Chapter 21 - Buffers and neutralisation Flashcards
What is a buffer solution?
A system that minimises PH changes when small amounts of an acid or base are added
What 2 components does a buffer solution contain?
Weak acid and it’s conjugate base
How can a weak acid buffer solution be prepared from a weak acid and its salt?
- When the weak acid (e.g. ethanoic acid ) is added to water, the acid partially dissociates and the number of ions is very small
- The salt (e.g. sodium ethanoate) completely dissolve into ions when added to water
How can a weak acid buffer solution be prepared by partial neutralisation?
Adding an aqueous solution of an alkali (e.g. NaOH(aq)) to an excess of the weak acid is partially neutralised forming the conjugate base. Some of the weak acid is left over unreacted. Solution contains mixture of the salt of the weak acid and any unreacted weak acid
What controls the PH in an acid buffer solution?
Conjugate acid-base pair (HA(aq)/A-(aq)) - using le chatelier’s principle
HA(aq) H+(aq) + A-(aq)
What factors determine the PH of a buffer solution?
- Ka/ acid strength
- Temperature
- Concentrations/ratio of weak acid and conjugate base/salt
In a buffer solution, what happens on addition of an acid?
HA(aq) H+(aq) + A-(aq)
- ) [H+(aq)] increases
- ) H+(aq) ions react with the conjugate base, A-(aq)
- ) Equilibrium position shifts left, removing most of the H+(aq) ions - increasing PH again
In a buffer solution, what happens when an alkali is added?
- ) [OH-(aq)] increases
- ) small concentrations of H+(aq) ions react with the OH-(aq) ions:
H+(aq) + OH-(aq) -> H2O(l)
3.) HA dissociates, shifting the equilibrium position to the right to restore most of H+(aq) ions - decreasing PH again
HA(aq) H+(aq) + A-(aq)
–> added alkali
When is a buffer solution most effective?
Equal concentrations of the weak acid and it’s conjugate base
Why is a buffer solution most effective at equal concentrations of the weak acid and its conjugate base?
- PH of the buffer solution is the same as the pKa value of HA
- Operating PH is typically over about 2 PH units, centred at the PH of the pKa value
Used to find operating PH range and ratio of conc of weak acid and conjugate base are adjusted to fine-tune the PH of the buffer solution
How do you calculate PH of a buffer solution?
[H+(aq)] = Ka x [HA(aq]/[A-(aq)]
How is PH calculated if [A] = [Ha]?
Ka = [H+]
pKa = PH
Calculate the PH when the buffer solution contains 0.1 moldm-3 CH3COOH and 0.3 moldm-3 CH3COONa.
Ka (CH3COOH) = 1.74 x 10-5 moldm-3
[H+(aq)] = Ka x [HA(aq)]/[A-(aq)]
[H+(aq)] = Ka x [CH3COOH(aq)]/[CH3COO-(aq)]
[H+(aq)] = 1.74 x 10-5 x 0.1/0.3
= 5.8 x 10-6 moldm-3
PH = -log[H+(aq)] = -log(5.8 x 10-6) = 5.24
150 cm3 of 0.1 moldm-3 HCOOH is mixed with 100 cm3 0.750 moldm-3 HCOONa. Calculate the PH of the buffer solution formed
Ka (HCOOH) = 1.78 x 10-4 moldm-3
(Calculating the PH of a buffer solution made from a weak acid and its salt)
calculate the mol of HCOOH and HCOO-:
n(HCOOH) = 1.00 x 150/1000= 0.150 mol
n(HCOO-) = 0.75 x 100/1000 = 0.075 mol
calculate concentrations of HCOOH and HCOO- in the buffer solution:
total volume = 250cm3
[HCOOH(aq)] = 1000 x 0.150/250 = 0.6 moldm-3
[HCOO-(aq)] = 1000 x 0.075/250 = 0.3 moldm-3
Calculate [H+(aq)] using buffer equation:
[H+(aq)] = Ka x [HCOOH(aq)]/[HCOO-(aq)]
= 1.78 x 10-4 x 0.6/0.3
= 3.56 x 10-4 moldm-3
PH = -log(3.56 x 10-4) = 3.45
100 cm3 of 0.75 moldm-3 NaOH(aq) is added to 150 cm3 of 1.5 moldm3 HCOOH. Calculate the PH of the buffer solution formed.
Ka (HCOOH) = 1.78 x 10-4moldm-3
(calculating the PH of a buffer solution made by partial neutralisation)
write equation: HCOOH(aq) + NaOH(aq) → HCOONa + H2O
Calculate moles of acid and base:
n(NaOH) = 0.75 x 100/1000 = 0.075 mol
n(HCOOH) = 1.5 x 150/1000 = 0.225 mol
Acid in excess, so all the base reacts. 0.075 moles of NaOH at start of reaction, so 0.075 moles of moles of salt formed. As 1 mole of acid reacts with 1 mole of base to form 1 mole of salt:
n(HCOOH) remaining = 0.225 - 0.075 = 0.150 mol
Calculate concentration of acid and salt in buffer solution:
[HCOOH] = 1000 X 0.150/250 = 0.6 moldm-3
[HCOO-] = 1000 X 0.075/250 = 0.3 moldm-3
Calculate [H+(aq)] using buffer equation:
[H+(aq)] = Ka x [HCOOH(aq)]/[HCOO-(aq)]
= 1.78 x 10-4 x 0.6/0.3
= 3.56 x 10-4 moldm-3
PH = -log(3.56 x 10-4) = 3.45