MCAT content library / Chemistry & physics / Acid–base strength: distinguish concentration from dissociation
Follow the proton, compare the correct equilibrium constants, and keep the acid’s strength separate from the acidity of a particular sample.
About 15 minutes, including the worked examples and checks.
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Concept: identify the proton transfer
A Brønsted–Lowry acid donates a proton in a specified reaction; a base accepts it. After donation, the acid becomes its conjugate base. After acceptance, the base becomes its conjugate acid. Each conjugate pair differs by exactly one H+, so its members also differ in charge by one unit.2 These relationships, acid–base strength, water ionization, and pH are included in AAMC category 5A.1
For NH3 + H2O ⇌ NH4+ + OH−, ammonia accepts a proton and water donates one. The pairs are NH4+/NH3 and H2O/OH−. Pairing ammonia with hydroxide merely because both can act as bases would miss the chemical transformation.
Charge alone does not assign a role. Neutral ammonia can accept a proton; positively charged ammonium can donate one. A species can also have both roles in different reactions. HCO3− can gain H+ to become H2CO3 or lose H+ to become CO32−. “Acid” and “base” describe the transfer under consideration, rather than simply the sign printed beside a formula.
Pattern: keep strength, concentration, and pH separate
Strength describes the tendency to transfer a proton in a stated solvent. Concentration describes how much of a substance is present per volume. For a strong monoprotic acid in dilute water, proton transfer is essentially complete. A weak acid establishes an equilibrium with appreciable undissociated acid. Its fraction ionized depends on concentration as well as its equilibrium constant.3
For HA + H2O ⇌ H3O+ + A−, the usual concentration form is Ka = [H3O+][A−]/[HA]. All bracketed quantities are equilibrium concentrations. The analytical acid concentration counts both forms, C = [HA] + [A−]; it is not automatically the remaining [HA] or the hydronium concentration.
Use water at 25°C and ideal dilute-solution approximations for the numerical examples here. Then Kw ≈ [H3O+][OH−] = 1.0 × 10−14, and pH ≈ −log[H3O+] when the concentration is entered in mol/L. The logarithm is base ten. A tenfold increase in hydronium lowers pH by one unit.4
Neutrality means equal hydronium and hydroxide concentrations. At 25°C they are each about 10−7 M in neutral water, giving pH 7.00. The number seven is temperature-dependent because Kw changes with temperature. An acidic solution still contains hydroxide; it contains less hydroxide than hydronium. At pH 4.00, for example, [H3O+] ≈ 10−4 M and [OH−] ≈ 10−10 M.
Application: compare equal concentrations, then equal pH
The following are original illustrative equilibrium values for monoprotic acids, with no other appreciable source of hydronium. Treat the tabulated numbers as rounded ideal-solution values.
| Solution | Analytical acid concentration | Equilibrium [H3O+] | pH |
|---|---|---|---|
| A: strong acid | 0.0100 M | 0.0100 M | 2.00 |
| B: weak acid HQ | 0.0100 M | 0.000100 M | 4.00 |
| C: strong acid | 0.000100 M | 0.000100 M | 4.00 |
| D: strong acid | 0.0000100 M | 0.0000100 M | 5.00 |
A and B start with the same analytical concentration. Their hydronium concentrations differ one hundredfold because their proton-transfer equilibria differ. In B, about 0.000100 M has become Q−, leaving 0.00990 M as HQ. Only about 1% of the dissolved acid has ionized. These values imply Ka ≈ (0.000100)2/0.00990 ≈ 1.01 × 10−6.
B and C have the same pH but very different acid inventories. C contains about one hundredth as much analytical acid, nearly all transferred to water. B contains a much larger reservoir of undissociated HQ. Equal pH therefore establishes equal hydronium activity under matched conditions, not equal acid identity, concentration, or response to later neutralization.
B has a lower pH than D even though D contains a strong acid. More extensive ionization of a much smaller starting amount can still produce less hydronium. A pH reading describes the sample in front of you; by itself, it is not a label for the acid’s intrinsic strength.
Rule: compare constants for the correct conjugate partners
A larger Ka means a stronger acid in the same solvent at the same temperature. Because pKa = −log Ka, a smaller pKa means a stronger acid. For a particular conjugate pair, Ka(HA)Kb(A−) = Kw. This inverse relation makes the stronger acid’s conjugate base weaker; it does not make every weak acid’s conjugate base a strong base.3
| Acid | pKa | Ka | Conjugate base Kb |
|---|---|---|---|
| HX | 4.0 | 10−4 | X−: 10−10 |
| HY | 7.0 | 10−7 | Y−: 10−7 |
| HZ | 10.0 | 10−10 | Z−: 10−4 |
HX is the strongest of these acids, while Z− is the strongest of these bases. Notice that Ka(HX) must be paired with Kb(X−), not Kb of whichever base happens to be mixed with HX.
For HX + Y− ⇌ X− + HY, combine the dissociation of HX with the reverse dissociation of HY. Their hydronium terms cancel, giving K = Ka(HX)/Ka(HY) = 10−4/10−7 = 103. Equivalently, K = 10pKa(HY) − pKa(HX). The equilibrium favors products containing the weaker acid and weaker base.
“Favors” does not mean that every reactant molecule disappears. Nor does K alone determine the current direction of change for an arbitrary mixture; the existing amounts also matter. The useful prediction here is the relative equilibrium preference. Before calculating, write the two conjugate pairs and identify which acid is on each side. That keeps the ratio and the pKa subtraction in the right direction.
ACTIVE PRACTICE · ACID–BASE CHEMISTRY
Compare Equal pH With Different Acid Inventories
Use the lesson’s original ideal 25°C examples. A strong monoprotic acid is treated as fully dissociated at the listed dilute concentrations; weak acid HQ has Ka ≈ 1.01 × 10⁻⁶ and analytical concentration initially 0.0100 M. The simplified weak-acid equilibrium neglects water autoionization where the acid dominates.
Analytical acid concentration: M.
Calculated equilibrium [H₃O⁺]: M; approximate pH .
Fraction of analytical acid ionized:. Undissociated acid remaining: M.
The numerical display uses the exact quadratic root x = (−Ka + √(Ka² + 4KaC))/2 for weak HQ, with negligible water contribution. At these concentration ranges this approximation remains appropriate. A strong-acid concentration is modeled as its hydronium concentration where it dominates water.
Check your reasoning
Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.
Check 1
Consider H2PO4− + NH3 ⇌ HPO42− + NH4+. Which listed pair consists of a conjugate acid and its conjugate base?
A. H₂PO₄⁻ and HPO₄²⁻
Correct. Removing one H⁺ from H₂PO₄⁻ leaves HPO₄²⁻. Both the hydrogen count and the one-unit change in charge fit a conjugate pair.
B. H₂PO₄⁻ and NH₃
Review this choice. These are the proton donor and acceptor in the forward reaction. They do not become each other by gaining or losing one proton.
C. HPO₄²⁻ and NH₄⁺
Review this choice. These are the two products, but their atomic frameworks differ. Membership on the same side of an equation does not define a conjugate pair.
D. NH₃ and HPO₄²⁻
Review this choice. Both can act as proton acceptors, but they are not conjugate partners. Ammonia’s partner is NH₄⁺.
The principle to carry forward
Match the same chemical framework before and after a one-proton transfer; roles or positions alone do not identify conjugate partners.
Check 2
Two hypothetical acids in the same solvent at the same temperature have pKa(HA) = 3.0 and pKa(HB) = 6.0. What is the equilibrium constant for HA + B− ⇌ A− + HB?
A. 0.001
Review this choice. This is the reciprocal constant, corresponding to the reverse reaction. The acid on the reactant side has the larger Kₐ.
B. 0.010
Review this choice. The pKₐ values differ by three, not two, units, and the forward reaction favors products. Use Kₐ(HA)/Kₐ(HB).
C. 100
Review this choice. This would correspond to a two-unit pKₐ difference. A three-unit difference means a factor of one thousand.
D. 1000
Correct. K = 10^(6.0 − 3.0) = 1000. Equivalently, 10⁻³/10⁻⁶ = 10³. The proton-transfer equilibrium favors the weaker acid HB and the weaker base A⁻.
The principle to carry forward
The acid on the product side minus the acid on the reactant side gives the pKₐ exponent for this proton-transfer equilibrium.
Check 3
In ideal aqueous solutions at 25°C, sample P contains a strong monoprotic acid and has pH 5.0. Sample Q contains a more concentrated weak monoprotic acid and has pH 3.5. Which interpretation is supported?
A. P has more hydronium per liter because a strong acid transfers its protons more completely.
Review this choice. Complete ionization describes a fraction, not the total amount available. The lower pH of Q means Q has more hydronium per liter under the stated approximation.
B. Q has more hydronium per liter even though P contains the acid with the greater proton-transfer tendency.
Correct. A pH difference of 1.5 means Q has about 10^1.5, or 32, times more hydronium. Its larger analytical concentration can produce this result despite weaker acid strength.
C. Q must be almost fully ionized because its lower pH establishes a larger fraction converted to ions.
Review this choice. The ionized fraction requires comparison with the analytical acid concentration. Lower pH alone reports more hydronium, not a larger ionized fraction.
D. The acids have equal strength because both samples have reached equilibrium at the same temperature.
Review this choice. Equilibrium does not make different acids’ equilibrium constants equal. Temperature must be matched for a comparison, but the chemical identities still matter.
The principle to carry forward
Compare pH to describe the samples’ hydronium levels; compare appropriate equilibrium constants to describe acid strength.
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From solutions to buffers
Build a quantitative acid–base explanation and apply it to amino acid charge.
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- Aqueous ions and concentration: keep amount and volume separate
- Acid–base strength: distinguish concentration from dissociation · You are here
- Weak acids, bases, and salts: choose a defensible pH calculation
- Buffers: explain pH resistance and its limits
- Amino acid and peptide charge: count the groups that can change
Related lessons
- Aqueous ions and concentration: keep amount and volume separate
- Weak acids, bases, and salts: choose a defensible pH calculation
- Buffers: explain pH resistance and its limits
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Sources
2. OpenStax Chemistry 2e — Brønsted–Lowry Acids and Bases
3. OpenStax Chemistry 2e — Relative Strengths of Acids and Bases