9.1 Brønsted-Lowry Acids/Bases, Measuring Acidity & Strengths

Brønsted-Lowry Acids and Bases

Quick Answer: A Brønsted-Lowry acid is any species that donates a proton (H+); a Brønsted-Lowry base is any species that accepts a proton. Every acid-base reaction transfers a proton from one side to the other, producing a conjugate base (what is left of the acid) and a conjugate acid (the base after it has gained H+).

The PA-CAT Bulletin of Information, rev. 20240815, lists Acids and Bases first among the ten chemistry subtopics in Table 5, and a released sample item hinges on the fact that ionization of a carboxylic acid increases water solubility because the conjugate base forms a salt. Master the proton-transfer frame before any math.

Conjugate Pairs

Consider ammonia reacting with water:

NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq)

• Base (accepts H+): NH3 → NH4+ (its conjugate acid)
• Acid (donates H+): H2O → OH− (its conjugate base)

The two members of a conjugate pair differ by exactly one H+. The weaker the acid, the stronger its conjugate base, and vice versa.

Acid vs Base Strength Is Proton-Affinity, Not Concentration

Do not confuse strong (complete dissociation) with concentrated (large amount per volume). 0.001 M HCl is dilute but still strong; 12 M acetic acid is concentrated but still weak.

Measuring Acidity: pH and pOH

The self-ionization of water gives the anchor equation:

Kw = [H+][OH−] = 1.0 × 10−14   (at 25 °C)

Taking −log of both sides yields pH + pOH = 14 at 25 °C, where pH = −log[H+] and pOH = −log[OH−].

Worked pH Example

A 0.025 M solution of nitric acid (strong acid, fully dissociates) is prepared. Find pH and pOH.

1. [H+] = 0.025 M (HNO3 → H+ + NO3−, 1:1)
2. pH = −log(0.025) = −log(2.5 × 10−2) = 2 − log 2.5 = 2 − 0.40 = 1.60
3. pOH = 14 − 1.60 = 12.40
4. Check: [OH−] = Kw / [H+] = 1.0 × 10−14 / 0.025 = 4.0 × 10−13 M; −log(4.0 × 10−13) ≈ 12.40 ✓

For weak acids, [H+] is not the initial concentration — solve using Ka.

Strong vs Weak: Ka and Kb

Strong acids you should memorize (PA-CAT expects you to recognize them):

• HCl, HBr, HI (hydrohalic, except HF)
• HNO3, HClO4
• H2SO4 (first proton only; second is weak)

Strong bases: Group 1 hydroxides (LiOH, NaOH, KOH, RbOH, CsOH) and the soluble Group 2 hydroxides Ca(OH)2, Sr(OH)2, Ba(OH)2.

For weak acids (e.g., acetic acid CH3COOH), dissociation is partial:

CH3COOH ⇌ H+ + CH3COO−      Ka = [H+][CH3COO−] / [CH3COOH] = 1.8 × 10−5

For weak bases (e.g., ammonia):

NH3 + H2O ⇌ NH4+ + OH−       Kb = [NH4+][OH−] / [NH3] = 1.8 × 10−5

The key relation for any conjugate pair:

Ka × Kb = Kw = 1.0 × 10−14      →      pKa + pKb = 14

Smaller pKa = stronger acid; larger pKa = weaker acid with a stronger conjugate base.

Estimating [H+] from Ka (small Ka, weak acid shortcut)

For a weak acid HA with initial concentration C and Ka, when Ka ≪ C:

[H+] ≈ √(Ka × C)

Example: 0.10 M acetic acid, Ka = 1.8 × 10−5 → [H+] ≈ √(1.8 × 10−6) ≈ 1.34 × 10−3 M → pH ≈ 2.87.

Buffers (Overview)

A buffer resists pH change on addition of small amounts of acid or base. It is built from a weak acid + its conjugate base (e.g., CH3COOH / CH3COO−) or a weak base + its conjugate acid (e.g., NH3 / NH4+) in comparable concentrations.

The Henderson–Hasselbalch equation gives the working pH:

pH = pKa + log([A−] / [HA])

• When [A−] = [HA], pH = pKa (the buffer's best resistance point).
• Adding strong acid converts A− → HA (ratio falls, pH drops slightly).
• Adding strong base converts HA → A− (ratio rises, pH rises slightly).

Buffer capacity is maximum when pH ≈ pKa and when both components are present in appreciable concentration. This is the physiological principle behind the bicarbonate/carbonic-acid buffer in blood (pKa ≈ 6.35) — relevant context for the PA-CAT's biomedical framing.

Tie to the Bulletin Sample Item

The Bulletin's chemistry sample item notes that carboxylic acids (R–COOH) become more water-soluble upon ionization because deprotonation gives R–COO−, which forms salts with cations like Na+ or K+. The conjugate base is charged, so it interacts far more strongly with water dipoles than the neutral parent acid. This is exactly the Brønsted-Lowry picture: R–COOH donates H+ to water (or to a base), and the resulting carboxylate is the conjugate base.