Two things decide acid strength
An acid HA gives a proton to water: HA + H₂O ⇌ H₃O⁺ + A⁻. The easier this step, the larger Ka. Look at two things:
- Whether the H–A bond breaks easily: the weaker and more polar the bond, the more easily H⁺ leaves.
- Whether the conjugate base A⁻ is stable: the more the negative charge is spread out (over a larger atom, over several atoms by resonance, or pulled away by nearby electron-withdrawing atoms), the more stable A⁻ is, and the stronger the acid.
For bases, look at the ability to accept H⁺: the more concentrated and available the lone pair, the stronger the base.
Binary acids: bond length within a group, electronegativity across a period
Going down a group, atoms get larger and the H–X bond gets longer and weaker. The H–F bond enthalpy is 567 kJ/mol, while H–Cl is only 431 kJ/mol (average bond enthalpies from Brown et al., Chemistry: The Central Science), and the charge on Cl⁻ is spread over a larger volume. So HF is a weak acid, while HCl, HBr and HI are strong; likewise H₂S < H₂Se < H₂Te.
Going across a period, the atoms are similar in size, but the electronegativity of the central atom increases, so the H–X bond becomes more polar and X⁻ holds the negative charge better. So CH₄ < NH₃ < H₂O < HF.
Oxoacids and carboxylic acids
In an oxoacid, the H that is given up is bonded to O: H–O–X.
- The more electronegative the central atom X, the stronger the acid. X pulls electrons away from the O–H bond, so the bond breaks more easily and XO⁻ is more stable.
- With the same central atom, the more O atoms not bonded to H, the stronger the acid. For example, HNO₂ < HNO₃ and H₂SO₃ < H₂SO₄. These O atoms withdraw electrons, and resonance共振 spreads the negative charge of the anion over more O atoms.
In the conjugate base of a carboxylic acid羧酸 (R–COOH), resonance spreads the negative charge of R–COO⁻ equally over two O atoms, so carboxylic acids are much more acidic than alcohols. The group attached to the carboxyl group matters too: the electron-donating CH₃ makes CH₃COOH (Ka = 1.8 × 10⁻⁵) weaker than HCOOH (Ka = 1.8 × 10⁻⁴). This withdrawing or donating of electrons through bonds is the inductive effect诱导效应.
Basicity of amines
An amine is basic because of the lone pair on N. Alkyl groups (such as CH₃) donate electrons to N, making the lone pair bond to H⁺ more readily, so CH₃NH₂ is stronger than NH₃. In aniline, C₆H₅NH₂, resonance spreads the lone pair on N into the benzene ring, so aniline is a much weaker base than NH₃.
Choose a set of substances, compare their Ka (or Kb) values, then click each substance to see how its structure differs:
Example 1: why is HF weak but HCl strong?
F is more electronegative (3.98) than Cl (3.16), so the H–F bond is more polar. Why, then, is HF the weaker acid?
- What matters within a group. F and Cl are in the same group and differ greatly in size (covalent radii 57 pm and 102 pm, Cordero et al. 2008), so bond strength is the main factor.
- Bond strength. The H–F bond enthalpy, 567 kJ/mol, is much larger than that of H–Cl (431 kJ/mol), so the H–F bond is hard to break.
- Conjugate base. F⁻ is small, so its negative charge is concentrated; Cl⁻ is larger, so its charge is more spread out and it is more stable.
- Conclusion. The electronegativity trend applies to comparisons within a period; within a group, the longer and weaker the bond, the stronger the acid. So HF < HCl.
Example 2: why are carboxylic acids more acidic than alcohols?
Formic acid, HCOOH, has Ka = 1.8 × 10⁻⁴, while methanol, CH₃OH, barely ionizes in water. Both give up a proton from an O–H bond, so why is the difference so large?
- Write the conjugate bases. HCOOH → HCOO⁻; CH₃OH → CH₃O⁻.
- HCOO⁻ has resonance. The two C–O bonds are equivalent, and the negative charge is shared equally by the two O atoms, each carrying about half of it.
- CH₃O⁻ has no resonance. The whole negative charge sits on one O atom, which is much less stable.
- Conclusion. The more stable the conjugate base, the more readily HA gives up H⁺. The O of the C=O group also withdraws electrons, making the O–H bond more polar. So HCOOH is far more acidic than CH₃OH.
- HCl or HF
Same group: the H–Cl bond is weaker, and Cl⁻ is larger and more stable.
- H₂O or H₂S
Going down a group, the H–X bond gets weaker.
- HBrO or HIO
Br is more electronegative than I.
- HNO₂ or HNO₃
HNO₃ has one more O atom not bonded to H.
- H₂SO₄ or H₂SO₃
H₂SO₄ has one more O atom not bonded to H.
- HCOOH or CH₃COOH
CH₃ donates electrons, making CH₃COO⁻ less stable.
- NH₃ or H₂O (as acids)
Same period: O is more electronegative than N.
- CH₃OH or HCOOH
HCOO⁻ has resonance, which spreads out the negative charge.
Common mistakes
Mistake 1: “The more H atoms in a molecule, the stronger the acid.” Acid strength depends on how easily an H leaves, not on how many H atoms there are. H₃PO₄ has 3 H atoms but is much weaker than HNO₃, which has 1.
Mistake 2: “Higher electronegativity means a stronger binary acid, so HF is the strongest.” The electronegativity trend applies only within a period. Within a group, atomic size changes a lot and bond strength dominates, so HF is actually the weakest.
Mistake 3: “Acid strength depends on intermolecular forces such as London dispersion forces.” Ionization breaks the H–A bond within the molecule and forms a stable A⁻; intermolecular forces affect boiling point and solubility, not Ka.
Quick check
Which of these hydrides is the strongest acid?
Going down a group, the central atom gets larger and the H–X bond weaker, so acidity increases. H₂Te has Ka₁ = 2.3 × 10⁻³ (OpenStax), the largest of the four.
For an oxoacid of the form H–O–X, which change makes it a stronger acid?
The more electronegative X is, the more it pulls electrons from the O–H bond, and the more stable XO⁻ is. D has the opposite effect.
Ka for CH₃COOH is smaller than for HCOOH mainly because:
The inductive effect: CH₃ donates electrons, making the conjugate base less stable and the acid weaker. A, B and D don't determine Ka.
Which base is the weakest? (Kb: NH₃ 1.8 × 10⁻⁵, CH₃NH₂ 4.4 × 10⁻⁴, (CH₃)₂NH 5.9 × 10⁻⁴, C₆H₅NH₂ 4.3 × 10⁻¹⁰)
In aniline, resonance spreads the lone pair on N into the benzene ring, so it bonds to H⁺ less readily and Kb is smallest.
The acidity of the period 2 hydrides increases CH₄ < NH₃ < H₂O < HF mainly because:
Atoms in the same period are similar in size, so the more electronegative the central atom, the more polar the H–X bond and the better X⁻ holds the negative charge.
Which best explains why carboxylic acids are much more acidic than alcohols?
The more stable the conjugate base, the stronger the acid. In the conjugate base of an alcohol, RO⁻, the negative charge sits on one O atom.
AP-style free response
Methylamine, CH₃NH₂, has Kb = 4.4 × 10⁻⁴, and aniline, C₆H₅NH₂, has Kb = 4.3 × 10⁻¹⁰ (OpenStax, 25 °C).
(a) Write the equation for the reaction of methylamine with water, and identify the conjugate acid.
(b) Calculate the pH of 0.10 M methylamine. Explain why the approximation x ≈ √(Kb·c) can’t be used here.
(c) Calculate the pH of 0.10 M aniline.
(d) Explain the difference in Kb in terms of molecular structure.
(e) Which is the stronger acid, C₆H₅NH₃⁺ or CH₃NH₃⁺? Calculate Ka for C₆H₅NH₃⁺.
Model answer and scoring
(a) CH₃NH₂(aq) + H₂O(l) ⇌ CH₃NH₃⁺(aq) + OH⁻(aq); the conjugate acid is CH₃NH₃⁺.
(b) The approximation gives x ≈ √(4.4 × 10⁻⁵) = 6.6 × 10⁻³, so x ÷ c ≈ 6.6% > 5%, and it isn’t valid. Solving x² + 4.4 × 10⁻⁴x − 4.4 × 10⁻⁵ = 0 gives x = 6.4 × 10⁻³ M, so pOH = 2.19 and pH = 11.81.
(c) x ≈ √(4.3 × 10⁻¹⁰ × 0.10) = 6.6 × 10⁻⁶ M, so pOH = 5.18 and pH = 8.82.
(d) Both are basic because of the lone pair on N. In methylamine, CH₃ donates electrons to N, so the lone pair bonds to H⁺ readily; in aniline, resonance spreads the lone pair into the benzene ring, so it bonds to H⁺ less readily and Kb is much smaller.
(e) The weaker the base, the stronger its conjugate acid, so C₆H₅NH₃⁺ is the stronger acid. Ka = Kw ÷ Kb = 1.0 × 10⁻¹⁴ ÷ 4.3 × 10⁻¹⁰ = 2.3 × 10⁻⁵.
Scoring: (a) equation and conjugate acid; (b) approximation check, quadratic, pH; (c) pH; (d) both the electron-donating and the resonance factors; (e) choice and Ka.
Summary
- Acid strength depends on two things: whether the H–A bond breaks easily, and whether the conjugate base A⁻ is stable.
- Binary acids: going down a group the bond weakens and acidity increases; going across a period electronegativity increases and so does acidity.
- Oxoacids: the more electronegative the central atom and the more O atoms not bonded to H, the stronger the acid. The conjugate base of a carboxylic acid has resonance. Electron-donating groups make acids weaker and amines stronger; resonance that spreads out the lone pair makes amines weaker.
AP tip: When explaining acid strength, name the specific structural factor (electronegativity, bond strength, resonance, inductive effect) and say how it affects breaking the O–H (or H–X) bond or the stability of the conjugate base. Don’t use intermolecular forces, molar mass or the number of H atoms as explanations.