4.7 Acid/base Equilibria
From the timeline we can see the change in definition of acids through history. The main ones to know are:
Arrhenius definition – when acids/bases dissolve in water then completely/partially dissociate into charged particles (ions)
Brønsted–Lowry definition – an acid is a proton donor and a base is a proton acceptor; acids (proton donors) will never release a H+ on its own, it is always combined with H2O to form HYDROXONIUM IONS – H3O+
*NOTE: Acid-base equilibria involves the transfer of protons, either donated or accepted.
| example | reason | |
| Strong Acid
Base
|
HCl(g) —> H+(aq) + Cl–(aq)
NaOH(s) + H2O(l) —> Na+(aq) + OH–(aq) |
Strong acids and bases ionise almost completely in water.
*HCl has a pH of 0 = completely ionised |
| Weak Acid
Base |
CH3COOH(aq) CH3COO–(aq) + H+(aq)
NH3(aq) + H2O(l) |
Weak acids and bases only slightly ionise. Equilibrium is set up with mostly reactants (to the left) |
Conjugate acid base pairs
- HA and A– are conjugate pairs
- H2O and H3O+ are conjugate pairs
WATER is special – it can behave as a base and an acid. You can work out the equilibrium constant in the same manner as we did before e.g. 
However, the equilibrium is very far left and so the equilibrium constant for this reaction is said to have a constant value;
At 298K/1atm, the Kc of water is 1.0 x 10-14 mol2dm-6
(We often define this with its own notation – Kw)
Kw = Kc x [H2O] = the ionic product of water = [H+][OH–] with UNITS: mol2dm-6
pH – “power of hydrogen” – is a measure of the hydrogen ion concentration
pH = – log [H+]
CALCULATION: finding the pH of a strong acid
- Calculate the pH of 0.05 moldm-3 of nitric acid.
pH = – log[H+] pH = – log[0.05]
= 1.3 (pH value is small – expected for a strong acid)
- An acid has a pH of 2.45, what is the hydrogen ion concentration?
pH = – log[H+] [H+] = 10-pH
= 3.55 x 10-3 moldm-3
*NOTE: H2SO4 dissociates to give 2[H+] and you will have to divide the final answer by 2 to find your hydrogen ion concentration
CALCULATION: finding the pH of a weak acid
Weak acids do not fully dissociate so it isn’t as straight forward as above. Another constant called Ka is introduced. There are some assumptions to make first:
- a) Only a tiny amount of product dissociates so initial concentration of reactant = equilibrium concentration of reactant
- b) All H+ ions come from the acid i.e. concentration of product 1 = concentration of product 2
1) Calculate the hydrogen ion concentration and the pH of a 0.02 moldm-3 solution of propanoic acid (CH3CH2COOH). The Ka of propanoic acid is 1.3 x 10-5 moldm-3.
Ka = [H+]2/[CH3CH2COOH]
[H+] = 5.09 x 10-4
pH = -log[5.09 x 10-4]
pH = 3.29
CALCULATION: finding the pH of a strong base
One OH– ion fully dissociates per mole of base so the concentration of OH– ions and concentration of the base is the same. However to work out pH from the formula, we need [H+]. Therefore, we use our knowledge of (@ 298K), Kw = 1.0 x 10-14 mol2dm-6
- Find the pH of 0.1 moldm-3 of NaOH at 298K.
[H+] = = = 1.0 x 10-13 mol dm-3
Therefore,
pH = -log [1.0 x 10-13]
= 13.0 (ph value is large – expected for a strong alkali)
CALCULATION: finding the pKa
pKa = – log [Ka]
- Calculate the pH of 0.05moldm-3 of methanoic acid (HCOOH). Methanoic acid has a pKa of 3.75.
3.75 = -log[Ka] Ka = 1.78 x 10-4
1.78 x 10-4 = [H+]2/[0.05] [H+] = 2.98 x 10-3
pH = -log[2.98 x 10-3] pH = 2.53
Lastly, you should be aware that;
- diluting a strong acid (e.g. HCl) by a factor of 10 increases the pH by 1
- diluting a weak acid (e.g. CH3COOH)by a factor of 10 increases the pH by 0.5
- add a measure of acid (with known concentration) to burette
- rough titration; swirl conical flask for approximate end point*
- accurate titration; drop by drop
- record amount of base needed to neutralise the acid
- Repeat for more accurate readings
*end point: when the solution changes colour (also known as equivalence point – see below)
pH Curves
Equivalence point = where a tiny amount of alkali causes a sudden big change in pH, where the acid is JUST neutralised. Equivalence point will vary depending on acid/alkali used. For the last graph between a weak acid and weak alkali, a pH meter is the best thing to use to find the equivalence point as the colour change is gradual and unclear.
CALCULATION: finding the Ka of a weak acid using a pH curve
The half equivalence point is the point where half the acid has been neutralised, where half the volume of strong base has been added the weak acid before equivalence.
At the half equivalence point [HA] = [A–]
Therefore
Thus, we can say that the half equivalence point is also the pKa of the weak acid, then we can use Ka = 10-pKa
Buffers
- RESIST changes in pH when small amounts of acid/alkali are added
- Doesn’t stop the pH from changing completely
- They only work when small amounts of acids/alkalis are added
| ACIDIC BUFFERS | ALKALINE BUFFERS |
| Weak acid + Salt | Weak base + Salt |
| CH3COO–Na+(aq)—> CH3COO–(aq) + Na+(aq)
This fully dissociates; therefore mostly ethanoate ions
CH3COOH(aq) This only slightly dissociates; therefore mostly ethanoic acid |
NH4Cl(aq)—> NH4+(aq) + Cl –(aq)
This fully dissociates; therefore mostly ammonium ions
NH4+(aq) This only slightly dissociates; therefore mostly ammonium |
| ADDING ACID: (small amount)
[H+] increases which combines with the CH3COO– to form CH3COOH so equilibrium shifts to left, no change in pH.
ADDING ALKALI: (small amount) [OH–] increases which combines with the H+ to form H2O which removes the H+ ions from solution, so more H+ dissociate from CH3COOH so equilibrium shifts to right, no change in pH. |
ADDING ACID: (small amount)
[H+] increases which combines with the NH3 to form NH4 so equilibrium shifts to left, no change in pH.
ADDING ALKALI: (small amount) [OH–] increases which combines with the H+ to form H2O which removes the H+ ions from solution, so more H+ dissociate from NH4+ so equilibrium shifts to right, no change in pH. |
Biological environments (don’t need to learn but be aware of)
| Example | Cells – need constant pH for biochemical reactions to take place | Blood – need to be kept at pH 7.4 | Food products – changes in pH occur due to fungi and bacteria |
| Buffer | Controlled by the equilibrium between dihydrogen phosphate and hydrogen phosphate
H2PO4–
|
Carbonic acid (H2CO3)
H2CO3 Lungs – by breathing out CO2, levels of H2CO3 decrease and so equilibrium moves to the right
H2CO3 Kidneys control this equilibrium |
Sodium citrate
Citric acid Or Phosphoric acid Or Benzoic acid
|
CALCULATION:
A buffer solution of 0.4 moldm-3 of methanoic acid and 0.6
moldm-3 sodium methanoate. For methanoic acid Ka= 1.6 x 10-4 moldm-3.
What is the pH of the buffer?