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Home Notes Chemistry Acids, Bases and Aqueous Solutions

Acids, Bases and Aqueous Solutions (AAMC 5A)

Almost every biological reaction happens in water, and the behavior of protons in water sets the charge on amino acids, the shape of enzymes and the pH of blood. This page covers acid-base definitions, pH arithmetic, Ka and pKa, buffers and the Henderson-Hasselbalch equation, titration curves, amino acid isoelectric points, solubility equilibria and colligative properties. Expect these ideas in Chem/Phys calculations and inside Bio/Biochem passages about proteins and physiology.

AAMC content category 5A. Subtopics: acid-base definitions, pH and Kw, weak acids and pKa, buffers, titration curves, amino acid pI, solubility and Ksp, and colligative properties. Le Chatelier's principle and K in general are on Equilibrium, Buffers and Solubility; vapor pressure and hydrogen bonding in water are on Liquids, Vapor Pressure and Water.

Acid-Base Definitions

Three definitions of acids and bases
ModelAcidBaseExample
ArrheniusReleases H+ in waterReleases OH− in waterHCl, NaOH
Brønsted-LowryProton (H+) donorProton acceptorNH3 is a base: it accepts H+ from water
LewisElectron-pair acceptorElectron-pair donorBF3, metal ions such as Fe3+ are Lewis acids

Each definition is broader than the last: every Arrhenius acid is a Brønsted-Lowry acid, and every Brønsted-Lowry acid-base reaction is also a Lewis reaction. When an acid HA donates a proton it becomes its conjugate base A−; when a base B accepts one it becomes its conjugate acid BH+. The stronger the acid, the weaker its conjugate base. Species that can act as either acid or base, such as water, HCO3− and amino acids, are amphoteric (amphiprotic when the exchange is specifically a proton).

Strong acids and bases to know

What makes an acid stronger

Anything that stabilizes the conjugate base strengthens the acid. Down a group, larger atoms spread the negative charge (HI > HBr > HCl). Across a period, more electronegative atoms hold charge better. Resonance (carboxylates, phenoxides) and nearby electron-withdrawing groups (trichloroacetic acid is far stronger than acetic acid) also stabilize the anion. For oxyacids, more oxygens means a stronger acid (HClO4 > HClO3 > HClO2 > HClO).

pH, pOH and Kw

Core relationships at 25°C

pH = −log[H3O+]; pOH = −log[OH−].
Kw = [H3O+][OH−] = 1.0 × 10−14, so pH + pOH = 14.
Each pH unit is a tenfold change in [H+]: pH 3 is 100 times more acidic than pH 5.

Kw increases with temperature because water's autoionization is endothermic. At 37°C neutral water has a pH of about 6.8, yet it is still neutral because [H+] = [OH−]. Neutral means equal concentrations, not "pH 7".

Fast log math

For [H+] = a × 10−b, pH is between b − 1 and b, approximately b − log a. Example: [H+] = 2 × 10−4 M gives pH ≈ 4 − 0.3 = 3.7. Useful values: log 2 ≈ 0.30, log 3 ≈ 0.48, log 5 ≈ 0.70.

Weak Acids, Ka and pKa

A weak acid ionizes only partly: HA + H2O ⇌ H3O+ + A−, with Ka = [H3O+][A−]/[HA]. A larger Ka (smaller pKa = −log Ka) means a stronger acid. For a conjugate pair, Ka × Kb = Kw and pKa + pKb = 14.

pH of a weak acid solution

If ionization is small (initial concentration C much larger than Ka), [H+] ≈ √(Ka · C). Example: 0.10 M acetic acid, Ka = 1.8 × 10−5: [H+] ≈ √(1.8 × 10−6) ≈ 1.3 × 10−3 M, so pH ≈ 2.9. The same approach with Kb gives [OH−] for a weak base.

Salts and hydrolysis

The conjugate of a weak partner reacts with water. NaCH3COO solution is basic (acetate is a weak base); NH4Cl solution is acidic (ammonium is a weak acid); NaCl solution is neutral (both ions come from strong partners).

Polyprotic acids

Acids such as H3PO4 and H2CO3 lose protons one at a time, and each step is weaker: Ka1 > Ka2 > Ka3, because removing H+ from an already negative ion is harder.

Buffers and the Henderson-Hasselbalch Equation

A buffer is a weak acid with its conjugate base (or a weak base with its conjugate acid) in similar amounts. Added H+ is captured by A−; added OH− is neutralized by HA. The ratio of the two forms changes only a little, so pH changes only a little.

Henderson-Hasselbalch

pH = pKa + log([A−]/[HA])
When [A−] = [HA], pH = pKa. A 10:1 ratio of base to acid gives pH = pKa + 1; a 1:10 ratio gives pH = pKa − 1. A buffer works well within about ±1 pH unit of its pKa, and its capacity grows with the total concentration of the pair.

Worked example

A buffer contains 0.20 M NaH2PO4 and 0.020 M Na2HPO4 (pKa2 = 7.2). pH = 7.2 + log(0.020/0.20) = 7.2 − 1 = 6.2. If the question gives moles of strong base added instead, first do the stoichiometry (OH− converts HA to A− one for one), then plug the new amounts into the equation.

Blood: the bicarbonate buffer

CO2 + H2O ⇌ H2CO3 ⇌ H+ + HCO3− (effective pKa about 6.1). Blood pH stays near 7.4 because the lungs adjust CO2 within minutes and the kidneys adjust HCO3− over hours to days. Hypoventilation raises CO2 and causes respiratory acidosis; hyperventilation lowers CO2 and causes respiratory alkalosis.

Common trap. Diluting a buffer with water barely changes its pH because the ratio [A−]/[HA] is unchanged, but it does lower the buffer capacity. A strong acid with its conjugate base (HCl with NaCl) is never a buffer: Cl− has essentially no tendency to accept a proton.

Titration Curves

In a titration, a solution of known concentration (the titrant) is added until the analyte is exactly neutralized. The equivalence point is where moles of added base equal moles of acid protons originally present (for a monoprotic acid: MaVa = MbVb). The endpoint is where the indicator changes color; a good indicator has a pKa close to the equivalence-point pH.

Reading titration curves
TitrationpH at equivalenceShape notes
Strong acid + strong base7Very steep jump from about pH 3 to 11; no buffer region
Weak acid + strong baseAbove 7 (A− is basic)Flat buffer region; at the half-equivalence point pH = pKa
Weak base + strong acidBelow 7 (BH+ is acidic)At half-equivalence, pOH = pKb, so pH = pKa of BH+
Polyprotic acid + strong baseOne equivalence point per acidic protonSeveral buffer plateaus, one centered on each pKa

The half-equivalence point is the flattest part of the curve, where the solution resists pH change most. The equivalence point is the steepest part, where pH changes most per drop.

Amino Acid Charge and the Isoelectric Point

Amino acids carry an acidic carboxyl group (pKa about 2) and a basic amino group (pKa of the ammonium about 9 to 10). At physiological pH both are ionized, so a simple amino acid exists as a zwitterion: –NH3+ and –COO−.

Charge rule

For each ionizable group: if pH < pKa, the group is mostly protonated; if pH > pKa, mostly deprotonated. Protonated carboxyl is neutral, deprotonated is −1. Protonated amino is +1, deprotonated is neutral. Add up the charges.

Isoelectric point (pI)

The pH at which the molecule's average net charge is zero. Average the two pKa values that bracket the neutral form.
· Neutral side chain (glycine: 2.3 and 9.6): pI = (2.3 + 9.6)/2 ≈ 6.0.
· Acidic side chain (aspartate: 2.0, 3.9 side chain, 9.9): average the two lowest, pI ≈ 3.0.
· Basic side chain (lysine: 2.2, 9.0, 10.5 side chain): average the two highest, pI ≈ 9.8.

Below its pI a protein is net positive; above its pI it is net negative. This is the basis of isoelectric focusing and ion-exchange chromatography, covered on Separation and Spectroscopy.

Solubility and Ksp

For a sparingly soluble salt in equilibrium with its saturated solution, the solubility product is the product of ion concentrations, each raised to its coefficient. The solid is left out.

Compare Ksp values directly only for salts that dissociate into the same number of ions; otherwise compare molar solubilities.

Factors that change solubility

Solubility rules worth knowing: salts of group 1 metals, NH4+, nitrate and acetate are soluble; most chlorides are soluble except those of Ag+, Pb2+ and Hg22+.

Colligative Properties

Colligative properties depend on the number of dissolved particles, not on their identity. The van 't Hoff factor i counts particles per formula unit: about 1 for glucose, 2 for NaCl, 3 for CaCl2. (Real values are slightly lower because of ion pairing.)

The four colligative properties
PropertyEquationEffect of adding solute
Vapor pressure loweringP = XsolventP° (Raoult's law)Vapor pressure falls
Boiling point elevationΔTb = i Kb mBoiling point rises
Freezing point depressionΔTf = i Kf mFreezing point falls
Osmotic pressureΠ = i M R TWater moves toward the more concentrated side

Note the units: boiling and freezing point changes use molality (mol solute per kg solvent), while osmotic pressure uses molarity. Osmotic pressure is the physiological one: a cell in a hypotonic solution swells (red cells can lyse), and in a hypertonic solution it shrinks (crenates).

Shortcut. To rank solutions by any colligative effect, multiply i by the concentration and compare the products. The largest i × m has the highest boiling point, the lowest freezing point, the lowest vapor pressure and the highest osmotic pressure.

Worked MCQs

Five MCQs on the patterns the exam returns to most often. Read every explanation, including the ones you got right.

Q1. What is the pH of a 0.010 M HCl solution at 25°C?

  • 1
  • 2
  • 7
  • 12

HCl is a strong acid and ionizes completely, so [H+] = 0.010 M = 10−2 M and pH = 2. A pH of 12 would be the answer for 0.010 M NaOH.

Q2. A 0.10 M solution of a weak acid HA has Ka = 1.0 × 10−5. What is its approximate pH?

  • 1
  • 2
  • 3
  • 5

For a weak acid with small ionization, [H+] ≈ √(Ka · C) = √(1.0 × 10−5 × 0.10) = √(10−6) = 10−3 M, so pH ≈ 3. A pH of 1 would require complete ionization, and 5 is the pKa, which applies only when [A−] = [HA].

Q3. A buffer is made from a weak acid with pKa = 6.8, with the conjugate base at 10 times the concentration of the acid. What is the pH?

  • 7.8
  • 5.8
  • 6.8
  • 8.8

pH = pKa + log([A−]/[HA]) = 6.8 + log 10 = 6.8 + 1 = 7.8. Reversing the ratio would give 5.8.

Q4. Aspartic acid has pKa values of 2.0 (α-carboxyl), 3.9 (side chain) and 9.9 (α-amino). What is its isoelectric point?

  • 3.9
  • 5.95
  • 6.9
  • 2.95

The neutral form exists between the first and second deprotonations (both carboxyls), so pI = (2.0 + 3.9)/2 = 2.95. Averaging 2.0 and 9.9 (5.95) is the classic mistake of ignoring the acidic side chain.

Q5. Which aqueous solution has the lowest freezing point?

  • 0.10 m glucose
  • 0.10 m NaCl
  • 0.10 m CaCl2
  • 0.25 m sucrose

Freezing point depression is proportional to i × m. Glucose gives 0.10, NaCl gives 0.20, sucrose gives 0.25, and CaCl2 gives 3 × 0.10 = 0.30 m of particles, the largest depression and so the lowest freezing point.

Quick Recap

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