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Biochemistry Unit 1: Water, pH and Buffers
Water's polarity and hydrogen bonding shape every biomolecule, and pH and buffers control the charge, structure and activity of proteins. This unit gives you the acid-base tools used throughout biochemistry.
Covers: Hydrogen bonding and properties of water, Noncovalent interactions and the hydrophobic effect, pH, Kw and strong acids and bases, Weak acids, Ka and pKa, Henderson-Hasselbalch equation, Buffers and titration curves, Physiological buffers
Study guide
Water is a bent, polar molecule. Its oxygen pulls electron density from the hydrogens, leaving the oxygen partially negative and the hydrogens partially positive. Each water molecule can donate two hydrogen bonds and accept two, forming a shifting network. This network explains water's high boiling point, high heat capacity and high surface tension. Water's high dielectric constant also weakens attractions between ions, so salts and charged biomolecules dissolve easily as hydration shells form around them. Molecules that can hydrogen bond or carry charges are hydrophilic; molecules made mostly of C–H bonds are hydrophobic.
Biomolecules are held in shape by noncovalent interactions: hydrogen bonds, ionic interactions (salt bridges), van der Waals contacts, and the hydrophobic effect. Each is weak alone, but thousands together are strong. The hydrophobic effect is mostly about entropy. Water forms ordered cages around nonpolar surfaces; when nonpolar groups cluster, that water is released and entropy rises. This drives protein folding, membrane formation and micelle formation by amphipathic molecules, which have both polar and nonpolar parts. Remember that hydrogen bonds inside a protein mostly replace hydrogen bonds the same groups would make with water, so their net contribution is modest; burying nonpolar surface is the larger driving force.
Water ionizes slightly: H₂O ⇌ H⁺ + OH⁻, with Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C. pH = −log[H⁺], and pH + pOH = 14. Strong acids such as HCl dissociate completely, so [H⁺] equals the acid concentration. Weak acids dissociate only partly, described by Ka = [H⁺][A⁻]/[HA] and pKa = −log Ka. A lower pKa means a stronger acid. For a weak acid alone in water, [H⁺] ≈ √(Ka × C). For example, 0.10 M acetic acid (Ka 1.74 × 10⁻⁵) has [H⁺] ≈ 1.3 × 10⁻³ M and pH 2.88, far higher than the pH 1.0 a strong acid would give.
The Henderson-Hasselbalch equation, pH = pKa + log([A⁻]/[HA]), links pH to the ratio of base to acid forms. When pH = pKa, the two forms are equal. Each pH unit above the pKa multiplies the base/acid ratio by 10. A buffer is a mixture of a weak acid and its conjugate base that resists pH change: added H⁺ is taken up by A⁻ and added OH⁻ is neutralized by HA. On a titration curve the flat region centered at the pKa is the buffering zone, and buffers work best within about one pH unit of their pKa. Capacity also grows with total buffer concentration.
In the body, the bicarbonate system keeps blood near pH 7.4: CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻, with an effective pKa of 6.1. It works far from its pKa because the lungs control CO₂ and the kidneys control HCO₃⁻. Retaining CO₂ causes respiratory acidosis; losing HCO₃⁻ causes metabolic acidosis, partly offset by faster breathing. Inside cells, phosphate (pKa 6.86) and protein side chains, especially histidine (pKa near 6), provide buffering. Because histidine's pKa is close to physiological pH, its charge can switch with small pH changes, which is why it often appears in enzyme active sites and pH-sensitive proteins.
Key ideas
- Water's polarity and hydrogen bonding make it an excellent solvent for ions and polar molecules.
- The hydrophobic effect is driven mainly by the entropy gained when ordered water is released.
- pH = −log[H⁺], and each pH unit is a tenfold change in [H⁺].
- pKa is the pH at which an ionizable group is half protonated.
- Henderson-Hasselbalch connects pH, pKa and the base/acid ratio.
- Buffers work best within about ±1 unit of their pKa and better at higher concentration.
- Blood pH is controlled by the bicarbonate system, regulated by lungs (CO₂) and kidneys (HCO₃⁻).
Formulas and rules
Common mistakes
- Flipping the Henderson-Hasselbalch ratio: it is base over acid, [A⁻]/[HA].
- Treating a weak acid as if it dissociates completely when finding its pH.
- Forgetting to update both acid and base amounts when strong acid or base is added to a buffer.
- Confusing pOH with pH, or forgetting that pH + pOH = 14.
- Thinking the hydrophobic effect comes from strong attraction between nonpolar molecules rather than water entropy.
Before the test, make sure you can
- Explain how hydrogen bonding gives water its solvent and thermal properties.
- Explain the hydrophobic effect in terms of entropy.
- Calculate pH, pOH, [H⁺] and [OH⁻] for strong acids and bases.
- Calculate the pH of a weak acid solution from Ka and concentration.
- Use Henderson-Hasselbalch to find pH, pKa or the base/acid ratio.
- Calculate the pH of a buffer after adding strong acid or base.
- Choose a suitable buffer for a target pH and compare buffer capacities.
- Predict how changes in CO₂ or HCO₃⁻ affect blood pH.
Words to know
- Hydrogen bond
- An attraction between an H bonded to N, O or F and a lone pair on another N, O or F; about 10–30 kJ/mol.
- Dielectric constant
- A measure of how well a solvent weakens attraction between charges; water's is high (about 80).
- Hydrophobic effect
- The tendency of nonpolar groups to cluster in water, driven by increased water entropy.
- Amphipathic
- Having both a polar (hydrophilic) region and a nonpolar (hydrophobic) region, like a fatty acid or detergent.
- Micelle
- A sphere of amphipathic molecules with tails in the core and polar heads facing water.
- Van der Waals interaction
- A weak attraction from transient induced dipoles between closely packed atoms.
- Salt bridge
- An ionic interaction between oppositely charged groups, such as Lys⁺ and Glu⁻ side chains.
- Kw
- The ion product of water, [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C.
- pH
- The negative log of the hydrogen ion concentration.
- Ka
- The acid dissociation constant, [H⁺][A⁻]/[HA]; larger Ka means a stronger acid.
- pKa
- −log Ka; the pH at which an acid is half dissociated.
- Henderson-Hasselbalch equation
- pH = pKa + log([A⁻]/[HA]); relates pH to the base/acid ratio.
- Buffer
- A weak acid plus its conjugate base that resists changes in pH.
- Buffer capacity
- The amount of acid or base a buffer can absorb with little pH change; greatest at pH = pKa and higher concentration.
- Conjugate base
- The species left after an acid loses H⁺, such as acetate from acetic acid.
- Bicarbonate buffer
- The CO₂/H₂CO₃/HCO₃⁻ system that controls blood pH near 7.4.
- Respiratory acidosis
- Low blood pH from retained CO₂, often due to hypoventilation.
- Metabolic acidosis
- Low blood pH from loss of HCO₃⁻ or excess acid production; compensated by hyperventilation.
Sample practice questions
30 more questions like these, from level 1 to 5, are in the practice set.
Which property of water is mainly responsible for its unusually high boiling point compared with H₂S?
- Its low molar mass relative to H₂S
- Strong covalent O–H bonds that must break on boiling
- Extensive hydrogen bonding between molecules
- Its linear molecular geometry
Show the answer and steps
Answer: C) Extensive hydrogen bonding between molecules
- Boiling separates molecules from each other; it does not break covalent bonds.
- Water's polar O–H bonds let each molecule hydrogen bond to up to four neighbors.
- Extra energy is needed to break that network, so the boiling point is high. Low mass would lower it, and water is bent, not linear.
What is the ion product of water, Kw, at 25 °C?
- 55.5 M²
- 1.0 × 10⁻²⁸ M²
- 1.0 × 10⁻¹⁴ M²
- 1.0 × 10⁻⁷ M²
Show the answer and steps
Answer: C) 1.0 × 10⁻¹⁴ M²
- Kw = [H⁺][OH⁻].
- In pure water at 25 °C, [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ M, so Kw = 1.0 × 10⁻¹⁴ M².
- 1.0 × 10⁻⁷ is just [H⁺]; 55.5 M is the molarity of water itself.
Phosphate buffer has pKa 6.86 (H₂PO₄⁻/HPO₄²⁻). At pH 7.40, what is the ratio [HPO₄²⁻]/[H₂PO₄⁻]?
- 1.08
- 3.47
- 0.54
- 0.29
Show the answer and steps
Answer: B) 3.47
- log([A⁻]/[HA]) = pH − pKa = 7.40 − 6.86 = 0.54.
- Ratio = 10⁰·⁵⁴ = 3.47.
- 0.54 is the log, not the ratio; 0.29 is the inverted ratio.