Buffers and Henderson-Hasselbalch
A buffer is a weak acid and its conjugate base. It absorbs added acid or base, because they only shift the ratio of the two partners.
Formulas and rules
- pH = pKa + log(c(base) / c(acid))
- Equal amounts of acid and base: pH = pKa
- Adding strong acid: base falls, acid rises (by the same amount)
- Adding strong base: acid falls, base rises
- Good buffering roughly within pKa ± 1
Your turn
Three problems on this topic. Type numbers like 12.5 or 1,250 and elasticities with their sign.
Task
A carbonic acid/hydrogen carbonate buffer (pKa = 6.35) is to have a pH of 6.55. In what ratio c(HCO₃⁻) / c(H₂CO₃) must base and acid be present?
Ratio base / acid (2 decimal places)
Which component must be in excess?
Numbers like 12.5 or 0.025, with as many decimals as asked. Masses and amounts count within 1%, pH values within ±0.02.
Worked example
A dihydrogen phosphate/hydrogen phosphate buffer contains c(H₂PO₄⁻) = 0.2 mol/L and c(HPO₄²⁻) = 0.1 mol/L (pKa = 7.21). Work out the pH.
Worked solution
- 1.Henderson-Hasselbalch: pH = pKa + lg(c(base) / c(acid))
- 2.pH = 7.21 + lg(0.1 / 0.2) = 7.21 + (−0.301) = 6.91
Typical mistake: Swapping acid and base in the fraction. The conjugate base goes on top: more base makes the buffer more basic.
1 L of ammonium/ammonia buffer (pKa = 9.25) contains 0.3 mol NH₄⁺ and 0.15 mol NH₃. 0.05 mol of hydrochloric acid (HCl) are added (ignore the change in volume). Work out the pH before and after.
Worked solution
- 1.pH = 9.25 + lg(0.15 / 0.3) = 8.95
- 2.The strong acid reacts with the base: NH₃ becomes NH₄⁺. n(base) = 0.15 − 0.05 = 0.1 mol, n(acid) = 0.3 + 0.05 = 0.35 mol
- 3.pH = 9.25 + lg(0.1 / 0.35) = 8.71
Typical mistake: Putting the added acid or base straight into pH = −lg c. In a buffer it is used up and only shifts the ratio of acid to base.
Common questions
Can I use moles instead of concentrations?
Yes, if both are in the same volume: the volume cancels in the fraction.