Acids and Bases for the MCAT: Everything You Need to Know
/Learn key MCAT concepts about acids and bases, plus practice questions and answers
(Note: This guide is part of our MCAT General Chemistry series.)
Table of Contents
Part 1: Introduction to acids and bases
Part 2: Definitions
a) Lewis and Bronsted-Lowry acids and bases
b) Important acids and bases
c) Amphoteric species
d) Equilibrium constants
e) pH and pOH
Part 3: Buffer systems
a) Buffer systems
b) Henderson-Hasselbalch equation
Part 4: Titrations
a) Titrating a strong acid and strong base
b) Titrating weak acids and bases
c) pKa of multiprotic acids
Part 5: High-yield terms
Part 6: Passage-based questions and answers
Part 7: Standalone questions and answers
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Part 1: Introduction to acids and bases
Maintaining equilibrium in acids and bases is a key component of maintaining homeostasis. Understanding the behavior of acidic and basic environments is key to understanding the basis of many organic reactions and biological systems.
This guide will cover properties that define acids and bases, buffer systems, and the fundamentals of titrations. By the end of this guide, you’ll be prepared for whatever the MCAT throws at you on acids and bases come test day. At the end of this article, there are also several MCAT-style practice questions for you to test your knowledge with.
Let’s get started!
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Part 2: Definitions
a) Lewis and Bronsted-Lowry acids and bases
The Lewis and Bronsted-Lowry definitions describe different properties of acids and bases. Lewis acids and bases are based on electron movement. A Lewis acid accepts an electron pair, while a Lewis base donates a pair of electrons.
All acids and bases have conjugates, which are formed after losing or gaining a hydrogen ion. A conjugate acid is a Bronsted-Lowry base that has acquired a proton. A conjugate base is a Bronsted-Lowry acid that has lost a proton. Strong acids have weak conjugate bases, and strong bases have weak conjugate acids. This is because a strong acid will not form a species that readily accepts a proton. If it did, then it would not be a strong acid.
b) Important acids and bases
- Potassium hydride: KH
- Lithium hydride: LiH
- Lithium hydroxide: LiOH
- Sodium hydroxide: NaOH
- Potassium hydroxide: KOH
- Rubidium hydroxide: RbOH
- Caesium hydroxide: CsOH
- Magnesium hydroxide: Mg(OH)2
- Calcium hydroxide: Ca(OH)2
- Strontium hydroxide: Sr(OH)2
- Barium hydroxide: Ba(OH)2
Elements from Group 17, the halogens, typically form strong acids. These compounds form very stable conjugate bases upon dissociation due to the large atomic radius of the resulting ion. Hydrobromic acid (HBr) dissociates into a hydrogen ion and bromide ion. The bromide ion’s larger size gives the negative charge more room to disperse, thus stabilizing the ion.
The exception to this rule is hydrofluoric acid (HF). The weakness of HF is a result of the instability of the dissociated products. The negative charge of the fluoride anion is very unstable due to the small atomic radius of the ion. Thus, the dissociation of HF is not very favorable.
The list below summarizes several commonly found strong acids. Any acids not on this list can be assumed to be weak.
- Hydroiodic acid: HI
- Hydrobromic acid: HBr
- Hydrochloric acid: HCl
- Perchloric acid: HClO4
- Sulfuric acid: H2SO4
- Nitric acid: HNO3
c) Amphoteric species
Amphoteric species are compounds that can act as either Bronsted-Lowry acids or bases. In an acidic environment, such a species acts as a base, while in a basic environment, it acts as an acid.
Amino acids are an example of amphoteric species. Recall that amino acids are classified as zwitterions. These are molecules that simultaneously have a positive and negative charge. Amino acids can act as acids through the amino group (or N-terminus), which donates protons. Additionally, they can act as bases through their carboxylate group (or C-terminus), which accepts protons.
Water is another very common amphoteric species. It undergoes different reactions when in the presence of a base or acid. Below are examples of water in the presence of a protonated acid (HA) and a protonated base (HB).
H2O + B- ⇋ OH- + HB
Note that when water is in an acidic environment, it accepts a proton to form the hydronium ion. However, when it is in a basic environment, it donates a proton to form a hydroxide ion. Water is also able to engage in a special type of reaction called autoionization. During this process, water reacts with itself. This is outlined by the following reaction:
d) Equilibrium constants
The autoionization of water also has an equilibrium constant. (For more information on equilibrium constants, be sure to refer to our guide on chemical equilibrium and kinetics.)
Two additional constants to be familiar with are the acid dissociation constant (Ka) and the base dissociation constant (Kb). These are the equilibrium constants of the dissociation of an acid (HA) and the dissociation of a base (BH). The formulas for both Ka and Kb are shown below.
B + H2O ⇋ BH+ + OH- Kb = [BH+][OH-]/[B]
Since strong acids dissociate completely, strong acids typically have a larger Ka value. Weak acids have a Ka less than 1.0. Similarly, a larger Kb value corresponds with a strong base, and weak bases have a Kb less than 1.0. The negative logarithms (base 10) of the acid dissociation constant and base dissociation constants provide several important values.
pKb = -log(Kb)
The pKa is the pH at which half of the species in a solution are protonated, and the other half are deprotonated. If the pH is higher than the pKa, the species will be mostly deprotonated. Strong acids tend to have low pKa values, and strong bases tend to have low pKb values. Let’s take a look at an example.
Find the concentration of hydronium in a 3.0 M aqueous solution of the weak acid, benzoic acid (C6H5COOH) (Ka = 6.46 x 10-5).
First, write out the chemical reaction.
Using this, the Ka expression can be written:
Solving this equation is very difficult. To make the equation solvable, a simplifying assumption must be made. Since the value of x value is very small, it is negligible. Thus, the expression can be rewritten as:
Lastly, the provided Ka value can be substituted and used to solve for x:
0.0001938 = x2
x = 0.014
Thus, the concentration of hydronium in the solution is equal to 0.014 M.
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