## Contents

I. Introduction

II. Autoprotolysis of Water

III. Autoprotolysis of Solvents

IV. The Relationship between Ka (KaH), Kb, and Kw (Ksolv)

V. The Relationship between Ka (KaH), Kb, and Kw (Ksolv) with Acid-Base Strengh

VI. Conclusion

## I. Introduction

We have previously mentioned in Acid-Base IV that solvent plays a role in determining the acidity of a species. Certain species may be a strong acid in one solvent but becomes a weak acid in another. We’ve said before that this depends on how good the solvent solvates the species, but there is more to it than just solvation capability.

Just like reactants and products, solvents are also chemicals and they can react with other chemicals. Water, for example, can be protonated by an acid (e.g. HCl) or deprotonated by a base (e.g. NH2). We have talked about this exhaustively in Acid-Base I and Acid-Base II.

What this actually means is that water reacts with both HCl and NH2 to form H3O+ and OH respectively. Therefore, it is implied that water limits the strength of certain acids and bases by reacting with them. This is called the Levelling Effect. We shall have a look at what this is in Acid-Base X.

A lot of students find it hard to wrap their head around the solvent levelling effect. I did too. That is why in this post and in the next post, I would like to talk thoroughly about the fundamentals that will be necessary for you to fully understand this important phenomenon.

In an organic chemistry lab, you would rarely use water as your solvent as most organic compounds aren’t soluble in it. There are lots of other choices for solvent besides water, known as the non-aqueous solvents. Some commonly used solvents in the lab are dimethylsulfoxide (DMSO), dimethylformamide (DMF), methanol (MeOH), ethanol (EtOH), tetrahydrofuran (THF), acetic acid (HOAc), ethyl acetate (EtOAc), acetone, toluene, chloroform, dichloromethane (DCM), and hexane.

There are also other non-aqueous solvents that are not commonly used as solvent, but have been used as one when necessary, such as pyridine, triethylamine (TEA), and ammonia (NH3). Of course, these lists are not exhaustive, as there are still plenty other compounds that can be used as solvents.

In this post we shall talk about several fundamental concepts related to the acid-base chemistry of solvents.

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## II. Autoprotolysis of Water

Water may react with itself in a reaction we know as the dissociation of water. A more specific term for this reaction is autoprotolysis, where a proton is taken from one molecule of water and transferred to another molecule of water. Some also refer to this process as autoionisation. They’re all basically saying the same thing.

I believe that everyone is familiar with the equilibrium constant of this reaction, we often call it Kw, short for Kwater. Its value at 25 oC is 10–14.

When an acid (HA) is dissolved in water, the acid protonates the water. This reaction has an equilibrium constant known as Ka of HA(aq). This Ka is what we have always been talking about and listed in the pKa table, and is also what students are most familiar with.

And when a base (A) is dissolved in water, the base deprotonates the water. The equilibrium constant for this reaction is known as Kb of A(aq).

We can combine the Ka of HA(aq) and Kb of A(aq) equations to obtain Kw.

Hence, we can write the formula of Kw as follows:

In other words:

$K_{\text{w}} = K_{\text{a}}\text{ (HA}_{\text{(aq)}}\text{)} \times K_{\text{b}}\text{ (A}^{-}_{\text{(aq)}}\text{)}$
$K_{\text{w}} = K_{\text{aH}}\text{ (A}^{-}_{\text{(aq)}}\text{)} \times K_{\text{b}}\text{ (A}^{-}_{\text{(aq)}}\text{)}$
$\text{p}K_{\text{w}} = \text{p}K_{\text{aH}}\text{ (A}^{-}_{\text{(aq)}}\text{)} + \text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(aq)}}\text{)}$

You get it? This shouldn’t be hard to follow, after all, this is the basic of acid-base chemistry.

Note: Almost all textbooks will simply write the second formula above as:

$K_{\text{w}} = K_{\text{a}} \times K_{\text{b}}$

Which has confused many students from time to time. It certainly confused me the first time I learnt about this! Why? Let me show you. Let’s use NH3 for this example.

Students who see that formula will think that Ka (NH3) multiplied by Kb (NH3) equals to Kw. And it is wrong! Prove it yourself as a practice. Use the method I’ve described above.

The correct expression is KaH (NH3) multiplied by Kb (NH3) equals to Kw. Now this is right. Again, prove it yourself as a practice. Use the method I’ve described above.

This is why I emphasised the importance of using KaH instead of just using Ka for everything. So for those of you reading this, please be very careful with this.

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## III. Autoprotolysis of Solvents

Now, let’s use what we already have in the previous section here.

As I mentioned in the Introduction, there are other choices of solvents that we can use in the lab. Water is not always applicable because there are a large number of compounds that can’t dissolve in water. Here we’ll just use HSolv to represent those solvents in general.

Please pay attention because I also add some important notes in between each explanation that I did not put in water. I put it here because this section talks about solvents in general.

Just like water, the autoprotolysis of HSolv can be depicted in the following chemical equation. Let’s call the equilibrium constant of this reaction Ksolv.

Note: the subscript ‘(solv)’ in the chemical equation indicates that a species is dissolved in solvent HSolv. Recall that we usually use the subscript ‘(aq)’ to indicate that a species is dissolved in water.

An acid (HA) dissolved in HSolv will protonate HSolv. Of course, this reaction has an equilibrium constant that we also shall call Ka of HA(solv).

Before we go to the base dissolved in HSolv, let’s talk a bit about this formula.

We have talked about strong and weak acids/bases many times through the series of Acid-Base posts. We defined the strength of acids/bases through their ability to accept or donate proton. But what about a more mathematical definition? I promise you it’s just super basic math.

Look at the Ka (HA(solv)) formula above. An acid dissolved in HSolv is defined as a strong acid if the value of Ka (HA(solv)) > 1 or pKa (HA(solv)) < 0. This means that there is more H2Solv+ than there is HA. This in turn implies that HA has given most of its proton to HSolv so that there’s only a few of HA left in the equilibrium while a large number of H2Solv+ has formed. According to the formula above, small number of HA and large number of H2Solv+ give large Ka (HA(solv)). All of this is in line with our definition of a strong acid!

What about weak acid? Using the logic above, we can say that an acid dissolved in HSolv is defined as a weak acid if the value of Ka (HA(solv)) < 1 or pKa (HA(solv)) > 0. Try to elaborate this statement and prove that it is in line with our definition of a weak acid!

A base (A) dissolved in HSolv will deprotonate HSolv. This reaction has an equilibrium constant that we shall call Kb of A(solv).

Now, before we go ahead and combine both formulas, let’s talk a bit about this formula first. Let’s do the same thing we’ve done before: define the strength of a base using the formula above.

Look at the Kb (A(solv)) formula above. A base dissolved in HSolv is defined as a strong base if the value of Kb (A(solv)) > 1 or pKb (A(solv)) < 0. This means that there is more Solv than there is A. This in turn implies that A has taken most of HSolv’s proton so that there’s only a few of A left in the equilibrium while a large number of Solv has formed. According to the formula above, small number of A and large number of Solv give large Kb (A(solv)). All of this is in line with our definition of a strong acid!

What about weak base? Using the logic above, we can say that a base dissolved in HSolv is defined as a weak base if the value of Kb (A(solv)) < 1 or pKb (A(solv)) > 0. Try to elaborate this statement and prove that it is in line with our definition of a weak base!

So it’s time to continue now!

Similar with the reactions in water, we can combine the Ka of HA(solv) and Kb of A(solv) equations to obtain Ksolv.

Hence, we can write the formula of Ksolv as follows:

In other words, and these are the formulas we will be working with in the next section and the next posts:

When we’re talking about an acid and its conjugate base:

$K_{\text{solv}} = K_{\text{a}}\text{ (HA}_{\text{(solv)}}\text{)} \times K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$
$\text{p}K_{\text{solv}} = \text{p}K_{\text{a}}\text{ (HA}_{\text{(solv)}}\text{)} + \text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$

When we’re talking about a base and its conjugate acid:

$K_{\text{solv}} = K_{\text{aH}}\text{ (A}^{-}_{\text{(solv)}}\text{)} \times K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$
$\text{p}K_{\text{solv}} = \text{p}K_{\text{aH}}\text{ (A}^{-}_{\text{(solv)}}\text{)} + \text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$

I hope you can see just how similar it is, whether we use water as a solvent or other type of solvents.

There is another thing that I need to say about the final formulas, but we will do this in the next section.

Several examples of common solvents (ammonia, methanol, DMSO, and acetic acid) and their autoprotolysis reactions:

Unfortunately, I could only find the pKsolv value of NH3 and DMSO. Combined with water’s pKw, here’s a short list of several pKsolv:

 Solvent Name Molecular Formula pKsolv Water H2O 14 Ammonia NH3 33 Dimethylsulfoxide (CH3)2SO 37

We shall see the application of pKsolv in the next post.

One last comment I would like to say is that it doesn’t always have to be proton that’s transferred. Here’s the autoionisation reaction of bromine trifluoride as a solvent:

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## IV. The Relationship between Ka (KaH), Kb, and Kw (Ksolv)

In this section we’ll be talking mainly about the formulas that we came up with earlier:

$K_{\text{solv}} = K_{\text{a}}\text{ (HA}_{\text{(solv)}}\text{)} \times K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$
$\text{p}K_{\text{solv}} = \text{p}K_{\text{a}}\text{ (HA}_{\text{(solv)}}\text{)} + \text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$

$K_{\text{solv}} = K_{\text{aH}}\text{ (A}^{-}_{\text{(solv)}}\text{)} \times K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$
$\text{p}K_{\text{solv}} = \text{p}K_{\text{aH}}\text{ (A}^{-}_{\text{(solv)}}\text{)} + \text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$

I believe everyone is familiar with this so at first I thought I didn’t need to talk about this. But then I feel that this is very important to understand the next section and the next post, so I’m still going to talk about it anyway.

Those formulas above enable one to obtain the Kb of a base if one knows the Ka of its conjugate acid (KaH) and Ksolv of the solvent. Or in other words, if one knows the Ka of a particular acid, then the formulas above will enable one to obtain the Kb of its conjugate base, providing one knows what the Ksolv value of the solvent is:

All of these formulas also show that Kb and KaH are inversely proportional to each other. This is in line with the conclusion we had at the end of Acid-Base II:

• Small Kb (large pKb) means large KaH (small pKaH). This means that a weak base has a strong conjugate acid or a strong acid has a weak conjugate base.
• Large Kb (small pKb) means small KaH (large pKaH). This means that a strong base has a weak conjugate acid or a weak acid has a strong conjugate base.

This serves as a mathematical proof of what we have previously talked about in Acid-Base II.

Of course, we can also derive the pK formula too.

The pK formula would be more useful since most of the time we talk about pK rather than K itself.

I don’t always trust formulas right away so let’s put it to a test. We know from the pKa table that the pKaH of NH3 (pKa of NH4+) in water is about 9.2. Using the formula above, we can obtain the approximate value of pKb of NH3 in water. The value of pKsolv for water is pKw = 14.

$\text{p}K_{\text{b}}\text{ (NH}_{3\text{(aq)}}\text{)} = \text{p}K_{\text{w}} - \text{p}K_{\text{aH}}\text{ (NH}_{3\text{(aq)}}\text{)}$
$\text{p}K_{\text{b}}\text{ (NH}_{3\text{(aq)}}\text{)} = 14 - 9.2 = 4.8$

The pKb of NH3 in water according to that calculation is approximately 4.8.

I have mentioned in Acid-Base II that the pKb value of NH3 in water is 4.74. You can see how the calculation gives you a value good enough and close enough to the experimental value!

We’ll be using this method to obtain several pK values in the next post: Acid-Base IX.

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## V. The Relationship between Ka (KaH), Kb, and Kw (Ksolv) with Acid-Base Strength

Ok, now that we’ve agreed that those formulas are correct, let’s look at the imporant consequences that they have. Recall the logic we talked about in Section III for this discussion.

Acid Strength: Strong Acid

An acid dissolved in HSolv is defined as a strong acid if:

• Ka (HA(solv)) > 1, in other words
• pKa (HA(solv)) < 0

According to this formula:

$\text{p}K_{\text{solv}} = \text{p}K_{\text{a}}\text{ (HA}_{\text{(solv)}}\text{)} + \text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$
$\text{p}K_{\text{a}}\text{ (HA}_{\text{(solv)}}\text{)} = \text{p}K_{\text{solv}} - \text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$

pKa (HA(solv)) < 0, therefore:

$\text{p}K_{\text{a}}\text{ (HA}_{\text{(solv)}}\text{)} < 0$
$\text{p}K_{\text{solv}} - \text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)} < 0$
$\text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)} > \text{p}K_{\text{solv}}$

I am not sure if we’ll ever find this expression useful, I just want to show you this for thoroughness.

Acid Strength: Weak Acid

An acid dissolved in HSolv is defined as a weak acid if:

• Ka (HA(solv)) < 1, in other words
• pKa (HA(solv)) > 0

According to this formula:

$\text{p}K_{\text{solv}} = \text{p}K_{\text{a}}\text{ (HA}_{\text{(solv)}}\text{)} + \text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$
$\text{p}K_{\text{a}}\text{ (HA}_{\text{(solv)}}\text{)} = \text{p}K_{\text{solv}} - \text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$

pKa (HA(solv)) > 0, therefore:

$\text{p}K_{\text{a}}\text{ (HA}_{\text{(solv)}}\text{)} > 0$
$\text{p}K_{\text{solv}} - \text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)} < 0$
$\text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)} < \text{p}K_{\text{solv}}$

Again, I am not sure if we’ll ever find this expression useful, I just want to show you this for thoroughness.

Although we may not find those expressions useful for acids, we will find it useful for base, mainly because most of the time we look at a base through its conjugate acid.

Base Strength: Strong Base

An acid dissolved in HSolv is defined as a strong base if:

• Kb (A(solv)) > 1, in other words
• pKb (A(solv)) < 0

According to this formula:

$\text{p}K_{\text{solv}} = \text{p}K_{\text{aH}}\text{ (A}^{-}_{\text{(solv)}}\text{)} + \text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$
$\text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)} = \text{p}K_{\text{solv}} - \text{p}K_{\text{aH}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$

pKb (A(solv)) < 0, therefore:

$\text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)} < 0$
$\text{p}K_{\text{solv}} - \text{p}K_{\text{aH}}\text{ (A}^{-}_{\text{(solv)}}\text{)} < 0$
$\text{p}K_{\text{aH}}\text{ (A}^{-}_{\text{(solv)}}\text{)} > \text{p}K_{\text{solv}}$

Therefore, we can also say here that a base dissolved in HSolv is considered strong if its pKaH is greater than pKsolv.

Base Strength: Weak Base

An acid dissolved in HSolv is defined as a weak base if:

• Kb (A(solv)) < 1, in other words
• pKb (A(solv)) > 0

According to this formula:

$\text{p}K_{\text{solv}} = \text{p}K_{\text{aH}}\text{ (A}^{-}_{\text{(solv)}}\text{)} + \text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$
$\text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)} = \text{p}K_{\text{solv}} - \text{p}K_{\text{aH}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$

pKb (A(solv)) < 0, therefore:

$\text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)} > 0$
$\text{p}K_{\text{solv}} - \text{p}K_{\text{aH}}\text{ (A}^{-}_{\text{(solv)}}\text{)} < 0$
$\text{p}K_{\text{aH}}\text{ (A}^{-}_{\text{(solv)}}\text{)} < \text{p}K_{\text{solv}}$

Therefore, we can also say here that a base dissolved in HSolv is considered weak if its pKaH is less than pKsolv.

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## VI. Conclusion

In this post we have seen the autoprotolysis of solvents in general. By doing so, we have also defined what constitutes a strong acid and a strong base in any solvents.

We have also come up with several important formulas that will help us:

When we’re talking about an acid and its conjugate base:

$K_{\text{solv}} = K_{\text{a}}\text{ (HA}_{\text{(solv)}}\text{)} \times K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$
$\text{p}K_{\text{solv}} = \text{p}K_{\text{a}}\text{ (HA}_{\text{(solv)}}\text{)} + \text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$

When we’re talking about a base and its conjugate acid:

$K_{\text{solv}} = K_{\text{aH}}\text{ (A}^{-}_{\text{(solv)}}\text{)} \times K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$
$\text{p}K_{\text{solv}} = \text{p}K_{\text{aH}}\text{ (A}^{-}_{\text{(solv)}}\text{)} + \text{p}K_{\text{b}}\text{ (A}^{-}_{\text{(solv)}}\text{)}$

Through these formulas, we have concluded that Kb and KaH are inversely proportional to each other. We have also seen that they are in line with the conclusion we had in Acid-Base II:

• Small Kb (large pKb) means large KaH (small pKaH). This means that a weak base has a strong conjugate acid or a strong acid has a weak conjugate base.
• Large Kb (small pKb) means small KaH (large pKaH). This means that a strong base has a weak conjugate acid or a weak acid has a strong conjugate base.

Those formulas will enable one to obtain the pKb of a base, if one knows the pKa of its conjugate acid (pKaH) and the pKsolv of the solvent, and vice versa.

Combining everything above, we have now fully defined the acid-base strength in any solvents by using pK values as a measure of strength.

For acids:

• An acid dissolved in HSolv is defined as a strong acid if:
• pKa (HA(solv)) < 0
• pKb (A(solv)) > pKsolv
• An acid dissolved in HSolv is defined as a weak acid if:
• pKa (HA(solv)) > 0
• pKb (A(solv)) < pKsolv

For bases:

• A base dissolved in HSolv is defined as a strong base if:
• pKb (A(solv)) < 0
• pKaH (A(solv)) > pKsolv
• A base dissolved in HSolv is defined as a weak base if:
• pKb (A(solv)) > 0
• pKaH (A(solv)) < pKsolv

What we have concluded in this post will be crucial to understand the next post, so make sure that you do understand this.

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Page last updated: 10ii17
RP

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