TABLE OF CONTENTS:
The first question is: how do you define an acid? There are three definitions tested on the MCAT. Each successive one is more inclusive, like the situation for different kinds of furniture:
Venn diagram of various types of furniture.
Not every piece of furniture is a couch. For example, a table would fit only into the blue area. Similarly, a sectional or three-seater are types of couches that are not loveseats, fitting into the green area.
Venn diagram of the Acid/Base Definitions
Some example compounds are included in the most restrictive class to which they belong.
Note that water, an important amphoteric species, is both a Bronsted-Lowry acid and Base.
Note that the conjugate acid of many Arrhenius Acids are Bronsted-Lowry Bases
Also, note that every Arrhenius Acid is a Bronsted-Lowry Acid, and every Bronsted-Lowry Acid is a Lewis Acid.
Arrhenius was a 19th-century chemist who dropped solids into water, and observed whether they dissociated to create a proton (acid) or hydroxide ion (base) in solution. The images show how to understand the Arrhenius definition. Notice that Arrhenius acids are intuitive, creating a dissolved proton. Arrhenius bases are limited by their definitions to just being metal hydroxides [ NaOH, KOH, Ca(OH)2, Fe(OH)3, for example].
The Bronsted-Lowry definition switched the focus to the movement of protons. This is the definition most people understand intuitively from undergrad. Acids donate protons, and bases accept protons. The diagram above shows a general acid-base reaction. The “gas cloud” is to remind you that the Bronsted-Lowry definition has nothing to do with solutions. Gaseous HCl and NH3 would still transfer one proton between them. The colored ellipses show what would happen if you did dissolve the acid and base. The acid would still dissociate, losing its protonated state. The remnant molecule is called the conjugate base. Since our example acid was neutral, losing a proton left the conjugate base with a net charge of -1. Both acids and bases can be positive, neutral, or negative. The key here is that a charge of (+1) gets transferred because that’s the charge of the proton.
The Lewis Acid definition is concerned with the movement of electron pairs. The diagram above shows the reaction between a proton and ammonia (NH3) with this analysis. The lone pair on the Nitrogen in ammonia is given as a donation by the Nitrogen to fill the empty 1s orbital around the proton. Once the electron pair is in the σ (single) bond between the N and H, it will orbit around both nuclei. In this sense, it is a covalent bond. The term coordinate covalent bond refers to a bond in which both electrons originally came from one atom. This is synonymous with a dative bond. I like this name better. “Data” in Latin means “things having been given,” or “gifts.” The N is “gifting” a net charge of (-1) to the H. By formal charge analysis, since each nucleus gets the electron pair ½ of the time, the H gets ½ the total charge of -2, so the proton becomes neutral, and the Nitrogen becomes positive. The image below shows how the same process applies to molecules with incomplete octets, another class of molecules that have an empty orbital, and thus, room for a lone pair.
All Lewis Bases have Lone Pairs, but the number of lone pairs does not correlate with the strength of the species as a base.
Some species in a basic series can accept more than one proton, e.g.: OH- → H2O → H3O+
Some students get confused by all the details, so to summarize:
- Arrhenius cares about dissociating into protons and hydroxides in SOLUTION.
- Bronsted-Lowry is the “intuitive definition” about TRANSFERRING PROTONS.
- This should be your home base for discussing acid chemistry. When making statements about Lewis or Arrhenius Acids, the MCAT will usually be explicit. This document uses the same assumption. Always remember: Keep it Simple!
- Lewis cares about the TRANSFER OF LONE PAIRS of electrons
- Mnemonic: “LALA” = Lewis Acid is a Lone-pair Acceptor
- All Bronsted-Lowry acids and bases are Lewis ones, because of the process shown above: To grab a proton (and thus be called a Bronsted base), a molecule must donate a lone pair.
- In organic chemistry, the Electrophile (often written as E+) is a Lewis Acid, and the Nucleophile (Nu:) is a Lewis base, which is why we explicitly show the lone pair.
For a generic acid’s dissociation reaction: HA ↔ H++ A-,
HA = The undissociated acid. This is the acid in its protonated state.
H+ = a Hydrogen ion = a proton = a naked, single positive charge (naked because there is no electron cloud)
A- = The conjugate base. What is left after the proton uncouples (Conjugate means linked together). See the pictures in the Acid Definition section to see how this is related to the undissociated acid.
Ethanoic Acid = CH3COOH = Acetic Acid
Ethanoate = CH3COO- = Acetate
The above images show Acetic acid and its conjugate base, Acetate, in ball-and-stick and electron density map forms. When looking at electron density maps, red areas show where in the electron clouds the electrons spend most of their time, and thus where the negative charge density resides. As you can see, they are structurally related. Two important differences between the conjugate species:
- Notice that in Acetic Acid, the Hydrogen (the atom on the right) has an electron cloud that is blue. The Hydrogen nucleus (a proton) is already partially denuded, making it almost a fully naked positive charge. So, while the universe abhors a naked (unpaired) charge, requiring a lot of energy to separate a proton from the center of its electron cloud, this proton is already most of the way to that high-energy transition state.
- Notice that in the Acetate molecule, the Hydrogen is gone, and the little electron density that used to be around the proton is now distributed evenly across both Oxygens. See the section on Predicting Acid Strength for a discussion of the stability of this arrangement.
Functionally, what defines a strong acid is the fact that it dissociates completely in water. The classic example is Hydrochloric Acid:
HCl(aq) + H2O(l) → Cl-(aq) + H3O+(aq)
If we write the equilibrium constant (Keq) for this reaction,
- Notice that H2O is a liquid, so it doesn’t appear in the Keq.
- Also, note that we can use H3O+ and H+ interchangeably, because they represent the same quantity, the protons that dissociated. When we talk about H3O+ (Hydronium), we assume that the reaction is being written with water on both sides. By cancelling the water out of both sides, we can talk about protons instead of hydronium.
Important: For the dissociation reaction of an acid, the acid is the reactant, and the dissociated proton and conjugate base are the products. Keep this in mind, and it will help you realize that acid-base chemistry is just a special case of equilibrium chemistry. Since we always write the reaction in this direction, the Keq is a specialized one that we call Ka. Like all specialized Keq’s, this one is directly related to temperature (in other words, higher Temp → more dissociation into protons and conjugate acids). Thus, for a generic acid HA:
Going back to the acid dissociation reaction for HCl:
HCl(aq) + H2O(l) → Cl-(aq) + H3O+(aq)
Since HCl is a strong acid, Ka should technically be infinite. Let’s say we put 3 moles of HCl into 1 L of water. The final state of the reaction should one in which all the HCl has split apart.
Yet, in real life, the Ka (HCl) = 1.3 x 106. If x is the amount of acid that reacted,
Solving this requires quadratic math (not used on the MCAT), so we won’t. The key here is that the ratio of x2 to (3-x) is roughly 1 million to 1. The x2 in the numerator implies that the single reactant breaks into two products, so we can take the square root of 106 to get the amount of “reactant equivalent mass” that exists in the dissociated form at equilibrium.
When you put HCl in water, it stops splitting apart when there is roughly 1000 times as much in the dissociated ionic form as there is in molecular form. So, while we say that strong acids dissociate completely in water, the truth is that they really just dissociate more than 99% completely. When dealing with problems where strong acids like HCl are added to solution, assume that all the acid mass is accounted for as dissociated ions. (Smart estimation). When dealing with problems involving the Ka, remember the truth is that it reaches an equilibrium, which means that the reaction is truly reversible. However, since this equilibrium picture so heavily favors the products, we can still talk about HCl as if it dissociated 100%. To elucidate this further, let’s talk about a weak acid, Acetic Acid. The dissociation reaction is:
CH3COOH(aq) + H2O(l) ⇌ CH3COO-(aq) + H3O+(aq) Ka = 1.8 x 10-5
Notice the two-headed reaction arrow, indicating that this reaction is reversible. The Ka is much less than 1, implying that the mass of the system is found more in the undissociated (reactant) form at equilibrium. Using the same analysis, the ratio of [Acetic Acid : Acetate] is ~[400:1] when the system comes to rest. Clearly, putting an equal amount of Acetic and Hydrochloric Acid into two separate beakers of water is going to produce vastly different amounts of dissolved protons, and this means we can use Ka information to gauge acid strength:
- Ka >> 1 ⇒ Strong acid (dissociates completely)
- Ka << 1 ⇒ Weak acid (stays mostly undissociated)
Notice that a Ka value of 1 is the dividing line here, as with most equilibria. The MCAT will not ask you to judge the strength of an acid with a Ka roughly around 1, so don’t worry about it.
For extra help visualizing acid strength, follow the external links to some Flash animations:
So far, we have been talking about acids, but what about bases? Well, there is a generic base dissociation reaction, too. It describes placing the base into H2O. The base makes water act like an acid, so H2O dissociates into its conjugate base (OH-) and a proton (H+). The proton is absorbed by the base. If you read the following equation from right to left, the conjugate acid of the base, HB+, is dissociating as any acid would. The charge state, and thus how you write the Kb, depends on whether your base is charged or neutral, like OH- and NH3, respectively.
B(aq) + H2O(l) → HB+(aq) + OH-(aq)
B-(aq)+H2O(l) → HB(aq) + OH-(aq)
Let’s say that we want to know the value of the Kb for a base. If we know the Ka of the conjugate acid, can we figure this out? Yes! The answer is based on some reaction addition:
CH3COOH(aq) + H2O(l) → CH3COO-(aq) + H3O+(aq)
Ka = 1.8 x 10-5
Kb = ?
2 H2O(l) → H3O+(aq) + OH-(aq)
Remember, if you have a two-step reaction pathway where A → B and B → C, the overall reaction is A → C. The Keq’s would be as follows:
You can figure out Koverall by multiplying K1 and K2,
which is consistent with the Keq you would predict if you only knew the overall reaction. Applying this concept to the net reaction equation for Acetic Acid and Acetate dissociation,
Again, the 2 H2O molecules are not included in the Koverall because they are liquid. Notice that the actual acid and the conjugate base are not represented in the net reaction equation. In fact, the sum of the dissociation reactions of ANY acid and its conjugate base will always be
2 H2O(l) → H3O+(aq) + OH-(aq)
This equation describes a reaction in which two water molecules bang into one another, transferring a proton in the process, creating a hydronium and hydroxide. This actually happens all the time in liquid water. This process is called the autoionization of water, because the neutral liquid water is turning into ions. Since we always write this equation with water on the left, it is another of those special Keq’s. This one is called Kw.
Like all the named Keq’s, this one is directly related to temperature. Which makes sense:
↑ Temp → ↑ # of collisions → ↑ autoionization. At 25∘ C, Kw = 10-14.
Thus, for the sum of the acid and conjugate base dissociation reactions:
which answers the question with which this section began.
Important point: The inverse mathematical relationship between Ka and Kb defines mathematically what is discussed in the section on Predicting Acid Strength: A strong acid has a weak conjugate base. An acid with a Ka≈ 10-7 is weak, but its conjugate base would have a Kb that is also roughly 10-7, so that is weak too. Using a Ka of 1 as the dividing line between strength and weakness, we can see that a strong acid’s conjugate base must have a Kb ≤ 10-14, which is so weak that it is functionally inert. Thus, only a very weak base (Kb ≤ 10-14) has a strong conjugate acid.
Table of Acid Strengths (for reference, if you want to review): http://depts.washington.edu/eooptic/links/acidstrength.html
The fact that Kw is a constant has implications for the relationship of the proton and hydroxide concentration in an aqueous solution. The auto-dissociation reaction is:
2 H2O(l) → H3O+(aq) + OH-(aq)
Or, if you view it as the splitting of a single water molecule and subtract a water molecule from both sides of the equation:
H2O(l) → H+(aq) + OH-(aq)
Either way, the Kw is:
In a solution of pure water, each water molecule that dissociates produces one proton and one hydroxide. The Hydrogen ion concentration, [H+], and the Hydroxide ion concentration, [OH-], both will equal 10-7. Remember that when the product of two variables equals a constant, for one to go up the other must go down. The same thing holds here. If you consider a solution with an acid dissolved in it, there will be two sources of H+: the small amount from the auto-dissociation of water, and the protons from the dissociation of the exogenous acid.
Example problem: What will the hydroxide concentration be if you add 1 mmol of HCl to 250 mL of water?
- Concept: We know that Kw links the proton and hydroxide concentrations of an aqueous solution. Thus, if we can figure out the proton concentration, we can answer the question.
- Equation: Kw= [H+][OH-]
- Algebra: [OH-] = Kw/[H+]
- Plug-n-chug: We need to know the [H+] first.
- [H+] = (#exogenous protons)/Volume + initial [H+] = (10-3 mol/.25 L) + (10-7M) = 4.0001 x 10-3 M ≈ 4 x 10-3 M.
- [OH-] = (10-14)/(4 x 10-3) = ¼ x 10-11 = 2.5 x 10-12
- Logic check: The [H+] went up by more than 4 orders of magnitude, so we would expect that the [OH-] went down by a similar ratio, and this checks out.
- Note: we could have ignored the contribution of the autoionization because the exogenous concentration is so much higher than 10-7. As a rule of thumb, if the exogenous acid would produce a concentration ≥ 10-5, you can ignore the autoionization.
Because the proton and hydroxide concentrations can vary so greatly, we use the p-scale to talk about them as a shorthand for orders of magnitude (i.e., powers of ten). We use the p-scale to focus on the bigger differences. Let’s say we have two solutions where:
- [H+]first solution = 7 x 10-3
- [H+]second sol’n = 2 x 10-6
Which difference is more important, 7 vs 2, or -3 vs -6? The answer is based in scientific notation. Any number can be represented as a significand S, where 1 ≤ S < 10, multiplied by an integer power of 10, leading to the standard notation: S x 10p. The p, or power, is clearly more important, as it describes the order of magnitude of the proton concentration. There is not a 5-unit difference as comparing the significands might indicate, but a roughly 103, or thousand-fold difference between the numbers. The way we discuss concentrations over such a huge range is the p-scale, which puts the power of ten up front.
p (Anything) = -log (that thing)
pH = -log [H+]
pOH = -log [OH-]
pKa = -log Ka
Important: You must know that the following statements mean the same thing. My recommendation is to repeat it, out loud, 3 times, and add it to a “memory dump” sheet for test day until it is so ingrained in your head that you can recite it when being woken up from sleep at 2 AM.
low pH = high [protons] = high acidity
Often, the MCAT will expect you to be able to interconvert between [H+] and the pH. Luckily, there is a math trick that is pure gold here. Follow the derivation once to understand it, but memorize the last line. Let’s begin by representing our proton concentration in scientific notation. Since a [H+] of 1 implies an acid solution strong enough to bore a hole through a lab bench (think about the scene in “Alien” where the blood dissolves the ship’s floors), pretty much every [H+] the MCAT cares about is smaller than 1. Therefore, the power to which we raise 10 will always be a negative integer.
For any [proton] in scientific notation:
[H+] = s x 10-p
By taking the log of both sides, we get:
log [H+] = log (s) + log (10-p)
log [H+] = log (s) - p
We aren’t quite yet at pH, so we multiply both sides by -1:
-log [H+] = - log (s) + p
pH = - log (s) + p
Rearranging terms on the right, we get:
pH = p - log(s)
Here’s where some knowledge about logarithm curves and scientific notation comes into play. Note that we chose “s” to be the significand, so: 1 ≤ s < 10. Looking at the graph at right, we can see that in this range for n, 0 ≤ log (s) < 1. Which means that by subtracting log(s) from p, we are subtracting a fractional number from an integer.
The second term, “- log (s)” can be approximated by the function “-s/10,” as you can see in the image at right.
By borrowing from the “1’s” column of the number, we can come to the following point:
pH = (p-1) + (1-log (s))
The term to the left is the integer part of the pH value. The term to the right is the fractional part of the pH value.
pH ≈ (p-1) + (1 - s/10)
Since 1 = 10 tenths, we can quickly estimate the pH as follows. This is the formulation you should memorize.
pH ≈ [p-1].[10 - s]
Example Problems: Time to test it out! Below are several [H+] values. Use the trick to estimate the answers, then scroll down to check how close you were to the actual value. But first, watch how it is put to use in a model problem.
Model problem: Let’s say you have a solution where [HCl] = 5.88 x 10-6 M. What is the pH?
- HCl dissociates completely, so you get an exogenous [H+] = [HCl] = 5.88 x 10-6 M.
- The autoionization produces 10-7 M protons.
- [H+]total = 5.88 x 10-6 M + 10-7 M = 5.98 x 10-6 M
- The power (p) here is 6
- The significand (s) is 5.98
- The pH ≈ [6-1].[10-5.98] ≈ [6-1].[10-6.0] ≈ 5.40
- Check: -log (5.98 x 10-6) = 5.22
4.37 x 10-3
3.80 x 10-9
1.95 x 10-13
The process works backwards, too. Just take the integer of pH value, add 1 and put it as the negative exponent value. Then subtract the 1/10 of the decimal part from 10 to get an estimate of the signficand, which despite the name, is far less important than the power of ten of the concentration. The formula makes this more clear than words.
For any pH value with with an Integer part and a Decimal part, where D is the
pH = I.Ddd
[H+] ≈ (10-D/10) x 10-(I+1)
(Insert Hold Music Here)
Error of estimate
4.37 x 10-3
3.80 x 10-9
1.95 x 10-13
Now that we have discussed p-scale measurements, we can strengthen our understanding of the Kw. We said earlier that the product of the proton and hydroxide concentrations had to equal 10-14, the Kw. In pure water under standard conditions, the [H+] = [OH-] = 10-7. This means that for neutral water, the pH = pOH = 7. Adding an acid to the water will increase the [H+], decreasing the pH. This will cause the pOH to go up. Specifically:
At 25∘c, pH + pOH = 14
A favorite question of the MCAT is to ask what happens to the pH of pure water at temperatures other than those of standard conditions. Since it is a named Keq, Kw ∝ T, so:
(Kw = [H+][OH-] ) > 10-14 at higher temperatures, and pH + pOH > 14.
Throughout this chapter so far, we have been asserting that hydrochloric acid (HCl) is a strong acid. How do we know? “Experience,” probably. But we can use our understanding of equilibrium chemistry to predict the activity of novel acids and bases, and take out the guesswork.
A strong acid is one with a stable conjugate base. The conjugate base will be happy to have gained a charge of -1 compared to the acidic version of itself. For a molecule, “happy” means being at a lower energy state. Once the acid dissociates, going back to the protonated state requires achieving the activation energy for the reverse reaction, which is greater than that of the forward one. So, it is likely to stay dissociated. The greater the energy difference (∆G), the more stable the conjugate base is. Chloride, the conjugate base of hydrochloric acid, is so stable it is effectively inert.
Below is a table discussing how to predict the strength of an acid or base. Following the table, we will discuss the specifics of what is meant by the groups listed there.
Hydro-Halic except HF
Every other Acid
Every other base
The Hydro-Halic Acids are characterized by a single Halogen bound to a Hydrogen.
The Electron density map for the undissociated form of Hydrochloric Acid is at right. notice how denuded of electron density (blue) the hydrogen is. This means that the chlorine atom is already holding most of the negative charge in the molecule. So, when the proton dissociates, leaving that negative charge behind, the chlorine will be able to take on that little extra bit of charge while actually losing potential energy. The enthalpy will increase a little due to the charge separation, but the entropic part of the Gibbs function is larger, allowing ∆G to be negative. This is true for all Hydro-Halic acids HX with the exception of HF.
Hydrofluoric acid is the exception to the rule because Fluorine is so electronegative that when bonded to Hydrogen, it can make H-bonds with surrounding molecules, shown as dashed red lines. In aqueous solution, these bonds stabilize the undissociated form, shifting the equilibrium of the acid dissociation to the left.
Okay, this is a term I made up. What I’m referring to is an acid that has a central atom bonded to a bunch of peripheral oxygen molecules. As a class, these acids can be monoprotic (bearing one H), or polyprotic (more than one H).
Phosphoric Acid, Carbonic Acid, and Nitric Acid are all Poly-oxic acids.
The reason for the exceptional acidity of the Polyoxic acids is best explained by looking at the Sulfuric Acid Series: H2SO4, HSO4-, SO42-.
Deprotonating Sulfuric Acid leaves a charge of -1 on the Oxygen to which it was attached. By resonance stabilization, this single extra electron spends an equal amount of time around each of the 3 Oxygens not bound to a Hydrogen. A similar process happens when bisulfate (HSO4-) loses a proton to become Sulfate (SO4-), as explained in the table below:
H2SO4: Notice how deshielded (blue) the protons are.
HSO4-: After losing one proton, there is a huge disparity between the highly negative Oxygens and the remaining Hydrogen (green)
SO42-: Fully deprotonated, all the Oxygen atoms are equally negatively charged. (The mapping software looks for differences, so it paints equality as green)
Charge on the Oxygens: All are neutral
Charge on O’s: 3 O’s each have a charge of -⅓
Charge on O’s: 4 O’s each have a charge of -½
Important fact about polyprotic Acids: Only the first proton is a strongly acidic proton. This truth comes from analyzing the Free Energy change ∆G. Deprotonating H2SO4 gives that first electron a lot of room to move, through the electron clouds of 3 oxygens. This increased “elbow room” is entropically favored. In other words, ∆ S > 0 for this process. However, the energy of the arrangement (enthalpy) of the charge distribution is higher in the product. 3 Oxygens in the bisulfate anion have a -⅓ formal charge, up from zero. This requires electrostatic work, and thus requires an investment of energy. Therefore, the enthalpy change is positive (∆H>0). For this first proton, the entropy term is more potent than the enthalpy term, so ∆G < 0. Deprotonating again will increase the charge on more Oxygens (∆H>0) while simultaneously decreasing the average elbow room (∆S<0) for the excess negative charge. An endothermic process that also decreases entropy like this will always be endergonic, and thus nonspontaneous. As a result, putting dry Sulfuric Acid into water leads to only slightly more than 1 proton per molecule going into solution.
*Carbonic Acid (H2CO3) has a Ka = 4.3 × 10-7, which means that it is technically a weak acid. However, it is most commonly discussed within the context of the body. As is stated in the second example problem of the section on buffer systems, Carbonic Acid mostly dissociates under typical body conditions. So, it functionally acts like a strong acid in the body. When the test asks you to calculate, use the Ka or pKa for a quantitative answer. For qualitative questions, especially in the body, you can treat it like it is strong.
Tip to reduce memorization: The MCAT expects you to know the charge on some anions. The charge on Halides is -1 and that on Oxides is -2. Some polyatomic anions that are biologically important are the fully-deprotonated forms of Polyoxic Acids. Rather than memorize this information twice, learn the molecular formula for the Acidic form. The charge on the anion = (-1) (# protons)
Fully Acidic Form
Now that we have explained the categories below, we can see how acids and bases that may be encountered on the MCAT fit into the table below.
Hydro-Halic except HF
Every other Acid
Every other base
Feared by medical students everywhere, the famous equation goes:
This Equation doesn’t need to be scary. Both in form and in what it tells you, this is just an application of equilibrium chemistry. In it’s most human-readable form, it is:
Not so bad, right? Before I can show how to arrive at that simple understanding, we need to define the component variables.
- pH refers to the pH of the solution in which the acid/conjugate base system is dissolved.
- pKa: Since the Ka describes the equilibrium picture of a specific acid molecule under standard conditions, and the pKa is just the -log(Ka), the pKa is a constant that is specific to that acid molecule.
- [HA] = concentration of the protonated acid.
- [A-] = concentration of the deprotonated conjugate base.
Okay, now on to the derivation!
We start with with the definition of the Ka:
Taking the log of both sides:
We can split apart the right-side logarithm:
Using the definitions of pH and pKa we get the final form of the equation:
This isn’t too hard to memorize. There are no minus signs, and the second term on the right is in the same form you learned in equilibrium chemistry:
Let’s take a look at the magic thing that happens when the pH of the acid solution is equal to the pKa of the acid dissolved within it:
Start with the Henderson-Hasselbalch Equation:
Let’s pull the pKa to the left side:
Let’s set the pH of the solution to the pKa:
We get this equality:
Get rid of the logarithm by setting both sides as a power of 10:
Simplifying the exponents:
Multiplying both sides by [HA]:
The magic thing that happens when the pH of the solution in which an acid is dissolved is set to the pKa is: the concentrations of the acid and its conjugate base are equal to one another. Read that sentence a few times until you believe it in your bones.
The human definition of the pKa: the pH at which the [conjugate base] = [weak acid].
Let’s examine the math of that:
pH = pKa + log (ratio of base to acid)
So, if that ratio = 1, [A-] = [HA], and the log (of the ratio) = 0, thus:
pH = pKa when (ratio of base to acid = 1)
If there is more base than acid, that ratio > 1, thus the log (of the ratio) > 0. (i.e., positive):
pH > pKa when [base] > [acid]
If there is more acid than base, that ratio < 1, thus the log (of the ratio) < 0. (i.e., negative):
pH < pKa when [acid] > [base]
These inequalities make sense when you think about applying Le Chatelier’s principle from a point where the pH = pKa, so that [Acid] = [Base]. Adding more protons will drive the equilibrium for the acid dissociation to the left:
HA ⇌ H+ + A-
Moving to a more basic pH than the pKa will remove protons from the system, driving it to the right, resulting in more base than acid.
Sometimes, a question will ask you to calculate the concentration of [A-] or [HA] given the pH and the pKa. Other times, you will be given the ratio of base to acid, then given either the pH or the pKa, and asked to calculate the other. The best way to explain this is through a series of problems.
Example Problem: The Ka of Acetic acid is 1.8 x 10-5. At what pH will the ratio Acetic Acid to Acetate be 100:1?
First, we need to convert the Ka to a pKa. Using the math trick for p-scale conversion:
pKa ≈ [5-1].[10-1.8] ≈ 4.82
Next, we need to plug into the Henderson-Hasselbalch equation:
pH = pKa + log ([base]/[acid])
But wait! Careful reading of the question stem reveals that we were given the ratio of acid to base, so we must plug in the reciprocal:
pH = 4.8 + log (1/100) = 4.82 + log (10-2) = 4.8 + (-2) = 2.8
Logic Check: We would expect that when pH < pKa, there would be more acid than base, so this checks out. More specifically, being 2 pH points more acidic implies that there are 102, or 100 times as many protons as there are at pKa.
Example Problem: Hypochlorous acid (HClO) has a Ka of 2.9 x 10-8. If the concentration of a solution of hypochlorous acid is 10-4 M when the pH is 10.7, what is the concentration of hypochlorite anion?
Solution: Again, we first need to get a pKa value to plug in.
pKa≈ [8-1].[10-2.9] ≈ 7.71
pH = pKa + log ([A-]/[HA])
10.7 ≈ 7.7 + log ([A-]/[HA])
3 ≈ log ([A-]/[HA])
103 ≈ [A-]/[HA]
103 x [HA] ≈ [A-]
103 x [HClO] ≈ [ClO-]
(103)(10-4) = 10-1 ≈ [ClO-]
Logic Check: At 3 pH points more basic than pKa, we would expect there to be 1000 times more conjugate base than acid.
The MCAT will often ask what the predominant species is at a given pH. Sometimes, the question will go one step farther and ask what the charge on the molecule will be. While in undergrad, I developed the following series of questions to quickly deal with this type of problem.
- What is the pKa?
- What is the current pH?
- Are there more or fewer protons hanging around than there would be at pKa?
- If there are more protons hanging around than there would be at pKa, the acid will be protonated, so the protonated form is predominant.
- If there are fewer protons hanging around than there would be at pKa, the acid will be deprotonated, so the deprotonated form is predominant.
- What is the charge on the protonated form? (Let it be Z)
- The charge on protonated form is Z.
- The charge on the deprotonated form is Z-1.
Example Problem: The pKa of the Ammonium ion is 9.3. What is the charge on the dominant species under neutral conditions?
Solution: Neutral pH is 7, so this is lower (more acidic) than Ammonium’s pKa. Thus, there are more protons hanging around than are necessary to have [conjugate base] = [acid]. So, the dominant form will be the protonated form. This is cationic (NH4+), so the charge is positive.
Useful Tip: Think of the pKa as the “Half-protonation point” of an acid/base molecule!
Aspartic Acid (left), Glycine (middle), and Lysine (right)
Where this system of analysis becomes super-useful is in understanding the isoelectric point of amino acids. Different amino acids can be grouped by the chemical functionality of their side-chains, also known as R (residue) groups. Normally, the pH in the blood is 7.4, so after amino acids are taken up from the gut, they are dissolved in the aqueous serum that has a roughly neutral pH.
For amino acids with a neutral R group (like glycine), the only functional groups which can take a charge are the amino and the carboxylic acid groups. Applying the algorithm: pH 7 is more basic than the pKa of a the acid, so there are fewer protons around than necessary to keep half of the acid groups protonated. Thus, the acid is deprotonated and negative. For the amino group, pH 7 is more acidic than the half-protonation point represented by it’s pKa, so acquires a proton and becomes charged. This doubly-charged form, with a net charge of zero, is called a zwitterion.
If you put Amino Acids into an electrophoresis gel with a pH gradient that is oriented as shown, then you can separate them according to their acid/base properties. At pH values lower than the lowest pKa, the excess protons will protonate the amino acids, making them positive. They will flee the positive anode, and move through the gel towards the negative cathode. However, as they move through the gel, they are surrounded by a decreasing concentration of protons, and thus become more negative. At some point between these two charge states, their charge must equal zero, at which point they will stop moving. You can put the amino acids into the basic end of the gel, and the same process happens in reverse. The pH point at which peptides come to rest is called the isoelectric point (pI), because this is where they have a net neutral charge. For amino acids with neutral R-groups, this is the pH at which the maximum concentration of the zwitterion exists, and is simply the average pKa of the amino and acid groups.
How to figure out the isoelectric point for basic amino acids:
How to figure out the isoelectric point for acidic amino acids:
So, to summarize, for amino acids with:
Neutral R Groups
Acidic R Groups
Basic R Groups
A buffer system is a solution that contains both a weak acid and its conjugate base in significant and roughly equal concentrations. Very strong acids will dissociate completely, and thus the ratio of base to acid will be almost infinite, so they cannot be used to make buffer systems. The point of a buffer solution is to take in or give up a big load of protons without resulting in a big change in the pH. This may seem surprising, but it all follows from the definition of a buffer system above, and from the Henderson-Hasselbalch system. This is best explained by running a few example problems.
Example Problem: Imagine you have a 1L solution of 0.1M Acetic Acid (pKa 4.75) and 0.1M Acetate. (a) What is the pH of this solution? (b) What will be the pH after adding 10-3 mol of Hydrochloric Acid?
Solution: (a) Since the concentration of Acetic Acid and its conjugate base Acetate are initially equal, the pH of the solution must equal the pKa by definition. Thus, the pH is initially 4.75.
(b) Together, the Acetic Acid and Acetate can make a buffer system. The Hydrochloric acid is the source of our exogenous acid, and since it is a strong acid, can be considered to dissociate completely, providing 10-3 mol of protons. When they join the solution, they will combine with the Acetate to make Acetic Acid by the following formula:
CH3COO-(aq) + H+(aq) ⇌ CH3COOH(aq)
So, for every Acetate that is consumed, one Acetic Acid molecule is created. Since the solution has a volume of 1L, the [H+]exogenous = 10-3M. We can plug that into the Henderson-Hasselbalch equation like so:
The conclusion we can draw is that a buffer system minimizes the effect of an acid or base load because the load doesn’t significantly change the ratio of ([base]/[acid]) away from 1. This mathematical truth is the basis of the definition: The ratio of [conjugate base]/[acid] must be nearly 1, so the [acid] ≈ [conjugate base]. Additionally, if the [acid] is not significantly higher than the concentration of exogenous acid or base loaded in, then the ratio can change greatly. For example, see the next problem:
Example Problem: A student tries to make a buffer solution using 10-3M Acetic Acid and an equal concentration of Acetate. What will the pH be after he adds 5x10-4 mol of HCl?
Solution: We can solve this using the same logic as the problem above:
This may not seem like a big difference in pH, but remember that pH is a semilog scale, so this ½ pH-point decrement means there are roughly 3 times as many protons floating around in the final solution. For the MCAT, recognizing that the buffer system was overwhelmed because the [buffer species] ≈ [acid load] is the key to choosing the right answer.
Clinical correlation: the pH of the body is very tightly regulated (normal pH = 7.4 ± 0.05). Such precise control is needed because of the way that acidic and basic side groups change their charge states depending on the pH of the solution. With a pH drop of “only” 0.5, the charge on many R groups would become more positive, and the way that the protein folds would change. This explains why a blood pH value of 6.9 represents a life-threatening acidosis that requires ICU management!
Now, let’s look at the way that the human body buffers its pH. Unsurprisingly, this is one of the most heavily-tested buffer systems, from the MCAT all the way to critical care Board exams.
Example Problem: The pH of the blood is maintained by the bicarbonate buffer system. The reaction governing it is:
H2O(l) + CO2(g) ⇋ H2CO3(aq) → HCO3-(aq) + H-(aq)
The first step is catalyzed by the enzyme Carbonic Anhydrase, found in high concentration in the cytosol of RBCs. Carbonic Acid, the intermediate, rapidly breaks down to Bicarbonate and a proton. Since it takes one Carbon Dioxide molecule to create each Carbonic Acid, we can use [CO2] to represent the [H2CO3-].
The pKa of Carbonic Acid is 6.4. Normal blood pH is 7.4 What is the ratio of Bicarbonate to Carbonic Acid under normal conditions?
There is 10 times as much base as acid under normal blood conditions. This excess of base is no accident. Acidosis (the creation of supernormal amounts of acid) is way more common than alkalosis (making too much base). Every time you exercise or go to sleep, you go into an acidosis. As a result, by resting on the basic side of the “ideal” base:acid ratio of 1:1 , the bicarbonate buffering system can take a HUGE acid hit, and actually become a better buffer.
One of the concepts that plagues premed students is the definition of the word “equivalent.” I have a way to keep it simple.
Equivalent definition: An equivalent is one functional unit of acid or base.
The reason we even need this definition is that there are such things as polyprotic acids. One mole of HCl produces 1 mole of protons when fully dissociated, thus:
1 mol HCl = 1 mol of acid units = 1 equivalent of HCl
This hold true for weak acids, too. Each mole of Acetic Acid (CH3COOH), if it were to dissociate completely, would produce 1 mole of protons. (Complete dissociation of a weak acid happens in a solution with lots of exogenous base. See the section on pKa.) Thus:
1 mol CH3COOH = 1 mol of acid units = 1 equivalent of CH3COOH
The critical question to ask yourself when using equivalents is: For each mole of substance X, how many moles of acid or base does it equal? This gets you a ratio of (eq/mol) for that substance.
Normality = a measure of concentration for Acids and Bases, parallel to Molarity, but adjusted for whether that species is polyprotic or not.
Thus, 1M solutions of HCl and H2SO4 are 1N and 2N, respectively. If this is unclear, go back and re-read the definition of equivalents.
Definition of Equivalence Point: The point in the titration when the reaction flask contains the same number of equivalents of acid and base.
At its heart, the equivalence point formula just restates the definition of the equivalence point:
(at the equivalence point)
Using the equivalence point formula often relies on substituting one or both sides of the equation with one of the following to match the available information in the question stem:
in shorthand, we often restate the equivalence point formula as the following. (Note that, based on the units, both sides of the equation still are measured in equivalents):
The “endpoint” of most titrations is the equivalence point of the titrand. Since you want to be absolutely sure of the number of equivalents of the titrant in the reaction flask, you must use a strong, monoprotic acid or base as a titrant. In practice, this means we usually titrate with HCl or NaOH.
Interpreting the titration graph of HCl:
- We plot the pH on the y-axis, and the Volume of titrant (NaOH) added on the x-axis
- because pH = -log([H+]), this is effectively a semilog graph.
- As we drip in a more NaOH, we increase the number of equivalents (moles) of base linearly. Each of these base moles neutralizes exactly 1 mole of HCl. Initially, this linear decrease in the [H+] doesn’t change its exponent very much, so you don’t get much of a change in the pH (i.e.: the graph stays flat initially).
- Imagine subtracting from 103 (a thousand) by 1’s. Until you have subtracted 900 times, you don’t change the integer value of the exponent. The next time the exponent loses another integer is when you have subtracted a mere 90 more times. Thus, the value of the pH accelerates upwards as the concentration of free protons approaches 10-7M.
- The part of the graph with the greatest slope is the equivalence point.
- For Strong Acids, the equivalence point is at pH 7, because all the moles of HCl have been neutralized by an equal number of moles of NaOH, so the only remaining source of protons is the autoionization of water.
- After the equivalence point, the pH shoots upwards, then flattens out because the NaOH is neutralizing the protons produced by the autoionization of water, which is a source that produces less and less as the [OH-] increases because Kw = [H+][OH-].
- There are fewer and fewer protons, thus it becomes harder for hydroxide ions to find them in solution and neutralize them. For this reason, the rate of change in the proton concentration (slope of the pH curve) decreases.
Interpreting the titration graph of NaOH: The exact same dynamics apply, but with protons being dripped in to neutralize titrand hydroxide ions. Effectively, the curve is just flipped vertically. The curve is initially flat (but at a high pH), changes suddenly, has a steep slope at the equivalence point, the pH of which is 7, and then the curve flattens out (but now has a low pH).
The image at the above left compares the titration curves of a weak acid and a strong acid. There are several important differences, some of which are detailed in the image at the above right:
- Initially, the weak acid “gives in” to the strong base titrant, resulting in an initial rise in the pH.
- The weak acid doesn’t produce as many protons per mole of acid as a strong acid, thus it is is easier for the strong base to “mop up” these protons and reduce the [proton] several orders of magnitude (powers of ten).
- The buffer zone “slows down” the rate of pH increase.
- When the pH rises to the pKa and the base continues to deprotonate the acidic form of the weak acid, turning it into the conjugate base, the ratio of [conjugate base]/[acid] changes minimally, so by the Henderson-Hasselbalch equation, the pH change is minimal.
- The equivalence point is more basic than pH 7
- At the equivalence point, (#moles of base added) = (#moles of weak acid)
- All of the acid (HA) has been converted to its conjugate base (A-)
- Because A- is the conjugate base of a weak acid, it is not inert. Rather, it is weakly basic. It reacts with water by the following reaction to produce hydroxide:
A−(aq) + H2O(l) ⇌ HA(aq) + OH−(aq)
- This endogenous source of hydroxide consumes some of the protons produced by the autoionization of water, raising the pH above 7.
- The slope of the curve around the equivalence point is less steep and less high of a step than that for a strong acid.
- Slope = rise/run = ∆ pH/(base added), and there is simply less of a pH change around the equivalence point, since you are stepping up from the pKa, rather than from a very acidic pH, as you would be with a strong acid.
Titration of a Generic Diprotic Acid.
Distribution curve for Fumaric Acid, a specific Diprotic Acid.
The image at the above left shows the titration curve for an unnamed generic diprotic acid (H2A).
- Notice that there are two equivalence points, one for each proton.
- The first proton in any polyprotic acid species is strongly acidic.
- Thus, the system (H2A / HA-) makes a poor buffer, which is why the curve is not flat when the pH = pKa1.
- The first equivalence point is steeply sloped, as would be expected for a strong acid.
- The second proton is weak (The third would be as well, if we were talking about a triprotic species like phosphoric acid, H3PO4).
- Thus, the system (HA- / A2-) makes a good buffer system, and the curve is flatter when pH = pKa2.
- The second equivalence point is not steep, as would be expected for a weak acid.
Notice about terminology: the pKa is sometimes referred to as the “half-equivalence point.” This name is based on the fact that you have added half the necessary base to reach the equivalence point. It does not mean that the pKa’s lie halfway between the pH’s of the equivalence points, as you can clearly see from the graph at above left. As noted previously, at a given proton’s pKa, half of the molecules have had that proton removed, and half still bear that proton. Thus, a more informative synonym for pKa is the “half-protonation point.”
One of the most commonly asked questions regards which species is the “dominant” one at a given pH. What the question is asking is when each species is found at the highest concentration. This can be determined by telling yourself the “story” of what happens to the acid species as the pH of the solution goes from very acidic to very basic. This is very easily done with the tools for intuitively understanding the relationship between pH and pKa described in an earlier section. While reading the steps below, refer to both of the above diagrams.
- Initially, all of the acid molecules are fully protonated. H2A (α2) is the only, and therefore, dominant species. This is true for any pH < pKa1.
- At pKa1, [H2A] = [HA-] (definition of pKa).
- At the first equivalence point, all of the H2A has been deprotonated once, becoming HA- (α1).
- At pKa2, [HA-] = [A2-]
- At the second equivalence point, all the HA- has been fully deprotonated. At pH > pKa2, A2- (α0) is essentially all that is left.
As mentioned in the beginning of this section on Titrations, when we are using titration to quantify the concentration of the Titrand, we need a way to know that we have reached the endpoint (equivalence point or pKa) of the titration. We could use a pH-meter, but the MCAT tests the older method, indicator dyes.
Indicator dye: A weak acid which, when deprotonated, undergoes a color change. This is often done by sharing the resultant lone pair as part of a large system of conjugated π bonds. The system needs to be large enough to absorb low-energy wavelengths of light. Benzene absorbs in the UV spectrum, and longer conjugated systems can absorb in the visible spectrum.
Methyl Red is a commonly-used indicator dye. Notice that because the pKa is 5.1, at pH=4, the ratio of red:yellow dye molecules will be 10:1, making the solution red. Similarly, at pH=6, there are 10 times as many yellow as red dye molecules, making the solution yellow. The solution appears orange when pH ≈ 5 ≈ pKa, because the ratio of red:yellow is ~1:1.
Choosing which indicator to use to monitor for the endpoint of a titration is based on the pKa of the indicator and the pH of the endpoint you are investigating. The image at the above left shows the pH ranges in which commonly used indicators change, and to which colors. This is just for reference. The image at the right shows the titration curve for Propanoic Acid, a weak acid. The pKa for Propanoic Acid is roughly 5, which is similar to the pKa for methyl red. Thus, we can use methyl red to monitor the titration, and stop when the solution turns orange. Notice that to monitor for the equivalence point of Propanoic Acid (~9), we can use a less precisely matched indicator dye, because from one drop of added base to the next, we will probably overshoot the exact pH of the equivalence point, turning the Phenolphthalein pink. Our error will only be ½ of one drop’s worth of base.
The following is a brief summary of the most important points regarding titrations:
- You always drip in a strong titrant that will dissociate completely.
- At the equivalence point, the #eq of acid = #eq of base.
- At the pKa for a specific proton, the concentrations of the protonated and deprotonated versions of the acid are equal.
- The equivalence points for strong acids are steep.
- The graph for a base is flipped vertically compared to that for an acid.
- Weak acid curves have a flat buffer regions around the pKa.
- Polyprotic acid curves can be read to determine the dominant species at a given pH.
- Indicators dyes are weak acids. Their pKa must be tightly matched to monitor for the pKa of the titrand as an endpoint. Monitoring for equivalence point can be done with a looser match.
The creation of this chapter was greatly facilitated by the images from the North Carolina School of Science and Mathematics’ “Teacher’s Instructional Graphics Educational Resource,” or TIGER. It is in the same spirit of free sharing of knowledge that motivates them that this site was created. As always, I am ever thankful for the string of teachers, from the grade school to the graduate level, who have shaped my understanding of this subject material, and allowed me to see with clarity so that I may share it with you.
Last update: 10-10-14