Acid Base Reactions
Acid-Base Reactions: Introducing Ka and pKa
Last updated: December 29th, 2022 |
The Acidity Constant Ka (And Its Negative Logarithm pKa ) Is An Extremely Useful Measure Of Acidity
So last time we went through all the different trends that affect acidity. The bottom line is that any factor which stabilizes the conjugate base will result in increased acidity.
Now, it’s great to know trends. It’s extremely important, in fact. (See article: Organic Chemistry Study Tips – Learn the Trends)
But what do you do when you want to compare the acidity of two molecules that aren’t connected by a trend?
For instance, resonance increases the acidity of alcohols, but by how much? More than you’d increase acidity by changing OH to SH? What if you added an electron withdrawing group?
It’s impossible to sort out all the opposing variables this way. Trends are only qualitative guides to this topic.
In order to answer this question we’ll need some hard data, and there’s only one way to get it in organic chemistry – through measurement.
And that’s what the acidity constant Ka (and its negative logarithm, pKa) are for.
Table of Contents
- The Acidity Constant Ka Represents The Equilibrium Constant For Dissociation Of An Acid Into Its Conjugate Base And A Proton
- Examples Of Acid Dissociation Constants Ka For A Strong Acid (HI), A Weak Acid (CH3OH), And An Extremely Weak Acid (CH4)
- The Negative Logarithm Of Ka, “pKa” Is A Far More Convenient Measure Of Acidity Than Ka
- The pKa Scale Encompasses Over Sixty Orders Of Magnitude
- Notes
1. The Acidity Constant Ka Represents The Equilibrium Constant For Dissociation Of An Acid Into Its Conjugate Base And A Proton
Let’s look at hydroiodic acid, H–I. In solution (let’s use water) H–I will protonate water to give H3O(+) and I(-). The reverse reaction also operates. [I (–) reacts with H3O(+) to give H-I back]. So this reaction is an equilibrium.
Here’s the useful part. We can measure the equilibrium constant of this reaction, and that will tell us about the relative acidity of H-I. The specifics of how we can measure this are another story, but for our purposes, let me assure you that these types of equilibrium constants can be measured accurately.
For H-I, it turns out that the equilibrium constant for this reaction is about 10,000,000,000. That means for every molecule of HI, there are 10 billion molecules of I(-). In other words, H-I donates its proton very readily to give its conjugate base, which is a prime example of a strong acid.
So what? Well, we can do this not just for H-I, but for ANY species with a hydrogen. This equilibrium is the “ruler” by which we can measure any acid – by its equilibrium constant. [Note 1]
We call this equilibrium constant Ka – the “acidity constant”.
2. Examples Of Acid Dissociation Constants Ka For A Strong Acid (HI), A Weak Acid (CH3OH), And An Extremely Weak Acid (CH4)
Let’s look at 3 examples. Note that these are somewhat oversimplified in that the solvent has been left out.*
Now comparing acidity between numbers with lots of exponents after them is not the most convenient way to do things. So instead, we’ve taken to using a logarithmic scale. These are common – the Richter scale is logarithmic, for instance – an 8.0 magnitude quake is 10 times more powerful than a 7.0 magnitude quake.
3. The Negative Logarithm Of Ka, “pKa” Is A Far More Convenient Measure Of Acidity Than Ka
For acidity, the number we use is called “pKa“. It’s obtained by taking the logarithm of the acidity constant Ka, and we arbitrarily decide to multiply it by negative one [the vast majority of Ka values are less than 1, and we’re generally more comfortable dealing with positive numbers].
So for H-I, the pKa is –10 (representing a very strong acid), that for methanol is 15 (a weak acid), and for methane, it’s 50 (an extremely weak acid).
4. The pKa Scale Encompasses Over Sixty Orders Of Magnitude
That scale comprises 60 orders of magnitude. That’s a huge number!!!!!
To give you an idea of the scale, of pKa, this is the range of the smallest value for length (the Planck length, 10–35 meters), to the width of the known universe (about 1026 meters). (See article: pKa Values Span 60 Orders of Magnitude)
Anyhow, these measurements have been done for thousands of different molecules now. The result is a big table that allows us to compare the acidity of all kinds of different functional groups. Here’s an example of a pKa table from a previous post.
The pKa table is your friend. In one document, it gives you information on the scope and magnitude of a wide range of chemical behavior – the strongest of the strong acids, and the weakest of the weak acids. And since the stronger the acid, the weaker the conjugate base, it also provides information about basicity.
Next Post: How to Use A pKa Table
Notes
Note 1 The choice of solvent for these measurements is important due to the “leveling effect”. For instance Water is commonly used as a solvent for molecules with a pKa value less than zero and no greater than about 14, since there can be no stronger acid in water than H3O(+) and no stronger base than HO(-).
To get some of the more extreme values other solvents must be used, such as dimethyl sulfoxide (DMSO). If you end up in a more advanced organic chemistry course, look at some of these values with a harder eye through that lens. Strongly recommended – Evans’ pKa table gives values in H2O and DMSO. And there is the heroic efforts of Hans Reich at Wisconsin in compiling pKa values.
Good post! Looks like you forgot to put in the link when you say ‘This website has a phenomenal animation of this’.
There’s something I don’t quite grasp in the HI reaction to give I-.
If the Ka value is 10^10, then doesn’t that mean that for every molecule of HI, the product of concentrations [I-] and [H3O+], instead of just [I-]? So shouldn’t there be the square root of 10^10 molecules of I-?
The link for the Evan’s pka table no longer exists. I searched a little and I think this is the link you want now: http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf
Yes, thank you!
Do the same principles apply to pkb as do pka? Meaning, does a lower pkb mean that the base is more reactive than a base with a higher pkb? Is there a pkb table as well?
I suggest not dealing with pKb. Instead I suggest thinking about the pKaH, the “pKa of the conjugate acid”. For example the pKaH of the ethoxide ion CH3CH2O(-) is 16.
in many websites and textbooks i found that pka of H2 is less that pka of NH3 or one can say amine comes before hydrogen, so which one is correct?
correction *less than
They are around the same.
I have a doubt:
when a polar protic solvent is used as a solvent for its conjugate base, will the conjugate base be the strongest base/nucleophile in that solvent?
Thanks!
Yes, this is known as the “leveling effect”. The strongest base possible in a solvent is the conjugate base of that solvent. So the strongest base that can exist in H2O is HO- (the conjugate base of water) since if one was to add a stronger base (e.g. NaNH2) the base would immediately deprotonate the solvent, generating its conjugate acid (NH3) and the conjugate base of the solvent (HO- )
Ah, Ohkay!
Thank you so much!
Sir ,I think tosic acid have -2.3 in water right ?it should not have higher than hydronium ion right?
That’s about the right number.
In the reactions of strong, weak, very weak acids. I think the direction of arrows for backward reactions are wrong.
Thanks for this website, I have exam in 2 months for PCM, and this is a lifesaver .
You are correct, fixed!
Hi! I was wondering about the equilibrium arrows in bullet point 2, Examples of Acid Dissociation Constants Ka For a Strong Acid (HI), A weak acid (CH3OH) and An Extremely Weak Acid (CH4). In the example, for weak acid and extremely weak acid, the arrows show the equilibrium lying to the right instead of the left. I am wondering if this is a mistake, or if I am missing something as to why equilibrium would shift to the carbon anion over the alkane? Thanks so much!
Great catch. Thank you. I fixed the image and uploaded a replacement. Thanks!!!
I just have an open ended question on something that confuses me. Why do esters have a lower acidity than ketones? Should they not have two effects working with them, the resonance of the oxygen and induction effects from another oxygen?
My guess would be that induction effects possibly only really matter with halogens, since they are highly electronegative. Or that the electron pair in the pi bond is already delocalized in resonance with the other single bonded oxygen, meaning the effects of resonance are less in the ester compared to the ketone. Or both? I would appreciate anyone with more knowledge than me helping me out here, every day it feels like I find something that goes against everything I thought I knew 🙏
Excellent question. The short version is, “experiments tell us resonance is more important for O and N”.
Acidity is all about the stability of the conjugate base, which for esters and ketones results in a species known as an “enolate”.
The more electron-poor the C=O bond, the better lone pair on carbon (i.e. the negative charge of the conjugate base) is able to donate into it, forming a new pi-bond and resulting in a (more stable) negative charge on oxygen.
Oxygen and nitrogen substituents have two effects which operate in opposite directions.
On one hand, these atoms are electronegative, which means that they will withdraw electron density through the single bond. On the other hand, they are capable of *donating* electron density through forming pi-bonds with adjacent electron acceptors (e.g. C=O) via resonance, an effect known as pi donation.
So what effect wins? It’s an easy question to ask, but hard to predict just from first principles alone. What we observe from experiments is that for oxygen and nitrogen, the pi-donation effect outweighs the inductive effect.
This means that the C=O is *less* electron poor in esters and (especially) amides because of donation from the adjacent oxygen and nitrogen. This means that the enolates of these species are less stabilized, which is reflected in their weaker acidity.
A similar effect is responsible for why O and N substituents are “activating” in electrophilic aromatic substitution reactions – the pi-donation of O and N outweighs the electron withdrawing effect.
(with chlorine, the electron-withdrawing effect outweighs the pi-donation effect).
I’d be happy to elaborate – text is a tough medium through which to explain these things :-)