[21.1] buffer solutions Flashcards
what is a buffer?
a system that minimises pH changes when small amounts of an acid or a base are added
what 2 components of buffers mean pH changes are minimised?
- weak acid, HA, removes added alkali
- conjugate base, A-, removes added acid
what 2 ways can buffers be prepared?
- mixture of weak acid and its salt
- mixture of an excess of WA and strong base (partial neutralisation of WA)
how can buffers be prepared from a weak acid and its salt? (based on ethanoic acid)
- when ethanoic acid is added to water, the acid partially dissociates and the amount of ethanoate ions in solution is very small
- when salts of weak acids are added to water, it completely dissolves
- ethanoic acid = source of weak acid component
- salts of weak acids = source of conjugate base
CH₃COOH (aq) ⇌ H⁺(aq) + CH₃COO⁻ (aq)
CH₃COONa (s) + aq -> CH₃COO⁻ (aq) + Na⁺ (aq)
how can control of pH be explained in terms of shifts in the equilibrium positions using le chatelier’s principle when alkali is added?
HA (aq) ⇌ H+ (aq) + A- (aq)
1. [OH-] increases
2. small conc of H+ ions react with OH- ions: H⁺ + OH⁻ -> H₂O
3. HA dissociates
4. POE shifts to right to restore most of H+ ions
how can control of pH be explained in terms of shifts in the equilibrium positions using le chatelier’s principle when acid is added?
HA (aq) ⇌ H+ (aq) + A- (aq)
1. [H+] increases
2. H+ ions react with conjugate base, A-
3. equilibrium position shifts to the left, removing most of the H+ ions
when is a buffer most effective at removing either added acid or alkali?
when there are equal concentrations of the weak acid and its conjugate base
how can buffers be prepared by partial neutralisation of the weak acid (ethanoic acid)?
- aqueous solution of an alkali + excess of weak acid
- weak acid is partially neutralised by alkali, forming the conjugate base
- resulting solution contains a mixture of salt of WA and any unreacted WA
- CH₃COOH and CH₃COO⁻ act as two resevoirs that are able to remove added acid and alkali by shifting the buffer’s equilibrium
CH₃COOH (aq) ⇌ H⁺ (aq) + CH₃COO⁻ (aq)
why is a buffer most effective when [HA] = [A-]?
- pH of buffer solution is same as pKa value of HaA
- operating pH is around 2 pH units, centred at pH of pKa value
how do you calculate the pH of a buffer solution?
[H+] = Ka × ([HA] / [A-])
how do you calculate the pH of a buffer solution when [HA] = [A-]?
- Ka = [H+]
- pKa = pH