both H3O+ and NH4+ lose a proton in the first step (step i), giving H2O and NH3, respectively. Both of them subsequently lose another proton in the second step (step ii), giving OH and NH2.

In the first step, both H3O+ and NH4+ are acids whilst H2O and NH3 are their conjugate bases, respectively. In the second step however, H2O and NH3 are acids and OH and NH2 are their conjugate bases, respectively.

Ok, because of their similarity, let’s just use H3O+, H2O, and OH as our primary example. I included the NH4+ series just as an example so you can see there are other species capable of doing the same thing.

Let’s consider each step (i and ii) in the H3O+ series separately.

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## II.3.1. Step i

First, let’s consider both the forward and the reverse reaction in each step separately.

In the forward reaction, H3O+ is an acid and H2O is its conjugate base. This reaction is the acid dissociation reaction of H3O+, and therefore its K (Kforward) is the Ka of H3O+, which has a value of 101.7.

In the reverse reaction, H2O is a base and H3O+ is its conjugate acid. The K of this reaction (Kreverse) is the reciprocal of Kforward.

Remember the title of this section? Seeing a base through its conjugate acid. So let me introduce you to KaH:

The KaH of a base is actually the Ka of its conjugate acid

We know that the Ka of the conjugate acid H3O+ is 101.7. If we want to talk about the base (H2O), we can say that the KaH of the base H2O is 101.7.

Again, this means that:

Ka of a conjugate acid = KaH of its base

So, the equilibrium constant of the forward and reverse reactions can be rewritten as follows:

We only say KaH when we’re talking about a base using the Ka of its conjugate acid. Small KaH value means weak conjugate acid, which means strong base. Conversely, large KaH value means strong conjugate acid, which means weak base.

For the same reason that has been explained before, pKaH is more useful than KaH itself due to the magnitude of KaH.

$\text{p}K_{\text{aH}} = - \log{K_{\text{aH}}}$

Therefore:

pKa of a conjugate acid = pKaH of its base

$\text{p}K_{\text{a}}(\text{H}_{3}\text{O}^{+}) = \text{p}K_{\text{aH}}(\text{H}_{2}\text{O}) = - \log{(10^{1.7})} = -1.7$

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## II.3.2. Step ii

We’ve basically covered all the basics about KaH and pKaH. This section is to give you another example for you to ponder about.

Let’s consider both the forward and the reverse reaction in each step separately.

In the forward reaction, H2O is an acid and OH is its conjugate base. This reaction is the acid dissociation reaction of H2O, and therefore its K (Kforward) is the Ka of H2O, which has a value of 10–15.7.

In the reverse reaction, OH is a base and H2O is its conjugate acid. The K of this reaction (Kreverse) is the reciprocal of Kforward.

Remember:

Ka of a conjugate acid = KaH of its base

Therefore, the equilibrium constant of the forward and reverse reactions can be rewritten as follows:

Since pKaH is more useful, then:

$\text{p}K_{\text{a}}(\text{H}_{2}\text{O}) = \text{p}K_{\text{aH}}(\text{OH}^{-}) = - \log{(10^{-15.7})} = 15.7$

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## II.3.3. Summary and Other Details

Now, you should already know that in the NH4+ series:

• In step i, pKa NH4+ = pKaH NH3
• In step ii, pKa NH3 = pKaH NH2

Organic chemists like to use pKa and pKaH interchangably, which often leads to confusion, especially to students who are not yet familiar with this. For example, at one time they may say the pKa of NH3 is 38, which is correct. However, they may also, at other times and depending on the circumstances, say that the pKa of NH3 is 9.2, which is actually the pKa of NH4+. I believe that to avoid confusion, they should’ve said that pKaH of NH3 is 9.2.

Therefore, in my opinion, I think the usage of pKaH should be popularised.

That being said, pKaH is still not a common term, so if you write it in an exam paper, your examiner may not accept it. My suggestion is to get used to using pKaH and only use it in your calculation. But when you actually write it in your exam paper, write ‘pKa of the conjugate acid’ instead.

You probably ask ‘why should we see a base through its conjugate acid and even go as far as to introduce pKaH?’. We’ll see the application and importance of using pKaH in Acid-Base V and Acid-Base VIII. For now, make sure you understand what pKaH is.

One last thing, I’d like to show you that this is also applicable to organic compounds. Have a look at the protonated methanol (CH3OH2+) and protonated diisopropylamine (a.k.a. diisopropyl ammonium or protonated DIPA, C6H14NH2+) series.

To conclude:

• Small pKaH means strong conjugate acid, which means the base is a weak base
• Large pKaH means weak conjugate acid, which means the base is a strong base

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## III. Conclusion

To conclude this post, let’s put everything we have got so far here:

• Strong acids give weak conjugate bases
• Weak acids give strong conjugate bases
• Strong bases give weak conjugate acids
• Weak bases give strong conjugate acids
• The smaller the pKa value of an acid is, the stronger the acid is, the weaker its conjugate base is
• Small pKaH means strong conjugate acid, which means the base is a weak base
• Large pKaH means weak conjugate acid, which means the base is a strong base

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## IV. Try It Yourself!

Let’s try to assign pKa and pKaH yourself.

Use the protonated methanol and the protonated DIPA series and assign the pKa and pKaH relationships. I’ll put the chemical equations again here:

Highlight text for answer:

1. In the protonated methanol series:

• In step i: pKa CH3OH2+ = pKaH CH3OH
• In step ii: pKa CH3OH = pKaH CH3O

2. In the protonated DIPA series:

• In step i: pKa C6H14NH2+ = pKaH C6H14NH
• In step ii: pKa C6H14NH = pKaH C6H14N

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Page last updated: 09ii17
RP

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