Unit 8 already covered the basics of acid-base chemistry and how to find the pH/pOH of solutions of strong acids and bases. However, the vast majority of acids and bases are, in fact weak, (remember, there are only seven strong acids as far as AP Chemistry is concerned)! In this section, we will discuss the difference between strong acids/bases and weak acids/bases, using equilibrium to find the pH of solutions with weak acids and bases.
Differentiating Strong Acids and Weak Acids
As we learned in Topic 8.1, there are two definitions of acids and bases: the Arrhenius definition, which states that acids yield H+ in water and bases yield OH-, and the Bronsted-Lowry definition, which broadens the scope to define acids and bases as proton "donors" and "acceptors" respectively. For the examples we discuss in this section, we will use the Arrhenius definition to exclude water as a sort of spectator to these reactions, but remember that behind the scenes, water is the backbone of why acids and bases can do what they do!
A strong acid or strong base is one that completely dissociates in water and does not return back to its product form. Or, at the very least, it does so to a negligible degree. For example, HCl is one of our seven strong acids that must be memorized. As a strong acid, it completely dissociates meaning that in solution, essentially all of the HCl molecules will dissociate into H+ (that is what makes it an acid!) and Cl-, the conjugate base. Strong bases to know are alkali and alkaline metal hydroxides, such as NaOH, Ca(OH)2, and Sr(OH)2.
It is also worth noting that when discussing strong acids and bases, although their conjugates are called conjugate bases and acids, respectively, they are not significant acids and bases. For example, although Cl- is the conjugate base of HCl, because HCl is such a strong acid, Cl- is an incredibly weak base to the point of not even being considered basic whatsoever. Similarly, when looking at a strong base like NaOH, it will fully dissociate into Na+ and OH-. Because NaOH is a strong base, Na+ is an incredibly weak acid, so weak that we do not even refer to it as an acid despite being a conjugate acid of NaOH.
On the other hand, weak acids and weak bases incompletely dissociate in water. That is to say, not all of the acid/base will dissociate to yield H+/accept OH-. For instance, CH3COOH, acetic acid, is a weak acid with a Ka = 1.8 * 10^(-5).
We will talk a bit more about Ka in a future section, but a lower Ka corresponds to a weaker acid (in reality, it is the equilibrium constant for the dissociation of a weak acid, but we'll get there, don't worry 😉). This means that when dissolved in water, some CH3COOH will dissolve to form CH3COO- (the conjugate base) and H+, but much more of it will remain CH3COOH at equilibrium.
Any acid that is not one of the seven strong acids is a weak acid. As far as bases go, the only weak base you should be familiar with off the top of your head is ammonia (NH3), but you could face others on the AP exam (you will just be made aware by the question that it is weak!)
Review of General Equilibrium
As we move into discussing the relationship between weak acids, weak bases, and equilibria, let's do a quick refresher on some big key concepts from Unit 7, which was all about equilibrium.
Equilibrium is the state in a reaction in which the forward reaction and backward reactions are proceeding at the same rate, meaning that the production of products and reactants perfectly cancel each other out.
Equilibrium does not mean nothing is happening. There is just no more change in the concentrations of reactants and products. The above image is a graphical representation of equilibrium occurring. To mathematically represent the extent to which a reaction goes forward, we use the equilibrium constant, K.
In general, K is the ratio of the concentrations of the products raised to their stoichiometric coefficients and the concentrations of the reactants raised to their stoichiometric coefficients:
When finding equilibrium concentrations, we oftentimes use an ICE (or RICE) box to represent changes in concentration at equilibrium and then plug it into our K expression to solve. This is a strategy that we will also employ when dealing with weak acids and weak bases.
Weak Acid-Base Equilibria
Weak Acid and Base equilibria are just further examples of using equilibrium! Many students get tripped up because of some new symbols. In reality, it is the same as applying equilibrium to any other reaction (minus taking a few extra steps to find things like pH).
Let's take a look at an example:
- Find the pH of a 2M solution of CH3COOH (Ka = 1.8 * 10^(-5))
First, we write out the equation for the dissolution of acetic acid:
This expression (generalized as HA <=> H+ + A) is sometimes called the Ka expression for the acid.
The reaction above will guide us as we use the data above to find our [H+] and then take the -log() of that concentration. We are told that the Ka of CH3COOH is 1.8 * 10^(-5), a number we can use to create an ICE box and solve for [H+]:
Finally, we plug in and solve for [H+]. Remember, because x is super tiny, we can assume that 2-x=2 to ease calculations:
Finally, taking the -log([H+]), we find that pH = 2.22.
Let's look at an example with a weak base:
- Find the pH of 1M ammonia (NH3). (Kb = 1.8 * 10^(-5)) (note: this is just a coincidence that this is the same as the Ka of Acetic Acid!)
Like before, we will start by writing out the Kb expression of Ammonia (the reaction of the weak base with water):
We know our Kb, so let's get ICE boxin! 💪
Same as before, we plug into our equilibrium expression to find x:
We may think we are home free. Just take the -log() of x, and we are done, right?! However, take a closer look. The -log() of x would not give us pH! Instead, it would give us pOH. To find pH, we first have to take pOH and then plug it into pH + pOH = 14.
pOH = -log([OH-]) = -log(0.0042) = 2.38
14 - 2.38 gives us a pH = 11.62.
Vocabulary
The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.
| Term | Definition |
|---|---|
| conjugate acid | The species formed when a base accepts a proton; the acid form in an acid-base conjugate pair. |
| conjugate base | The species formed when an acid donates a proton; the base form in an acid-base conjugate pair. |
| equilibrium | The state in which the forward and reverse reaction rates are equal, resulting in constant concentrations or partial pressures of reactants and products. |
| hydronium ion | The aqueous ion H3O+(aq) formed when a hydrogen ion bonds with a water molecule; represents the form of hydrogen ion in aqueous solution. |
| hydroxide ion | The negatively charged ion OH− produced when water autoionizes or when a base dissolves in water. |
| ionization | The process by which an acid or base separates into ions when dissolved in water. |
| Ka | The acid ionization constant that expresses the equilibrium between a weak acid and its conjugate base in water. |
| Kb | The base ionization constant that expresses the equilibrium between a weak base and its conjugate acid in water. |
| Kw | The ion product constant for water, equal to [H3O+][OH−] = 1.0 × 10−14 at 25°C, representing the equilibrium constant for water autoionization. |
| monoprotic weak acid | An acid that can donate one proton and only partially ionizes in water, establishing an equilibrium between the molecular acid and its conjugate base. |
| monoprotic weak base | A base that can accept one proton and only partially ionizes in water, establishing an equilibrium between the molecular base and its conjugate acid. |
| percent ionization | The percentage of weak acid or base molecules that ionize in solution, calculated from the equilibrium concentration of ions and the initial concentration of the acid or base. |
| pH | A logarithmic scale used to express the concentration of hydronium ions in a solution, calculated as −log[H3O+]. |
| pKa | The negative logarithm of the acid dissociation constant (Ka); used to compare the relative strength of weak acids and predict protonation state at different pH values. |
| pKb | The negative logarithm of the base dissociation constant (Kb); used to compare the relative strength of weak bases. |
| pKw | The negative logarithm of Kw; equals 14.0 at 25°C and represents the sum of pH and pOH in any aqueous solution at that temperature. |
| pOH | A logarithmic scale used to express the concentration of hydroxide ions in a solution, calculated as −log[OH−]. |
| un-ionized | The molecular form of a weak acid or base that has not separated into ions in solution. |
Frequently Asked Questions
What's the difference between a weak acid and a strong acid?
A strong acid fully (or nearly fully) ionizes in water—almost every HA molecule donates a proton, so [H3O+] ≈ initial [HA] and you can get pH directly from concentration. A weak acid only partially ionizes: it establishes an equilibrium HA ⇌ H3O+ + A− with Ka = [H3O+][A−]/[HA], so [H3O+] is much smaller than the starting [HA] and most acid remains as HA. Practically, for weak acids you use an ICE table or the Ka (or pKa) and the initial concentration to solve for pH and percent ionization; remember Ka × Kb = Kw for conjugate pairs (pKa + pKb = pKw). This distinction appears on the AP exam in Topic 8.3 (use Ka/pKa, ICE, percent ionization; see CED 8.3.A). For a quick review, check the Topic 8.3 study guide (https://library.fiveable.me/ap-chemistry/unit-8/weak-acid-base-equilibria/study-guide/J3ggVgoQIyMYhq5nEKuw), the Unit 8 overview (https://library.fiveable.me/ap-chemistry/unit-8), and try practice problems (https://library.fiveable.me/practice/ap-chemistry) to build skill with ICE tables and Henderson–Hasselbalch where relevant.
How do I calculate the pH of a weak acid solution?
Start with the dissociation: HA + H2O ⇌ H3O+ + A−, and Ka = [H3O+][A−]/[HA]. Do an ICE table: initial [HA] = C, [H3O+] ≈ 0, change: +x, +x, −x, equilibrium: C − x, x, x. Plug into Ka: Ka = x^2/(C − x). Solve for x (=[H3O+]): - If Ka ≪ C, use the approximation C − x ≈ C so x ≈ sqrt(Ka · C). Then pH = −log[H3O+] = −log x. - If x/C > 5% (check percent ionization = x/C ×100%), solve the quadratic: x^2 + Ka·x − Ka·C = 0 and take the positive root. For buffers or mixtures use Henderson-Hasselbalch: pH = pKa + log([A−]/[HA]). AP-style problems expect you to show the ICE work, justify any approximations, and report percent ionization when asked. For a topic review see the Fiveable study guide (https://library.fiveable.me/ap-chemistry/unit-8/weak-acid-base-equilibria/study-guide/J3ggVgoQIyMYhq5nEKuw) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
Why don't weak acids completely ionize in water like strong acids do?
Weak acids don’t completely ionize because their proton transfer to water is an equilibrium process, not a one-way reaction. For a monoprotic weak acid HA: HA + H2O ⇌ H3O+ + A−, the position of that equilibrium is given by Ka = [H3O+][A−]/[HA]. If Ka is small (large pKa), the equilibrium lies far left, so [HA] at equilibrium is much larger than [H3O+] or [A−]. Physically, that means the acid’s H–A bond and the relative stability of A− make losing a proton unfavorable for most molecules, so only a small percent ionizes (percent ionization depends on Ka and initial [HA]). This is exactly what Topic 8.3 emphasizes: weak acid solutions contain both un-ionized acid and its conjugate base at equilibrium. For practice calculating pH, percent ionization, and using ICE tables, see the Topic 8.3 study guide (https://library.fiveable.me/ap-chemistry/unit-8/weak-acid-base-equilibria/study-guide/J3ggVgoQIyMYhq5nEKuw) and try problems at (https://library.fiveable.me/practice/ap-chemistry). These skills are directly tested under Learning Objective 8.3.A.
What is Ka and how is it different from pKa?
Ka (acid dissociation constant) is the equilibrium constant for a weak acid (HA) ionizing in water: Ka = [H3O+][A−]/[HA]. It tells you, at equilibrium, how much of the acid has donated a proton—larger Ka means a stronger acid (more ionization). In AP terms this is exactly what you use in ICE tables and percent ionization calculations (CED 8.3.A.2, 8.3.A.5). pKa is just the negative base-10 logarithm of Ka: pKa = −log Ka. pKa turns the often tiny Ka numbers into a convenient scale: smaller pKa = stronger acid. Example: acetic acid has Ka ≈ 1.8×10^−5 so pKa ≈ 4.74. You’ll use pKa a lot when estimating pH from initial concentration (and in the Henderson–Hasselbalch form for buffers). For more review on weak-acid equilibria see the Topic 8.3 study guide (https://library.fiveable.me/ap-chemistry/unit-8/weak-acid-base-equilibria/study-guide/J3ggVgoQIyMYhq5nEKuw) and extra practice problems (https://library.fiveable.me/practice/ap-chemistry).
I'm confused about equilibrium expressions for weak acids - can someone explain the Ka formula?
Ka is the equilibrium constant for a weak acid HA reacting with water: HA + H2O ⇌ H3O+ + A−. Ka = [H3O+][A−]/[HA]. It tells you how far the acid ionizes—big Ka = more ionization, small Ka = less. For a monoprotic weak acid you usually start with an initial [HA]0, set x = [H3O+]eq produced, then use an ICE table: [HA] = [HA]0 − x ≈ [HA]0 (if x ≪ [HA]0), [H3O+] = x, [A−] = x, so Ka ≈ x^2/[HA]0 and pH = −log x. Use the “x small” shortcut when percent ionization < ~5% (or check it afterwards). pKa = −log Ka, and Ka × Kb = Kw for conjugate pairs (useful if you know pKb). For worked examples and AP-style practice on percent ionization, ICE tables, and pH calculations, see the Topic 8.3 study guide (https://library.fiveable.me/ap-chemistry/unit-8/weak-acid-base-equilibria/study-guide/J3ggVgoQIyMYhq5nEKuw) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).
How do you find the percent ionization of a weak acid?
Percent ionization tells you what fraction of a weak acid HA has donated a proton. Steps: 1. Write the equilibrium: HA + H2O ⇌ H3O+ + A− and Ka = [H3O+][A−]/[HA]. 2. Use an ICE table with initial [HA] = C, [H3O+] ≈ 0, change: +x for H3O+ and A−, −x for HA. At equilibrium [H3O+] = x, [HA] = C − x. 3. Solve Ka = x^2/(C − x). If x << C (common for weak acids), approximate x ≈ sqrt(Ka × C). Then [H3O+]eq = x. 4. Percent ionization = ([H3O+]eq / C) × 100%. So with the approximation: % ionization ≈ (sqrt(Ka × C) / C) × 100% = (sqrt(Ka/C)) × 100%. If you know pKa and C, convert pKa → Ka (10−pKa) and use the formula. If you measure pH, use [H3O+] = 10−pH and plug into percent ionization = [H3O+]/C ×100%. This procedure and use of ICE/Ka is exactly what AP Topic 8.3 expects (see the Topic 8.3 study guide: https://library.fiveable.me/ap-chemistry/unit-8/weak-acid-base-equilibria/study-guide/J3ggVgoQIyMYhq5nEKuw). For extra practice, try problems at Fiveable’s practice page (https://library.fiveable.me/practice/ap-chemistry).
What's the relationship between Ka and Kb for conjugate acid-base pairs?
For any conjugate acid–base pair (HA and A−), their ionization constants multiply to Kw: Ka × Kb = Kw (so pKa + pKb = pKw). That means if the acid (HA) has a large Ka (stronger weak acid → smaller pKa), its conjugate base (A−) has a correspondingly small Kb (weaker base → larger pKb). Conversely, a very weak acid (small Ka) has a relatively stronger conjugate base (larger Kb). Practically: knowing pKa lets you get pKb (pKb = pKw − pKa; at 25 °C pKw = 14). This is essential for calculating pH of solutions, percent ionization, and buffer behavior on the AP exam (Topic 8.3; see CED EK 8.3.A.6). For extra practice and worked examples, check the Topic 8.3 study guide on Fiveable (https://library.fiveable.me/ap-chemistry/unit-8/weak-acid-base-equilibria/study-guide/J3ggVgoQIyMYhq5nEKuw) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).
Why is the pH of a weak base solution not just based on the initial concentration?
You can’t get pH from the initial concentration alone for a weak base because the base doesn’t fully dissociate—it sets up an equilibrium with its conjugate acid and water. Kb = [OH−][HB+]/[B], so the equilibrium [OH−] (and pOH/pH) depends on both the initial [B] and the value of Kb (or pKb). In practice you solve an ICE table (or approximate with x^2/(C−x) ≈ Kb if x ≪ C) to find the equilibrium [OH−]. Also note percent ionization changes with concentration: dilute solutions ionize a larger fraction, so two solutions with the same initial [B] can have different pH if their Kb differs or if conditions change. This is exactly what CED 8.3.A.3–8.3.A.4 expects you to use on the exam—use Kb + ICE or the Henderson–Hasselbalch form for buffer cases. For worked examples and practice problems, see the Topic 8.3 study guide (https://library.fiveable.me/ap-chemistry/unit-8/weak-acid-base-equilibria/study-guide/J3ggVgoQIyMYhq5nEKuw) and thousands of practice questions (https://library.fiveable.me/practice/ap-chemistry).
How do I know if I should use an ICE table for weak acid problems?
Use an ICE table whenever the reaction goes to equilibrium—that’s the usual case for a monoprotic weak acid or base. Practically: if you’re given a weak acid HA (with Ka) and an initial [HA]0, set up HA ⇌ H3O+ + A− I: [HA]0, 0, 0 C: −x, +x, +x E: [HA]0 − x, x, x and solve Ka = x^2/( [HA]0 − x ). Do the algebra unless you can justify an approximation. Common shortcuts AP expects you to know: - If x/[HA]0 < 5% (or Ka ≤ 10−2 × [HA]0 roughly), you can approximate [HA] ≈ [HA]0 and use x ≈ sqrt(Ka[HA]0). - If percent ionization is large (>5%) or Ka is similar to [HA]0, you must solve the quadratic (no “x≈0” step). - For buffers or conjugate pairs use Henderson–Hasselbalch instead of a full ICE. For more worked examples and AP-aligned practice, see the Topic 8.3 study guide (https://library.fiveable.me/ap-chemistry/unit-8/weak-acid-base-equilibria/study-guide/J3ggVgoQIyMYhq5nEKuw) and Unit 8 resources (https://library.fiveable.me/ap-chemistry/unit-8). Need practice? Try problems at (https://library.fiveable.me/practice/ap-chemistry).
What does it mean when they say "only a small percentage of weak acid molecules ionize"?
It means that for a weak acid HA in water, only a small fraction of the original HA molecules break apart into H3O+ and A− at equilibrium. If you start with [HA]0 (say 0.10 M), the equilibrium [H3O+] will be much smaller than 0.10 M (often only a few percent or less)—most HA stays as the un-ionized molecule. Mathematically: Ka = [H3O+][A−]/[HA], and percent ionization = ([H3O+]eq / [HA]0) × 100%. For weak acids Ka is small, so [H3O+]eq << [HA]0; that’s why we often use the “x is small” approximation in ICE tables (acceptable when x < ~5%). This affects pH: you must solve an equilibrium (not just -log[acid] like a strong acid). Practicing these ICE/percent-ionization problems is exactly what AP Topic 8.3 expects—see the Topic 8.3 study guide for examples (https://library.fiveable.me/ap-chemistry/unit-8/weak-acid-base-equilibria/study-guide/J3ggVgoQIyMYhq5nEKuw) and more practice problems at (https://library.fiveable.me/practice/ap-chemistry).
Can someone explain the difference between Kb and pKb in simple terms?
Kb is the base dissociation constant—a number that tells you how strongly a weak base B reacts with water: Kb = [OH−][HB+]/[B]. A large Kb (>>10−7) means the base ionizes more (stronger); a small Kb means it ionizes less (weaker). pKb is just the negative base-10 log of Kb: pKb = −log Kb. That makes large Kb → small pKb and small Kb → large pKb. pKb is convenient because p-values are easier to compare and plug into calculations (like pKw = pKa + pKb). On the AP test you’ll use Kb (or pKb) plus the initial base concentration in an ICE table to find [OH−], then pOH and pH (learning objective 8.3.A). pKb is especially handy when using log math or comparing strengths. For a quick topic review, check the Fiveable weak-acid/base study guide (https://library.fiveable.me/ap-chemistry/unit-8/weak-acid-base-equilibria/study-guide/J3ggVgoQIyMYhq5nEKuw) and try practice problems at (https://library.fiveable.me/practice/ap-chemistry).
I don't understand why we need both Ka and Kb - what's the point?
You need both Ka and Kb because they describe the equilibrium behavior of different but related reactions—one for an acid losing a proton, one for a base gaining (or accepting) a proton—and each lets you calculate pH (or pOH) from initial concentrations. Ka = [H3O+][A−]/[HA] applies directly to a weak acid (only a small % ionizes, per CED 8.3.A.1–A.2), while Kb = [OH−][HB+]/[B] applies to a weak base (CED 8.3.A.3–A.4). For any conjugate pair Ka × Kb = Kw, so knowing one gives you the other (CED 8.3.A.6). Practically, use Ka/pKa when you start with HA, Kb/pKb when you start with B, and convert when you have the conjugate (useful for hydrolysis, percent ionization, and buffer math/Henderson–Hasselbalch). On the AP exam you’ll be expected to pick the right constant, set up an ICE table, and relate pH/pOH to species concentrations (Topic 8.3). For a focused review see the Topic 8.3 study guide (https://library.fiveable.me/ap-chemistry/unit-8/weak-acid-base-equilibria/study-guide/J3ggVgoQIyMYhq5nEKuw), the Unit 8 overview (https://library.fiveable.me/ap-chemistry/unit-8), and extra practice problems (https://library.fiveable.me/practice/ap-chemistry).
How do you calculate the pH of a weak base like ammonia?
Use the base ionization equilibrium and Kb with an ICE table, then convert [OH−] to pOH and pH. Steps (monoprotic weak base NH3 example): 1. Write equilibrium: NH3 + H2O ⇌ NH4+ + OH− 2. ICE (start with c0 = initial [NH3], 0 for products): I: [NH3] = c0, [NH4+] = 0, [OH−] = 0 C: −x, +x, +x E: [NH3] = c0 − x, [NH4+] = x, [OH−] = x 3. Use Kb: Kb = [NH4+][OH−]/[NH3] = x^2/(c0 − x). For most weak bases x ≪ c0, so approximate x ≈ sqrt(Kb · c0). 4. Calculate pOH = −log[OH−]; then pH = 14.00 − pOH (use pKw = 14 at 25°C). Worked example: c0 = 0.100 M NH3, Kb = 1.8 × 10−5 x ≈ sqrt(1.8×10−5 × 0.100) = sqrt(1.8×10−6) = 1.34×10−3 M = [OH−] pOH = −log(1.34×10−3) = 2.87 → pH = 14.00 − 2.87 = 11.13 Check: x/c0 = 1.34×10−3 / 0.100 = 1.34% (approximation valid). This follows the CED essentials for weak bases (Kb, pKb, pOH) in Topic 8.3. For more review and practice problems, see the Topic 8.3 study guide (https://library.fiveable.me/ap-chemistry/unit-8/weak-acid-base-equilibria/study-guide/J3ggVgoQIyMYhq5nEKuw) and Unit 8 resources (https://library.fiveable.me/ap-chemistry/unit-8). Fiveable also has lots of practice problems if you want extra drills (https://library.fiveable.me/practice/ap-chemistry).
What's the connection between Kw, Ka, and Kb that I keep seeing in problems?
Think of Kw, Ka and Kb as the same equilibrium family talking about different reactions. - Kw is water’s ion-product: Kw = [H3O+][OH−]. At 25°C Kw = 1.0×10−14 (pKw = 14). - Ka is acid dissociation for HA: Ka = [H3O+][A−]/[HA]. - Kb is base hydrolysis for A−: Kb = [OH−][HA]/[A−]. If you multiply the Ka and Kb expressions the [HA] and [A−] cancel and you get Ka·Kb = [H3O+][OH−] = Kw. So Ka × Kb = Kw and pKa + pKb = pKw. Practically: if you know Ka for a weak acid, Kb for its conjugate base = Kw/Ka (and vice-versa). That’s how you switch between pH (acid) and pOH (base) problems for conjugate pairs on the AP exam (CED equations listed). For more worked examples, see the Topic 8.3 study guide (https://library.fiveable.me/ap-chemistry/unit-8/weak-acid-base-equilibria/study-guide/J3ggVgoQIyMYhq5nEKuw) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
Why do weak acids and bases reach equilibrium instead of going to completion?
Because the acid (or base) and its conjugate are close enough in stability that proton transfer is reversible, only some molecules ionize and the system settles where the forward and reverse rates are equal—an equilibrium. Weak acids have small Ka values (Ka << 1), so [H3O+] is much less than the initial [HA] and most HA remains un-ionized. Mathematically Ka = [H3O+][A−]/[HA]; a small Ka means the numerator stays small compared to [HA]. The position of that equilibrium (and the percent ionization) depends on Ka (or pKa) and the initial concentration—that’s why you use an ICE table or the Henderson–Hasselbalch relation to find pH from [HA] and pKa, as required by the CED (8.3.A.1–A.2). For more practice and worked examples on percent ionization, ICE tables, and pH calculations, see the Topic 8.3 study guide (https://library.fiveable.me/ap-chemistry/unit-8/weak-acid-base-equilibria/study-guide/J3ggVgoQIyMYhq5nEKuw) and the AP practice problem bank (https://library.fiveable.me/practice/ap-chemistry).