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8.9Acids and Bases

HendersonHasselbalch Equation — AP CHEMISTRY Revision Notes & Topical Questions

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8.9 Henderson-Hasselbalch Equation — AP CHEMISTRY notes: Use the Henderson-Hasselbalch equation to calculate buffer pH and the effect of changing the ratio.Starting point — Ka = [H+][A−]/[HA]What the symbols mean: Ka = acid dissociation constant (mol dm−3) · [H+], [A−], [HA] = eq…

  • Take negative logs
  • Acidic buffer
  • Basic buffer
  • After reaction
  • Rearrange — [H+] = Ka × [HA]/[A−]
  • Data — 0.305 M CH3COOH, 0.520 M CH3COONa and pKa = 4.76

Section 1 of 9Derivation

Starting point
  • Ka = [H+][A−]/[HA]
    What the symbols mean: Ka = acid dissociation constant (mol dm−3) · [H+], [A−], [HA] = equilibrium concentrations (mol dm−3); larger Ka = stronger acid
Rearrange
  • [H+] = Ka × [HA]/[A−]
Take negative logs
  • Take negative logs pH = pKa + log([A−]/[HA])

Section 2 of 9Equations

Acidic buffer
  • Acidic buffer pH = pKa + log([salt]/[acid])
Basic buffer
  • Basic buffer pOH = pKb + log([salt]/[base])

Section 3 of 9Effect of the ratio

Ratio [A−]/[HA]pH
EqualpH = pKa
Greater than 1pH > pKa
Less than 1pH < pKa

Section 4 of 9Pattern: ethanoic acid buffer

Data
  • 0.305 M CH3COOH, 0.520 M CH3COONa and pKa = 4.76
Result
  • Result pH = 4.76 + log(0.520/0.305) = 4.99

Section 5 of 9Pattern: ethylamine buffer

Data
  • 0.0400 M C2H5NH2, 0.0865 M C2H5NH3Cl and pKb = 3.99
Result
  • Result pOH = 3.99 + log(0.0865/0.0400) = 4.32, so pH = 9.68

Section 6 of 9Conditions

Valid use
  • Valid use a buffer with appreciable amounts of both components
After reaction
  • After reaction use the moles remaining after neutralisation, since the volume cancels in the ratio

Section 7 of 9More from the specification

1Weak acids and conjugate bases equilibrium
  • Weak acids and conjugate bases equilibrium buffer systems combine weak acids with their conjugate bases to resist pH shifts upon adding strong acids.
2Acid-base proton exchange equilibria
  • Bronsted-Lowry bases accept hydronium ions, shifting equilibrium ratios between weak conjugate acids and conjugate base species.
3Solution
  • A mixture of a solute dissolved in a solvent.
4Property
  • A feature that describes what a substance is like and what it does.
5Buffer capacity
  • How much added acid or base a buffer can absorb before its pH changes significantly.
6Buffer solution
  • A solution that resists change in pH when small amounts of acid or alkali are added.
7Strong acid
  • An acid that is completely dissociated in water.
8Equilibrium position
  • Point of minimum free energy.
9Conjugate base
  • The species left after an acid donates a proton.
10Solubility
  • Mass of solute that dissolves in a given mass of solvent.
11Ethanoate
  • The salt of ethanoic acid.
12Hydrogen
  • A gas that gives a squeaky pop with a lighted splint.

Section 8 of 9Key terms

TermMeaning
Henderson-Hasselbalch equationpH = pKa + log([A−]/[HA]).
SaltSource of the conjugate base in a buffer.
pKb−log Kb.
Ratio[A−]/[HA].

Section 9 of 9Exam technique

Avoid these mistakes
Inverting the ratio of salt to acid.
Using pKa for a base buffer — use pKb and pOH.
Using it for a solution without both components.

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