7.1 Brønsted–Lowry Acids/Bases, Water Ionization & pH

Content category 5A of the Chemical and Physical Foundations section asks how water's unique chemistry shapes acids, bases, and aqueous equilibria in living systems. Mastery of Brønsted–Lowry definitions, Kw, and pH is the foundation for every buffer, solubility, and titration problem later in this chapter — and for clinical acid–base physiology passages that dominate the MCAT's biology-linked chemistry items.

Brønsted–Lowry Acids and Bases

The MCAT tests acid–base behavior primarily through the Brønsted–Lowry definitions, which describe acids and bases in terms of proton transfer rather than the older Arrhenius model (acids release H⁺ in water; bases release OH⁻). Under the Brønsted–Lowry framework:

• A Brønsted–Lowry acid is a species that donates a proton (H⁺) to another species.
• A Brønsted–Lowry base is a species that accepts a proton from another species.

This proton-transfer view is more general than Arrhenius because it does not require the reaction to occur in water, and it explains how ammonia (NH₃), which contains no hydroxide ion, still acts as a base. When ammonia dissolves in water, it accepts a proton from water:

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

Here, water acts as the acid (donates H⁺) and ammonia acts as the base (accepts H⁺). Because water can donate protons in some reactions and accept them in others, it is amphoteric (also called amphiprotic) — it behaves as either an acid or a base depending on its partner. The same idea reappears for biological ampholytes such as HCO₃⁻ and H₂PO₄⁻, which buffer blood and intracellular fluid by switching roles as the surrounding pH changes.

The MCAT occasionally also references Lewis acids and bases (electron-pair acceptors and donors). Brønsted–Lowry acids are a subset of Lewis acids, but complex-ion formation (Ag⁺ accepting a lone pair from NH₃) is pure Lewis acid–base chemistry without proton transfer. When a passage mentions coordinate covalent bonds or metal–ligand complexes, think Lewis; when it mentions H⁺ transfer or pH, think Brønsted–Lowry.

Conjugate Acid–Base Pairs

Every Brønsted–Lowry reaction produces two conjugate pairs. A conjugate acid–base pair consists of two species that differ by exactly one proton. When a base accepts a proton, it becomes its conjugate acid; when an acid donates a proton, it becomes its conjugate base.

The classic MCAT example is the ammonium/ammonia pair:

A strong acid has a weak conjugate base, and a strong base has a weak conjugate acid — the stronger the original species is at giving up or taking on a proton, the less reactive its conjugate partner will be. This inverse relationship is tested repeatedly, especially in buffer and titration questions. For example, Cl⁻ is an extremely weak base (the conjugate of strong HCl) and does not hydrolyze water, whereas CH₃COO⁻ is a moderately strong conjugate base of weak acetic acid and does raise solution pH through hydrolysis.

MCAT trap: "Conjugate" always means a one-proton difference. NH₂⁻ is not the conjugate acid of NH₃; NH₄⁺ is. Likewise, HPO₄²⁻ and H₃PO₄ are not a conjugate pair — they differ by two protons — even though they sit on the same polyprotic ladder.

Ionization of Water and Kw

Water undergoes a small but critical autoionization (self-ionization) reaction in which one water molecule donates a proton to another:

2 H₂O ⇌ H₃O⁺ + OH⁻

(often simplified to H₂O ⇌ H⁺ + OH⁻, treating the hydronium ion as equivalent to a solvated proton). This equilibrium has its own equilibrium constant, the ion-product constant of water (Kw):

Kw = [H⁺][OH⁻] ≈ 1.0 × 10⁻¹⁴ at 25°C and 1 atm

Because the stoichiometry produces one H⁺ for every OH⁻, in pure water [H⁺] = [OH⁻]. Solving Kw = [H⁺]² gives [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ M in pure water at 25°C.

High-yield temperature trap: autoionization of water is endothermic (heat is a reactant), so Kw increases as temperature rises. At higher temperatures, both [H⁺] and [OH⁻] increase above 10⁻⁷ M, which drops the numerical pH of pure water below 7. Students often assume this makes hot water "acidic," but it does not — pure water is neutral whenever [H⁺] equals [OH⁻], regardless of the numerical pH value. Neutrality is defined by that equality, not by pH = 7 specifically; pH = 7 is only the neutral point at 25°C. A passage that reports pH 6.6 for pure water at elevated temperature is describing a neutral, not acidic, sample if [H⁺] still equals [OH⁻].

Kw is also the bridge between every conjugate Ka and Kb later: for any conjugate pair, Ka × Kb = Kw, so if you know one you can always find the other at a stated temperature.

The pH Scale and No-Calculator Estimation

pH is a logarithmic measure of hydrogen ion concentration:

pH = −log[H⁺]

Equivalently, [H⁺] = 10^(−pH). A parallel scale exists for hydroxide:

pOH = −log[OH⁻]

Taking the negative log of both sides of the Kw expression at 25°C gives the identity used constantly on Test Day:

pH + pOH = 14 (at 25°C, 1 atm only)

Because pH is logarithmic, each one-unit drop in pH corresponds to a tenfold increase in [H⁺]. A solution at pH 3 has 10 times the [H⁺] of a solution at pH 4, and 100 times the [H⁺] of a solution at pH 5. The MCAT frequently tests this order-of-magnitude reasoning without requiring an exact logarithm — recognizing the tenfold relationship is often enough.

At 25°C: pH = 7 is neutral, pH < 7 is acidic ([H⁺] > [OH⁻]), and pH > 7 is basic ([OH⁻] > [H⁺]). Physiological fluids sit near neutrality but not exactly on it: arterial blood averages about pH 7.40, so [H⁺] ≈ 4 × 10⁻⁸ M — slightly basic relative to pure water, which is why the bicarbonate buffer (next section) must continuously manage metabolic acid load.

Worked Example: Calculating pH

Problem: A solution has a hydrogen ion concentration of 4.0 × 10⁻⁵ M. What is the pH, and is the solution acidic or basic?

Solution:

pH = −log(4.0 × 10⁻⁵) = −(log 4.0 + log 10⁻⁵) = −(0.60 − 5) = 4.40

Since 4.40 is below 7, the solution is acidic. As a check, [OH⁻] = Kw/[H⁺] = (1.0×10⁻¹⁴)/(4.0×10⁻⁵) = 2.5×10⁻¹⁰ M, which is far smaller than [H⁺]. You can estimate this pH quickly without a calculator: 4.0×10⁻⁵ sits between 10⁻⁴ (pH 4) and 10⁻⁵ (pH 5), and log(4.0) ≈ 0.6, so pH lands at roughly 4.4 — closer to 4 than to 5. Memorize useful log anchors (log 2 ≈ 0.30, log 3 ≈ 0.48, log 5 ≈ 0.70) so that for any [H⁺] = n × 10^(−m) you can write pH ≈ m − log n without touching a calculator.

Recognizing conjugate pairs and manipulating Kw, pH, and pOH quickly is foundational for every other topic in this chapter — buffers, hydrolysis, solubility, and titration curves all rest on these relationships.