Types of Acids and Bases
Arrhenius Definition
There are two main schools of thought for what should be the definitions of acids and bases. The Arrhenius definition categorizes an acid as any compound that increases the concentration of hydrogen ions ([H+]) in a solution. Conversely, the Arrhenius definition of a base is any compound that increases the concentration of hydroxide ions ([OH-]) in solution. Essentially, the Arrhenius definition of an acid/base is anything that yields H+ or OH- respectively in water.
For example, HCl (hydrochloric acid) can be described as an Arrhenius acid because, in water, the following reaction occurs:
HCl → H+ + Cl-
Thus, HCl yields an H+ ion in water, making it an Arrhenius acid.
The concentrations of hydronium ions and hydroxide ions are often reported as pH and pOH, respectively:
- pH = −log[H3O+]
- pOH = −log[OH−]
Let's see how this works! If we have a solution where [H3O+] = 1.0 × 10⁻⁵ M, then: pH = −log(1.0 × 10⁻⁵) = 5.0
Since we know Kw = [H3O+][OH−] = 1.0 × 10⁻¹⁴, we can find [OH−]: [OH−] = Kw/[H3O+] = (1.0 × 10⁻¹⁴)/(1.0 × 10⁻⁵) = 1.0 × 10⁻⁹ M
Therefore: pOH = −log(1.0 × 10⁻⁹) = 9.0
pH of Water
Water autoionizes, meaning a proton is transferred from one water molecule to another to produce a hydronium ion (H₃O⁺) and a hydroxide ion (OH⁻), with the equilibrium constant Kw:
Kw = [H3O+][OH−] = 1.0 × 10−14 at 25°C
In pure water, pH = pOH, making it a neutral solution because pH = pOH = 7.0. There's a really useful relationship here: since Kw = 1.0 × 10⁻¹⁴ at 25°C, we can take the negative log of both sides to get:
pKw = 14 = pH + pOH at 25°C
This means that in any aqueous solution at 25°C, the sum of pH and pOH will always equal 14! However, the value of Kw is temperature dependent, so the pH of pure, neutral water deviates from 7.0 at temperatures other than 25°C.
Brønsted-Lowry Definition
Meanwhile, the Brønsted-Lowry definition of acids and bases defines acids and bases in the form of a donation reaction. Basically, an acid/base is seen as an H+ donator or accepter (the acid donates, the base accepts):
HA + B- → HB + A-
In this example, HA is the acid, which donates an H+ ion to the B- ion (the base) to form HB and A-.
Get used to seeing HA and B- as sample acids and bases; it is a common notation! A and B could be stand-ins for any number of ions.
Quick thinkers may be wondering about the reverse reaction in the case that HA + B- is an equilibrium reaction:
HA + B- ⇄ HB + A-
The Hydronium Ion
A consequence of Bronsted Acids and Bases is that when an acid dissolves in water, it needs something to donate its H+ to! In this case, it donates it to water, creating H3O+ or hydronium. Thus, we can see the dissolution of an acid both in an Arrhenius sense as HA <--> H+ + A- and a Bronsted sense as HA + H2O <--> H3O+ + A-. Both of these essentially mean the same thing (and when we learn pH, we'll find that [H+] = [H3O+]), but they display the difference between Arrhenius and Bronsted Acids.
By the way, you'll see both "hydrogen ion" and "hydronium ion" used in chemistry, along with the symbols H+(aq) and H3O+(aq). They're often used interchangeably because they represent the same thing - a proton in water. While H3O+(aq) is technically more accurate (since bare protons don't exist in water), both H+(aq) and H3O+(aq) are accepted on the AP Exam, so don't stress about which one to use!
Conjugate Acids and Bases
How do we classify HB and A- then? Would A- be a base and HB be an acid if the reaction were reversed? The answer is short: yes. However, they are given a special name: HB is called the conjugate acid of B-, and A- is called the conjugate base of HA.
In the case that HA is a weak acid and B- is a weak base, HB and A- would be considered to have significant acidity/basicity. The conjugate acid/base of a strong base/acid does not have acidity/basicity. The weaker the acid/base, the stronger the conjugate base/acid.
Vocabulary
The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.
| Term | Definition |
|---|---|
| autoionization | The process by which water molecules react with each other to produce hydronium and hydroxide ions in equilibrium. |
| 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. |
| 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. |
| neutral solution | An aqueous solution in which pH = pOH = 7.0 at 25°C, meaning the concentrations of hydronium and hydroxide ions are equal. |
| pH | A logarithmic scale used to express the concentration of hydronium ions in a solution, calculated as −log[H3O+]. |
| 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−]. |
Frequently Asked Questions
What's the difference between pH and pOH?
pH and pOH are just different ways to report the concentrations of the two water ions. pH = −log[H3O+] (sometimes written −log[H+]) and pOH = −log[OH−]. At 25°C water’s ion-product Kw = [H3O+][OH−] = 1.0×10−14, so pKw = 14 and pH + pOH = 14. That means pure water (neutral) has pH = pOH = 7.0 at 25°C. If you know one, you can get the other: pOH = 14 − pH (at 25°C). Remember Kw changes with temperature, so neutral water won’t be pH 7 at other temperatures. On the AP exam you’ll be asked to calculate pH and pOH from [H3O+] or [OH−] and from Kw (CED 8.1.A). For extra practice and a quick topic review see the Topic 8.1 study guide (https://library.fiveable.me/ap-chemistry/unit-8/intro-acids-bases/study-guide/rG2ZBgD9evdBohz4a7Mj) and the Unit 8 page (https://library.fiveable.me/ap-chemistry/unit-8); over 1,000 practice problems are at (https://library.fiveable.me/practice/ap-chemistry).
How do you calculate pH from hydronium ion concentration?
Use the definition: pH = −log[H3O+]. Plug the hydronium concentration (in mol·L−1) into a base-10 logarithm and take the negative. Quick examples: - If [H3O+] = 1.0 × 10−3 M, pH = −log(1.0×10−3) = 3.00. - If [H3O+] = 2.5 × 10−6 M, pH = −log(2.5×10−6) ≈ 5.60 (use your calculator). Notes useful for the AP exam (Topic 8.1): use H3O+ and H+(aq) interchangeably, and remember at 25°C Kw = [H3O+][OH−] = 1.0×10−14 so pH + pOH = 14.00 for neutral water (pH = 7.00). For significant figures, the digits after the decimal in pH reflect log precision from the mantissa—keep one extra digit on the concentration when calculating pH, then report pH to the correct sig figs. For more review, see the Topic 8.1 study guide (https://library.fiveable.me/ap-chemistry/unit-8/intro-acids-bases/study-guide/rG2ZBgD9evdBohz4a7Mj) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
What does it mean when water autoionizes?
Autoionization (or self-ionization) of water means H2O molecules react with each other to form hydronium and hydroxide ions at equilibrium: 2 H2O(l) ⇌ H3O+(aq) + OH−(aq) At 25°C this process gives Kw = [H3O+][OH−] = 1.0 × 10−14, so pure water has [H3O+] = [OH−] = 1.0 × 10−7 M and pH = pOH = 7.0 (pH = −log[H3O+], pOH = −log[OH−], pKw = 14). Autoionization is a very small but important equilibrium that sets the baseline for what “neutral” means and is used in many AP problems (Topic 8.1) when you calculate pH/pOH from Kw. Kw changes with temperature, so neutral pH shifts away from 7.0 at other temperatures. For a focused review and practice, see the Topic 8.1 study guide (https://library.fiveable.me/ap-chemistry/unit-8/intro-acids-bases/study-guide/rG2ZBgD9evdBohz4a7Mj) and more practice problems at (https://library.fiveable.me/practice/ap-chemistry).
I'm confused about why pure water has a pH of 7 - can someone explain this?
Pure water “autoionizes”: two H2O molecules make H3O+ and OH− in equilibrium: 2 H2O ⇌ H3O+ + OH−. At 25°C the ion-product constant Kw = [H3O+][OH−] = 1.0×10−14. In pure water there’s no other source of acid or base, so [H3O+] = [OH−]. Let that common concentration = x, so x^2 = 1.0×10−14 → x = 1.0×10−7 M. pH is defined as −log[H3O+], so pH = −log(1.0×10−7) = 7.0. Because pH = pOH at neutrality (pKw = 14 at 25°C), pure water is “neutral” at pH 7 only at that temperature. Kw changes with temperature, so pure water’s pH shifts away from 7 at other temperatures (CED 8.1.A). For a quick topic review see the Topic 8.1 study guide (https://library.fiveable.me/ap-chemistry/unit-8/intro-acids-bases/study-guide/rG2ZBgD9evdBohz4a7Mj) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).
Why do we use negative log for pH calculations instead of just the regular concentration?
We use pH = −log[H3O+] because hydronium concentrations are tiny and span many orders of magnitude, so the negative logarithm turns very small numbers into easy-to-use, comparable values. For example, [H3O+] = 1.0×10−7 M becomes pH 7, and [H3O+] = 1.0×10−2 M becomes pH 2—much simpler to read and compare than the raw concentrations. The negative sign makes acids (higher [H3O+]) give smaller pH numbers, which matches how we talk about acidity. Logarithms also convert multiplicative relationships (like Kw = [H3O+][OH−]) into additive ones (pKw = pH + pOH), so at 25°C pKw = 14.00 gives pH + pOH = 14. That’s why pH/pOH are used in AP problems (CED 8.1.A: pH = −log[H3O+], Kw = 1.0×10−14 at 25°C). For more practice and review, see the Topic 8.1 study guide (https://library.fiveable.me/ap-chemistry/unit-8/intro-acids-bases/study-guide/rG2ZBgD9evdBohz4a7Mj) and the AP practice pool (https://library.fiveable.me/practice/ap-chemistry).
What's the difference between H+ and H3O+ ions?
H+ and H3O+ are often used interchangeably on the AP exam, but they’re not exactly the same chemically. H+ is a bare proton—it can’t exist freely in water because water’s lone pairs grab it. In aqueous solutions the proton is solvated, most simply as hydronium, H3O+(aq), or even more realistically as H5O2+ or larger proton-water clusters. For AP work use H3O+(aq) (preferred) or H+(aq)—they mean the same thing for pH/pOH calculations. Remember pH = −log[H3O+] and Kw = [H3O+][OH−] = 1.0×10^−14 at 25°C (CED 8.1.A.1–A.2). Practically: when you write equilibrium or autoionization of water, show H3O+ (or H+(aq)) to be safe. For a quick review, see the Topic 8.1 study guide on Fiveable (https://library.fiveable.me/ap-chemistry/unit-8/intro-acids-bases/study-guide/rG2ZBgD9evdBohz4a7Mj) and try practice problems at (https://library.fiveable.me/practice/ap-chemistry).
How do I remember the equation for Kw?
Think of Kw as “K for water”—the equilibrium constant for water autoionization. Write the autoionization: 2 H2O ⇌ H3O+ + OH−, then Kw = [H3O+][OH−]. At 25°C Kw = 1.0 × 10−14, so in pure water [H3O+] = [OH−] = √Kw = 1.0 × 10−7 → pH = −log[H3O+] = 7 (hence pKw = 14 and pH + pOH = 14 at 25°C). Memory tricks: (1) “H times OH” = Kw, (2) remember 10−14 → neutral pH 7, (3) convert to p-terms: pKw = 14 = pH + pOH. Always note AP language: H3O+ preferred but H+ accepted. Kw changes with temperature, so pKw ≠ 14 except at 25°C. For extra practice and quick review tied to the CED, check the Topic 8.1 study guide on Fiveable (https://library.fiveable.me/ap-chemistry/unit-8/intro-acids-bases/study-guide/rG2ZBgD9evdBohz4a7Mj) and hit the practice problems (https://library.fiveable.me/practice/ap-chemistry).
Why does the pH of pure water change when temperature changes?
Pure water’s pH changes with temperature because the water autoionization equilibrium (Kw = [H3O+][OH−]) is temperature dependent. Kw increases as temperature rises (autoionization is endothermic), so more H3O+ and OH− form. In pure water [H3O+] = [OH−] = √Kw, so pH = −log[H3O+] = −½ log Kw = ½·pKw. At 25°C Kw = 1.0×10−14 (pKw = 14), giving pH = 7.0, but at higher T Kw is larger (pKw smaller) so neutral pH drops below 7; at lower T Kw is smaller so neutral pH rises above 7. This is exactly why the CED notes that neutral water only equals pH 7 at 25°C (see 8.1.A.2–8.1.A.4). For a quick review of these ideas check the Topic 8.1 study guide (https://library.fiveable.me/ap-chemistry/unit-8/intro-acids-bases/study-guide/rG2ZBgD9evdBohz4a7Mj) and more practice on Fiveable (https://library.fiveable.me/practice/ap-chemistry).
What happens to pH and pOH values when you add them together?
If you add pH and pOH you get pKw—at 25°C pKw = 14, so pH + pOH = 14. That’s why neutral water at 25°C has pH = pOH = 7. More generally pH = −log[H3O+] and pOH = −log[OH−], and since Kw = [H3O+][OH−], taking −log gives pH + pOH = pKw. Remember Kw (and thus pKw) depends on temperature, so pH + pOH will still equal pKw but not always 14 if you’re not at 25°C. On the AP exam you should state pH + pOH = pKw and note the 25°C reference (CED 8.1.A.2–3). For a quick review, see the Topic 8.1 study guide (https://library.fiveable.me/ap-chemistry/unit-8/intro-acids-bases/study-guide/rG2ZBgD9evdBohz4a7Mj) and more practice problems at (https://library.fiveable.me/practice/ap-chemistry).
Can someone explain what pKw means and why it equals 14?
pKw is just the negative base-10 log of the ion-product constant of water: pKw = −log(Kw). At 25°C Kw = [H3O+][OH−] = 1.0×10−14, so pKw = −log(1.0×10−14) = 14. That’s why on the AP exam you’ll see pKw = 14 at 25°C and the neutral water condition gives pH = pOH = 7 (because pH + pOH = pKw). Important: Kw (and therefore pKw) changes with temperature, so pure water isn’t always pH 7 at other temperatures—only at 25°C. For AP Chem, use Kw = 1.0×10−14 and pKw = 14 when the problem states 25°C (CED EK 8.1.A.2–A.3). If you want a quick review or practice problems on this topic, check the Topic 8.1 study guide (https://library.fiveable.me/ap-chemistry/unit-8/intro-acids-bases/study-guide/rG2ZBgD9evdBohz4a7Mj) and Unit 8 resources (https://library.fiveable.me/ap-chemistry/unit-8). For more practice, try the AP problems collection (https://library.fiveable.me/practice/ap-chemistry).
How do you calculate pOH if you only know the hydroxide ion concentration?
Use pOH = −log[OH−]. That’s the direct definition from the CED (Topic 8.1). So if [OH−] = 3.2 × 10−4 M, pOH = −log(3.2 × 10−4) ≈ 3.49. If you need pH at 25°C, use pH + pOH = 14.00 (because Kw = 1.0×10−14 at 25°C), so pH = 14.00 − pOH. Remember: Kw is temperature dependent, so that 14.00 sum is only for 25°C—use pKw = −log Kw if you’re at another temperature. Keep sig figs appropriate: report pOH to the same number of decimal places as sig figs in the log input (usually matching the mantissa rules). For AP prep, that pOH formula and the pH+pOH relationship are exactly in the CED (Topic 8.1); see the Topic 8.1 study guide for a quick refresher (https://library.fiveable.me/ap-chemistry/unit-8/intro-acids-bases/study-guide/rG2ZBgD9evdBohz4a7Mj). For extra practice, try problems at (https://library.fiveable.me/practice/ap-chemistry).
I don't understand why Kw always equals 1.0 × 10^-14 at 25°C - what makes this constant?
Kw = [H3O+][OH−] is just the equilibrium constant for water autoionizing: 2 H2O ⇌ H3O+ + OH−. Its numerical value comes from thermodynamics and experiment—Kw = e^(−ΔG°/RT), where ΔG° (and therefore Kw) depends on temperature. At 25.0 °C the measured equilibrium of water gives Kw ≈ 1.0×10−14, so [H3O+]=[OH−]=1.0×10−7 M in pure water and pKw = 14.0 (thus pH = pOH = 7.0) as the CED states (8.1.A.2 and 8.1.A.3). If temperature changes, ΔG° changes and Kw shifts, so “neutral” pH won’t be exactly 7 at other temperatures (CED 8.1.A.4). For a quick refresher on these ideas and AP-style practice, see the Topic 8.1 study guide (https://library.fiveable.me/ap-chemistry/unit-8/intro-acids-bases/study-guide/rG2ZBgD9evdBohz4a7Mj) and extra problems (https://library.fiveable.me/practice/ap-chemistry).
What's the relationship between hydronium and hydroxide ion concentrations in neutral water?
In neutral pure water at 25°C the concentrations of hydronium and hydroxide are equal because water autoionizes: Kw = [H3O+][OH−] = 1.0×10−14. For neutrality [H3O+] = [OH−] so each = √Kw = 1.0×10−7 M. That gives pH = −log[H3O+] = 7.00 and pOH = −log[OH−] = 7.00, so pH = pOH and pKw = 14 (CED 8.1.A.2–3). Note Kw depends on temperature, so “neutral” pure water only has pH 7.00 at 25°C; at other temperatures [H3O+] and [OH−] remain equal but their numeric value (and pH) shifts. For a quick review of these ideas see the Topic 8.1 study guide (https://library.fiveable.me/ap-chemistry/unit-8/intro-acids-bases/study-guide/rG2ZBgD9evdBohz4a7Mj) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
Why is 25°C always used as the standard temperature for these calculations?
25°C is used because the AP CED defines Kw = 1.0×10−14 specifically at 25°C, which makes pKw = 14.0 and gives neutral water pH = pOH = 7.0. Using 25°C gives a standard, convenient benchmark so everyone uses the same Kw value and the nice round pKw = 14 in calculations (CED 8.1.A.2–A.3). Kw actually changes with temperature (CED 8.1.A.4), so pure water isn’t exactly pH 7.00 at other temperatures—you’d need the Kw for that temperature to get the correct pH/pOH. For AP problems, unless a different temperature or Kw is given, assume 25°C and Kw = 1.0×10−14. For a quick refresher, see the Topic 8.1 study guide (https://library.fiveable.me/ap-chemistry/unit-8/intro-acids-bases/study-guide/rG2ZBgD9evdBohz4a7Mj) and try practice questions (https://library.fiveable.me/practice/ap-chemistry).
How do I solve problems where I need to find both pH and pOH from just one ion concentration?
If you’re given one ion concentration, use the definitions and Kw. Steps at 25°C: - If you’re given [H3O+]: pH = −log[H3O+]. Then pOH = 14.00 − pH (because pKw = 14 at 25°C). If you need [OH−], use [OH−] = Kw / [H3O+] = (1.0×10−14)/[H3O+]. - If you’re given [OH−]: pOH = −log[OH−]. Then pH = 14.00 − pOH. And [H3O+] = Kw / [OH−]. Keep units straight (M), use a calculator for logs, and remember Kw changes with temperature so pKw ≠ 14 away from 25°C (then use pKw = −log Kw). This is exactly what Topic 8.1 expects (pH, pOH, Kw relationships). For a quick refresher, see the Topic 8.1 study guide (https://library.fiveable.me/ap-chemistry/unit-8/intro-acids-bases/study-guide/rG2ZBgD9evdBohz4a7Mj) and more practice problems at (https://library.fiveable.me/practice/ap-chemistry).