Understanding points
Reactivity 3.1.9—The pOH scale describes the [OH–] of a solution. pOH = –log10[OH–]; [OH–] = 10–pOH (AHL)
Reactivity 3.1.10—The strengths of weak acids and bases are described by their Ka, Kb, pKa or pKb values. (AHL)
Reactivity 3.1.11—For a conjugate acid–base pair, the relationship Ka × Kb = Kw can be derived
from the expressions for Ka and Kb. (AHL)
Reactivity 3.1.12—The pH of a salt solution depends on the relative strengths of the parent acid and base. (AHL)
Reactivity 3.1.13—pH curves of different combinations of strong and weak monoprotic acids and
bases have characteristic shapes and features. (AHL)
Reactivity 3.1.14—Acid–base indicators are weak acids, where the components of the conjugate
acid–base pair have different colours. The pH of the end point of an indicator, where it changes
colour, approximately corresponds to its pKa value. (AHL)
Reactivity 3.1.15—An appropriate indicator for a titration has an end point range that coincides
with the pH at the equivalence point. (AHL)
Reactivity 3.1.16—A buffer solution is one that resists change in pH on the addition of small
amounts of acid or alkali. (AHL)
Reactivity 3.1.17—The pH of a buffer solution depends on both:
•
the pKa or pKb of its acid or base
•
the ratio of the concentration of acid or base to the concentration of the conjugate base or acid. (AHL)
pOH
•
pOH = -log10 [OH-], [OH-] =10-pOH
•
pH + pOH = 14
◦
e.g. A solution that has [OH-]=0.001 mol dm–3 has a pOH = –log10(0.001) = 3
◦
pH = 14 – 3 = 11
pH, pOH calculations
•
[equilibrium] and Ka/Kb required
•
*approximation of [equilibrium] required
Find the pH of a solution of 0.75M ethanoic acid with Ka value of 1.8 × 10–5 (at a specific temp.)
(Write out ethanoic acid hydrolysis equilibrium reaction)
CH3COOH(aq) ⇌ CH3COO–(aq) + H+(aq)
(Determine [H+], [each component])
[H+] = x (unknown b/c of partial dissociation)Conc. (M)CH3COOHCH3COO–H+Initial0.7500Equilibrium0.75-xxx
(Setup Ka equation)
Ka =[CH3COO–][H+]/[CH3COOH]
We’re given
1. value of Ka
2. [H+]=[CH3COO-]
1.8 x 10-5 = x2/0.75-x
1.8 x 10-5 = x2/0.75 (assumption = [H+] is extremely small, [CH3COOH]initial = [CH3COOH]equilibrium)
x = 3.7 x 10-3 = [H+]
pH = 2.4 | Find pH of 0.2M ammonia with Kb value of 1.8 × 10–5 at a specific temp.
(Write out ammonia hydrolysis equilibrium reaction)
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)
(Determine [OH-], [each component])
[OH-] = x (unknown b/c of partial dissociation)Conc. (M)NH3NH4+OH–Initial0.200Equilibrium0.2-xxx
(Setup Kb equation)
Kb = [NH4+][OH–]/[NH3]
We are given,
1. value of Kb
2. [NH4+] = [OH–]
1.8 x 10-5 = x2/0.2-x
1.8 x 10-5 = x2/0.2 (assumption = [OH-] is extremely small, [NH3]initial = [NH3]equilibrium)
x = 1.9 x 10-3 = [OH-]
pOH = 2.72
pH = 14 - 2.72 = 11.28 |
Comparison of acid/base strength - pKa, pKb
•
pKa= -log(Ka) and pKb= -log(Kb)
•
The lower the pKa or pKb value, the higher the Ka or Kb value, and the stronger the acid or base respectively
Strength | [H+], Ka | pH, pKa | [OH-], Kb | pOH, pKb |
Acid ↑ | ↑ | ↓ | ||
Base ↑ | ↓ | ↑ | ↑ | ↓ |
As a weak acid only partially dissociates in water and the rate of dissociation is unknown, in order to calculate pH of a weak acid, the extent of acid dissociation needs to be known. ˙.˙ [acid]≠ [H3O+] and [base]≠[OH-] as the definition of a weak acid/base is that it only partially dissociates in solution
•
Dissociated [A-] and [H3O+]/ [BH+][OH–] are in 1:1 stoichiometry (monoprotic)
Hydrolysis of weak acid, HA
HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq)
Kc = [H3O+][A–]/[HA][H2O] (water is too large/treated as constant)
Ka = [H3O+][A–]/[HA] = Kc for weak acid
Examples:
CH3COOH(aq) ⇌ CH3COO–(aq) + H+(aq)
Ka =[CH3COO–][H+]/[CH3COOH]
H2CO3(aq) ⇌ HCO3–(aq) + H+(aq)
Ka =[HCO3–][H+]/[H2CO3]
*assume only 1 H+ dissociated | Hydrolysis of weak base, B
B(aq) + H2O(l) ⇌ BH+(aq) + OH–(aq)
Kc =[BH+][OH–]/[B][H2O] (water is too large/treated as constant)
Kb= [BH+][OH–]/[B] = Kc for weak base
Examples:
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)
Kb =[NH4+][OH–]/[NH3]
CO32-(aq) + H2O(l) ⇌ HCO3-(aq) + OH-(aq)
Kb=[HCO3-][OH-]/[CO32-] |
Ka, Kb calculations
•
[equilibrium] and pH required
1.
Write out weak acid/base dissociation equilibrium reaction
2.
Determine [H+] or [OH-], [each component]
3.
Determine [HA]eq as [HA]initial - 10-pH (approximation valid only if [HA]initial > 1 and pH > 4)
4.
Setup Ka or Kb equation
Calculate Ka at 298 K for 0.01M ethanoic acid with pH of 3.4
(Write out ethanoic acid dissociation equilibrium reaction)
CH3COOH(aq) ⇌ CH3COO–(aq) + H+(aq)
(Determine [H+], [each component])
[H+] = 10-3.4 (deduced from pH)
[H+] = 3.98 x 10-4MConc. (M)CH3COOHCH3COO–H+Initial0.0100Equilibrium0.01 - (3.98 x 10-4)
~0.013.98 x 10-43.98 x 10-4
(Setup Ka equation)
Ka =[CH3COO–][H+]/[CH3COOH]
Ka = (3.98 x 10-4)2/0.01
Ka = 1.58 x 10-5 | Calculate Kb at 25 °C of 0.1M NH3 with pH 11.80
.
(Write out ammonia dissociation equilibrium reaction)
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)
(Determine [OH-], [each component])
[OH-] = 10-2.2 (deduced from pOH = 14 - 11.8 = 2.2)
[OH-] = 6.3 x 10-3Conc. (M)NH3NH4+OH–Initial0.100Equilibrium0.1 - (6.3 x 10-3)
~ 0.16.3 x 10-36.3 x 10-3
(Setup Kb equation)
Kb = [NH4+][OH–]/[NH3]
Kb = (6.3 x 10-3)2/0.1 ([NH4+] = [OH–])
Kb = 4,22 x 10-4 |
Conjugate base of a strong acid, Conjugate acid of a strong base
•
HA + B ⇄ HB+ + A-
•
HA= strong acid, readily donates its proton as it is reactive in the protonated form
◦
Conjugate base A- = weak base ∵ A- is the more stable form and does not want to accept the proton to form the more reactive species
•
B= strong base, readily accepts a proton as it is reactive
◦
Conjugate acid HB+= weak acid ∵ HB+ is more stable and does not want to donate the proton to form the more reactive species
Ka, Kb of a conjugate acid/base pair
Hydrolysis | Acid | Conjugate Base |
Equilibrium rxn | HA + H2O ⇌ A- + H3O+ | A- + H2O ⇌ HA + OH- |
Equilibrium constant | Ka = [H3O+][A-]/[HA] | Kb = [HA][OH-]/[A-] |
Ka x Kb product | [H3O+][A-]/[HA] x [HA][OH-]/[A-] = [H3O+][OH-]
Ka x Kb = Kw/ [H+][OH-] = Kw
pKa + pKb = 14/ pH+pOH = pKw | |
Significance | *Predict strength of conjugate acid/base*
Strong acid (low pKa) = weak conjugate base (high pKb)
Strong base (low pKb) = weak conjugate acid (high pKa)
Weak acid (high pKa) = strong conjugate base (loww pKb)
Weak base (high pKb) = strong conjugate acid (low pKa) |
(*the same can be derived by using a base and its conjugate acid pair)
Titration Curves
•
Same amount of base is needed to neutralize equal amount of strong/weak acid but at different rates
e.g. NaOH + CH3COOH → CH3COONa + H2O
•
NaOH: CH3COOH = 1:1
•
nNaOH = nCH3COOH
•
cNaOH x vNaOH = cCH3COOH x vCH3COOH where
•
cNaOH, vCH3COOH are fixed
•
vNaOH will be measured (i.e. volume of NaOH added until equivalence point)
•
cCH3COOH can be calculated using the above values and equation
Constant ↓↑ in [H+] but sudden steep ↓↑ in pH for strong vs strong
50cm3 of 0.1M HCl (analyte) vs 0.1M NaOH (titrant) | |||||
v(base) | 0 | 0.025 | 0.049 | 0.05 | 0.051 |
n(acid) | 0.005 mol | ||||
n(OH-) | 0 | 0.0025 | 0.0049 | 0.005 | 0.0051 |
n(H+) | 0.005 | 0.0025 | 0.0001 | N/A | 0.0001 (OH-) |
v(mixture) | 0.05 | 0.075 | 0.099 | 0.1 | 0.101 |
[H+] | 0.1 | 0.0333 | 0.00101 | N/A | 0.00099[OH-] |
pH | 1 | 1.5 | 3.0 | 7 | 11 |
Equivalence point
1.
Acid and base neutralize each other to form a salt + water
2.
≠ neutral pH ˙.˙ salt may be acidic/basic (*see salt hydrolysis)
3.
Indicated on titration curve as point of inflection
a.
(big) pH jump at point of inflection
b.
pH change range
i.
Strong vs strong (8) > strong vs weak (4) > weak vs weak (v small)
ii.
pH change range not large enough to form point of inflection and define equivalence point
Half equivalence point
1.
Half of acid has been neutralized into conjugate base
a.
pH when half volume of base used to neutralize acid
2.
Composition mixture establishes buffer
3.
Buffer cancels out H+ addition/ maintains relatively constant pH around half-equivalence point, buffer region indicates strengths of weak acids
a.
[HA] = [A-] ˙.˙ half of the acid has been neutralized into conjugate base
b.
Ka = [H+][A]/[HA] = [H+]
c.
pKa = pH
d.
Half-equivalence at low pH = relative strong weak acid
e.
Half-equivalence at high pH = relative weak weak acid
Salt hydrolysis
1.
Recall that Ka x Kb = Kw/ pKa + pKb = 14 for a conjugate acid-base pair
a.
Strong acid/base → weak conjugate base/acid
b.
Weak acid/base → strong conjugate base/acid
2.
pH of salt solution determined by how it hydrolyses water
a.
Strong conjugate acid/bases hydrolyse
i.
i.e) the conjugate base of a weak acid is a strong conjugate base; therefore will hydrolyse
b.
Weak conjugate acid/bases do not hydrolyse
i.
i.e) the conjugate base of a strong acid is a weak conjugate base; therefore will not hydrolyse
Conjugate acid/base (strength) | Hydrolysis | Resulting pH |
CH3COO- (strong)
[C.B. of CH3COOH (weak acid)] | CH3COO- + H2O ⇌ CH3COOH + OH- | >7 |
NH4+ (strong)
[C.A. of NH3 (weak base)] | NH4+ + H2O ⇌ NH3 + H3O+ | <7 |
Cl- (weak)
[C.B of HCl (strong acid)] | Cl- + H2O ⇌ OH- + HCl
HCl= strong acid ∴ [HCl]=[OH-]=0
(fully dissociates) | 7 |
Na+ (weak acid) | Na+ can act as a weak Lewis acid (not strong enough to hydrolyse water) | 7 |
Explanations for differing equivalence point pH in titration curves:
Strong Acid | Weak Acid | |
Strong Base | weak conjugate acid/base in salt cannot hydrolyse water
.˙. no change in pH | Strong conjugate base in salt hydrolyse water (take away H+ from H2O)
.˙. OH- produced, pH ↑ |
Weak Base | Strong conjugate acid in salt hydrolyse water
.˙. H3O+ produced, pH ↓ | Both conjugates in salt are strong and hydrolyse water
.˙. H3O+, OH- produced almost cancel out, hard to define pH |
Indicator: weak acid/base where protonated and deprotonated forms have different colours
•
HIn(aq) (colour A) ⇌ H+(aq) (colourless) + In–(aq) (colour B)
In acidic soln | In basic soln |
[H+] ↑ shifting equilibrium to the left
→ more of colour A | [H+] ↓ shifting equilibrium to the right
→ more of colour B |
Indicator change colour when [HIn] = [In–], pH = pKa (half-equivalence point)
1.
Indicator has equally mixed color of A and B
2.
Indicator is in the middle of colour change
3.
Further addition of either small [HIn] or [In–] will lead to shift in equilibrium to the other side → color B or A respectively
4.
End-point (range) = HIn buffer region (equivalence pH range around pKa)
5.
Some indicators have 3 colors, e.g. methyl orange, b/c color A and B when mixed gives a distinct color.
6.
Most indicators have 2 colors, e.g. phenolphthalein, b/c color A and B when mixed doesn’t give a distinct color
End-point overlapping with pH change range at equivalence point (point of inflection) signals neutralisation point
1.
Different indicators have different pKa and pH range
2.
Different salts have different pH within a pH range at equivalence
Phenolphthalein | Methyl Orange | Phenol Red | Bromophenol Blue | |
pKa | 9.6 | 3.46 | 7.9 | 4.10 |
pH range | 8.2~10.0 | 3.2~4.4 | 6.8~8.4 | 3.0~4.6 |
Colour (acid → base) | Colourless → pink | Red → yellow | Yellow → red | Yellow → blue |
•
Most indicators are suitable for strong vs strong
•
Low pH range for strong acid and weak base and high for weak acid & strong base
•
No suitable indicator for weak vs weak as no significant pH change at equivalence
Buffer region
1.
Around half equivalence point, weak acid/base equilibrium has equal [HA] and [A-] / [B] and [BH+]
2.
Buffering effect can take place to counteract OH- / H+ addition until 1:1 ratio of [HA]:[A-] / [B]:[BH+] is lost
3.
pKa of weak acid used (or pKb of weak base) is the pH of buffer region
Buffers
Function/definition | Resist changes in pH when small quantities of acid or alkali are added. | |
Example | H2CO3 + H2O ⇌ HCO3- + H3O+ (blood pH maintenance) | |
Type | Acidic | Basic |
Composition | Weak acid + conjugate base (directly added as salt or made from neutralization with strong base)
Example: Mixture of CH3COOH and CH3COONa *
(Or neutralization of CH3COOH (excess) and NaOH **)
*high [CH3COOH], [CH3COO-] (ideally 1:1)
**CH3COOH + NaOH → CH3COONa + H2O | Weak base (NH3) + conjugate acid (directly added as salt or made from neutralization with strong acid)
Example: Mixture of NH3 and NH4Cl
(Or neutralization of NH3 (excess) and HCl**)
*High [NH3], [NH4+]
(ideally 1:1)
** NH3 + HCl → NH4Cl |
Mechanism | [acid/base], [conjugate base/acid] are used to react and neutralize H+/OH- added | |
H+ addition | H+ is removed via reaction with CH3COO- (equilibrium shifts towards CH3COOH)
CH3COO- + H+ ⇌ CH3COOH
Ka of CH3COOH <<1 | H+ neutralizes with OH-,
→ Le Chatelier’s principle: equilibrium shifts to counter [OH-]↓, regenerating OH-
NH3+ H2O ⇌ NH4+ + OH- |
OH- addition | OH- neutralizes with H+
→ Le Chatelier’s principle: equilibrium shifts to counter [H+]↓, regenerating H+ and CH3COO-
CH3COOH ⇌ CH3COO- + H+ | OH- is removed via reaction with NH4+
NH4++ OH- ⇌ NH3 + H2O
Kb of NH3 <<1 |
Making a buffer solution | 1. Add salt (ion = conjugate) to weak acid/base
2. Partially (half) neutralize with strong base/acid
*weak acid/base cannot be fully neutralized |
1.
In acidic buffer made via neutralization, strong base is not used to neutralize incoming [H+] but form a salt which dissociates into A- conjugate base. A- conjugate neutralize incoming [H+]
2.
Diprotic weak acid may also form buffer, e.g. H2CO3 and NaHCO3 where NaHCO3 is a soluble salt and H2CO3 + H2O ⇌ HCO3- + H3O+ is established. H2SO4 and HSO4- cannot form buffer ˙.˙ not strong acid