Acid/base chemistry Buffers Henderson-Hasselbalch equation.

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Presentation transcript:

Acid/base chemistry Buffers Henderson-Hasselbalch equation

Weak acid (eg. acetic acid) Acetic acid ↔ acetate + proton CH 3 COOH ↔ CH 3 COO - + H + Weak acidConjugate base

CH 3 COOH ↔ CH 3 COO - + H + Ratio holds true at equilibrium if you neutralize H + (decrease numerator), the concentrations will adjust to get back to K a equilibrium shifts to the right K a = [CH 3 COO - ][H + ] [CH 3 COOH] [Ac - ][H + ] [HAc] =

K a /pK a 1.Relative strength of an acid Higher Ka/lower pKa = stronger 2.pH of the pure acid 3.pH range at which the weak acid is an effective buffer pH range at which acid and conjugate base are both present in appreciable quantities

Dissolve a weak acid in water (acetic, K a = 1.74 x ) HAc → H + + Ac - Reaction proceeds until it reaches equilibrium [H + ] depends on K a, [HAc] equilibrium [H + ] = M pH = 2.9 Initial pH

“Buffer” – functional definition Solution that resists change in pH upon addition of small amounts of H + or OH - (strong acid or base) Resists change in pH upon dilution

HAc ↔ Ac - + H + Buffering region (~pKa ± 1 pH unit) Neutralizing H +

Buffering region (~pKa ± 1 pH unit)

What happens when pH ≈ pK a ? First: add strong acid to pure water 0.1 M HCl → pH? pH 7 → 1 Second: add strong acid to 0.5 M solution of acetic acid at pH 4.76 pH 4.76 → 4.38

Henderson-Hasselbalch K a = [A - ][H + ] [HA] [H + ] = K a [HA] [A - ] -log[H + ] = -log(K a ) - log [HA] [A - ] pH = pK a - log [HA] [A - ] pH = pK a + log [A-] [HA]

Henderson-Hasselbalch Calculate pH, given pKa and ratio of [conjugate base]/[weak acid] Calculate pKa, given pH and ratio **Calculate ratio, given pKa and pH

What’s the ratio of base/acid of aspartic acid’s side chain at neutral pH? (pK a = 3.65) What’s the ratio of base/acid of histidine’s side chain at neutral pH? (pK a = 6.00)