Buffer Solutions

Buffer Solutions

Definition

  • Aqueous solution with a weak acid/base and its conjugate.

  • Resists pH change upon addition of small amounts of acid or base.

  • Used to maintain constant pH in chemical applications.

Chemistry

  • Equilibrium between weak acid (HA) and conjugate base (A-): HA↔H++A−HA \leftrightarrow H^+ + A^-

  • Adding H+H^+ shifts equilibrium left; adding OH−OH^- shifts it right (Le Chatelier's principle).

  • Acid dissociation constant: Ka=[H+][A−][HA]Ka = \frac{[H^+][A^-]}{[HA]}

  • Rearranged: [H+]=Ka×([HA][A−])[H^+] = Ka \times (\frac{[HA]}{[A^-]})

  • Henderson-Hasselbalch equation: pH=pKa+log([A−][HA])pH = pKa + log(\frac{[A^-]}{[HA]}) or pH=pKa+log([conjugate base or salt][acid])pH = pKa + log(\frac{[conjugate \ base \ or \ salt]}{[acid]})

Making a Buffer

  • Use Henderson-Hasselbalch equation to determine the ratio of conjugate base to acid needed for a specific pH.

  • pHpH depends on the ratio of acid/salt concentrations, not actual concentrations.

Calculating pH

  • Use the Henderson-Hasselbalch equation.

  • Example: 0.20M CH<em>3COOHCH<em>3COOH and 0.30M CH</em>3COONaCH</em>3COONa solution, Ka=1.8×10−5Ka = 1.8 \times 10^{-5}, pH = 4.9.

Buffer Capacity

  • Measure of buffer's efficiency in resisting pH changes.

  • Indicates amount of acid/base a buffer can handle before losing its resistance.

  • Depends on:

    • Proximity of buffer pH to its pKa (within 1-2 pH units).

    • Total buffer concentration.

  • Higher concentrations provide greater buffer capacity.

Applications

  • Chemical manufacturing and biochemical processes.

  • Enzymes require precise pH; buffers prevent denaturation and maintain activity.

  • Carbonic acid and bicarbonate buffer in blood plasma (pH 7.35-7.45).

  • Industrial uses: fermentation, dyes for fabrics, chemical analysis, pH meter calibration.

  • Common buffer: PBS (phosphate buffer saline) at pH 7.4 for biological samples.

Demonstration

  • Calculate pH of buffer solution (e.g., 0.2 mol K<em>2HPO</em>4K<em>2HPO</em>4 and 0.1 mol KH<em>2PO</em>4KH<em>2PO</em>4, Ka=6.2×10−8Ka= 6.2\times 10^{-8}).

  • Calculate pH change after adding NaOH or HCl.