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1
Acids and Bases: Overview
• Definitions of acids and bases
• Equilibria involving acids and bases
• Conjugate acid‐base pairs
• Autoionisation of water
• The “p” scale (pH, pOH, pKa, pKb)
• Weak acids and bases
• Calculations involving pKa and pKb• Strong acids and bases
• Polyprotic acids
• Salts of acids and bases
• Buffers
• Indicators
• Titrations
Chem 1102 1B 1
Common Uses of Acids and Bases
Chem 1102 1B 2
Common Types of Acids
•Many acids are oxy acids where the proton is attached to an oxygen atom.
Chem 1102 1B 3
Organic acids Organic acids have a carboxyl group
Hydrohalic Acids H-F, H-Cl, H-Br and H-I
Oxy Acids
Common Types of Bases
•Organic bases have a nitrogen atom with a lone pair
Chem 1102 1B 4
Organic bases
LiOH, NaOH, K2O etc.Hydroxides and oxides
Acids and Bases
•Acids were originally recognised by their sour taste, e.g., lemon (citric acid).
•Bases usually by their bitter taste and slippery feel, e.g., drain uncloggers (caustic soda or sodium hydroxide).
Chem 1102 1B 5
Arrhenius was the first to recognisethe nature of acids and bases.
Strong acids and bases:
HCl (aq) + NaOH (aq) ∆H0 = -56 kJ mol-1
HI (aq) + NaOH (aq) ∆H0 = -56 kJ mol-1
HNO3 (aq) + NaOH (aq) ∆H0 = -56 kJ mol-1... etc
Why do these all have the same heat of reaction? Because the only species that actually react are:
H+ (aq) + OH- (aq) = H2O(l) ∆H0 = -56 kJ mol-1
Definitions•Arrhenius: H+ (aq) + OH‐ (aq) ⇌ H2O (l)
– ACID: H+ producer in aqueous solution i.e. AH– BASE: OH‐ producer i.e. MOH
•Brønsted ‐ Lowry:•H+ + A‐⇌ HA
– ACID: proton donor (H+) e.g. HCl– BASE: proton acceptor e.g. NH3
•Lewis•A + :B⇌ A:B
•ACID: electron pair acceptor e.g. BF3– BASE: electron pair donor e.g. NH3
Chem 1102 1B 6
2
NaOH and Dry Ice
•1 M NaOH pH ≈ 12
CO2 (s) ⇌ CO2 (g) ⇌ CO2 (aq) ⇌ H+ (aq) + HCO3‐ (aq)
•H+ (aq) + OH‐ (aq) → H2O (l)
Chem 1102 1B 7
Electron-pair acceptor (Lewis acid)
DEMO: 5.10 dry ice in water plus 1M HCL and NaOH + litmus paper
Brønsted‐Lowry Model•acid = proton donor
•base = proton acceptor
Chem 1102 1B 8Note that water can act as an acid or a base - it is amphoteric
Acids, Bases & Equilibrium
•In water: an acid (e.g., HCl) ionises to produce H+ (aq)
– NB: Actually H3O+ (aq), but we usually just write H+ (aq)
Chem 1102 1B 9
H3O+
+
Generally exists as [H(H2O)n]+ (aq)
Acids, Bases & Equilibrium
HA (aq) + H2O (l) ⇌ H3O+ (aq) + A- (aq)
•A STRONG acid has equilibrium to the right (HA mostly dissociated)•A WEAK acid has equilibrium to the left (HA mostly intact)
•Equilibrium Equation:
•Ka is the ACID DISSOCIATION CONSTANT
Note: it has the same form as if reaction was a simple dissociation: HA (aq) ⇌ H+ (aq) + A- (aq)
Chem 1102 1B 10
Conjugate Acid‐Base Pairs
•NH4+ is the conjugate acid of NH3
•NH3 is the conjugate base of NH4+
•A conjugate base has one less proton than its conjugate acid.
•HSO4–: conjugate base is SO4
2–
conjugate acid is H2SO4
•H2SO4 is a dibasic or diprotic acid:
Chem 1102 1B 11
H2SO4 (aq) + H2O (l) ⇌ H3O+ (aq) + HSO4– (aq)
HSO4– (aq) + H2O (l) ⇌ H3O+ (aq) + SO4
2– (aq)
(lies ~100% to right)
Questions:
Chem 1102 1B 12
Write the formula of the Write the formula of the conjugate bases conjugate acids
H3O+ OH-
H2SO4 H2O
HClO4 CN-
CH3COOH NH3
HPO42– HPO4
2–
3
Acid‐Base Reactions
Chem 1102 1B 13
The positions of these equilibria determine the strengths of the acids.
Autoionisation of Water
•H2O (l) ⇌ H+ (aq) + OH– (aq)
•Equilibrium constant given special symbol:Kw = [H+][OH–]
NB: [H2O (l)] = constant
At 25 °C: Kw = 1.0 × 10‐14 ← REMEMBER THIS!
– Neutral solution: [H+] = [OH– ] = 10‐7 M– Acidic solution: [H+] > 10‐7 M– Basic : [H+] < 10‐7 M
Chem 1102 1B 14
This is why pure water is a weak electrolyte
The pH Scale
Because the concentrations of acids and bases can vary over many orders of magnitude, it is convenient to define a logarithmic scale to compare them:
pH = −log10[H+]
e.g. If [H+] = 1.0 × 10‐6 M then pH = ‐log(10‐6) = ‐ (‐6) = 6.0
Chem 1102 1B 15
The ‘p’ ConventionpH = – log10[H
+]pOH = – log10[OH
– ]pKw= – log10[Kw] = 14 at 25 °C
Acid : pH < 7Neutral: pH = 7Basic: pH > 7
Since Kw = [H+][OH– ]:log10 Kw = log10 [H
+] + log10 [OH– ]
– log10 [H+] – log10 [OH
– ] = – log10 Kw
pH + pOH = 14pOH = 14 – pH
Chem 1102 1B 16
Note that an acidic solution contains some OH-, and a
basic solution contains someH3O+
Temperature Dependence of pH
Kw = 1.0 × 10‐14 only at 25 °C
For T > 25 °C, Kw > 10‐14 ⇒ pH + pOH < 14 if T > 25 °C
For T < 25 °C, Kw < 10‐14 ⇒ neutral pH > 7 if T > 25 °CChem 1102 1B 17
H2O
H+ + OH-
ΔH0 = 56 kJ mol-1Ea
The pH Scale
Chem 1102 1B 18
Remember, this is a log scale -so 0.1 M HCl is ~100 × more acidic than 0.1 M acetic acid.
4
Autoionisation of Water
•H2O (l) ⇌ H+ (aq) + OH– (aq)
Equilibrium constant given special symbol:
Kw = [H+][OH–]NB: [H2O (l)] = constant
At 25 °C: Kw = 1.0 × 10-14
– Neutral solution: [H+] = [OH– ] =10-7 M– Acidic solution: [H+] > 10-7 M– Basic : [H+] < 10-7 M
Chem 1102 1B 19
This is why pure water is a weak electrolyte
Acids, Bases & Equilibrium
•HA (aq) + H2O (l) ⇌ H3O+ (aq) + A‐ (aq)
A STRONG acid has equilibrium to the right (HAmostly dissociated)
A WEAK acid has equilibrium to the left (HAmostly intact)
Equilibrium Equation:
Ka is the ACID DISSOCIATION CONSTANT
– Note: it has the same form as if reaction was a simple dissociation:
HA (aq) ⇌ H+ (aq) + A‐ (aq)
Chem 1102 1B 20
Weak Acids
Most acids and bases are weak ‐ they do not completely ionise in water.
•The more positive the pKa, the weaker the acid (and the stronger the conjugate base).
Chem 1102 1B 21
pKa = – log10Ka
HA(aq) H+(aq) + A–(aq)
Acid dissociation constant
Acetic Acid: Vinegar
acetic (ethanoic) acid, CH3COOH (HAc for short):
Ka = 10‐4.7 = 1.99 x 10‐5
pKa = 4.7
pH of 0.1 M solution of acetic acid > 1
(pH would be –log(0.1) = 1
only if it were completely ionised)
Chem 1102 1B 22
Relationship Between Ka and pKa
The larger the value of Ka the stronger the acid and the lower the value of pKa.
•Ka = 1.02 × 10‐2 then pKa = ‐log10(1.02 × 10‐2) = 1.991
•pKa = 1.991 then Ka = 10‐1.991 = 1.02 × 10‐2
Chem 1102 1B 23
Examples of Ka Values
Chem 1102 1B 24
Note effect of changing the organic
part of carboxylic acids
5
Example
Find the pH of 0.1 M acetic acid, CH3COOH (HAc)
DATA: pKa = 4.7, Ka = 10‐4.7
Chem 1102 1B 25
HAc (aq) ⇌ H+ (aq) + Ac– (aq) Ka = 10-4.7
H2O (l) ⇌ H+ (aq) + OH– (aq) Ka = 10-14
Concns. (M) HAc (aq) ⇌ H+ (aq) + Ac– (aq)before eqm: 0.1 0 0
after: 0.1 – x x xmake x either
H+ or Ac–
Example (continued)
Chem 1102 1B 26
Since the equilibrium constant is very small we assume
x << 0.1, i.e. ( 0.1 – x ) ≈ 0.1 10-4.7 ≈ x 2 / 0.1
x2 ≈ 0.1 × 10-4.7 = 10-5.7
pH = – log10 [H+ ] = – log10 x = – log10 √10-5.7
= 2.9Check: x = 10-2.9 = 1.4 × 10-3 << 0.1,
( 0.1 – x ) = 0.0986 M
which is less than 5 % different to our assumption that 0.1 - x = 0.1, so our assumption was valid.
[Note that the exact answer found by solving a quadratic would be x = 0.098579]
Demo: Strong Acids versusWeak Acids
Chem 1102 1B 27DEMO: 5.9 strong and weak acid with metal
ExampleFind % ionisation of 0.50 M HF (pKa = 3.1).
Chem 1102 1B 28
Conc (M) HF (aq) ⇌ H+ (aq) + F– (aq)
before eqm.: 0.50 0 0
after: 0.50 – x x x
Hence Ka = 10-3.1 ≈ x2 / (0.50 - x)
If x << 0.50, 10-3.1 ≈ x2 / 0.50
x2 = 0.50×10-3.1
x = 2×10– 2 M (check: indeed x << 0.50 M)
% ionisation = x / 0.50 × 100 = 4 % (indeed a weak acid!)
Weak Bases
Ionisation of a weak base:NH3 (aq) + H2O (l) ⇌ NH4
+ (aq) + OH– (aq)
Equilibrium constant is called base ionisation constant, Kb :
•We can calculate pOH and hence pH, given Kb.
Chem 1102 1B 29
Weak Bases
Chem 1102 1B 30
Note effect of changing what the N is attached
to
6
pKa and pKbFor conjugate systems (Brønsted‐Lowry acid‐base pairs)
As acid (HA):HA (aq) ⇌ H+ (aq) + A– (aq)
As conjugate base (A– ):A– (aq) + H2O (l) ⇌ HA (aq) + OH– (aq)
•Hence only need values of pKa , since pKb = 14 – pKa
Chem 1102 1B 31
pKa + pKb = 14.0
ExampleFind the pH of 10–2 M NaHCO2 (pKa of formic acid (HCO2H) is 4.1)(check: it will be basic so expect pH > 7)
Chem 1102 1B 32
HCO2- (aq) + H2O (l) ⇌ OH- (aq) + HCO2H (aq)
before: 10-2 0 0
after: 10-2 – x x x
pKb = 14 – pKa = 14 – 4.1 = 9.9
Example (continued)
Chem 1102 1B 33
(indeed x << 10-2 )
x = [OH-] so pOH = 5.95 ; pH = 14.0 – 5.95 = 8.05
= 8.1 (to one significant figure)
When we say 10-2 M, experimentally we mean
0.95 × 10-2 M < conc < 1.05 × 10-2 M
Q. What would the pH be for 1.05 × 10-2 M NaHCO2 ?0.95 × 10-2 M NaHCO2 ?
Weak Acids and BasesHA (aq) ⇌ H+ (aq) + A– (aq)
Acid dissociation constantpKa = – log10Ka
More positive pKa⇒ weaker acid (and stronger conjugate base)
Chem 1102 1B 34
B (aq) + H2O (aq) ⇌ HB+ (aq) + OH– (aq) Base ionisation constant
pKb = – log10KbMore positive pKb⇒ weaker base (and stronger
conjugate acid)
Relative Strengths of Acids and Bases
Chem 1102 1B 35
Relative Strengths of Acids and Bases
Strongest acids lose their protons easily:For H-X bonds, larger X → weaker H-X bonds → stronger acids.For H-O-X bonds, more electronegative X → weaker H-O bonds → stronger acid
Chem 1102 1B 36
7
Relative Strengths of Acids and Bases
Strongest bases hold on to protons strongly:
An acid-base equilibrium always lies in the direction of the weaker acid and weaker base.
Chem 1102 1B 37
HCN (aq) + CH3COO– (aq) ⇌ CH3COOH (aq) + CN– (aq)
weaker acid & weaker base ← stronger acid & stronger base
pKa = 9.2 pKb = 9.3 pKa = 4.7 pKb = 4.8
Strong Acids and Bases
Completely ionise in water: e.g. HCl (aq) ⇌ H+ (aq) + Cl– (aq)
‐ equilibrium lies completely to right, Ka ≈ ∞Strong acids:
H2SO4 , HCl, HBr, HI, HNO3, HClO4
Strong bases:All hydroxides of Groups 1 & 2 (except Be):NaOH, Ca(OH)2, …
Chem 1102 1B 38
Learn these!
Example
What is the pH of a 0.1 M HCl solution?
Chem 1102 1B 39
What are the sources of [H+]?HCl (aq) → H+ (aq) + Cl- (aq) [H+] = 0.1 MH2O (l) ⇌ H+ (aq) + OH- (aq) [H+] = 10-7 M
The water equilibrium will be driven to the left by the high value of H+ from the HCl hence [H+] from auto-ionisation of water will be negligible and similarly [OH-] is negligible.
Thus [H+] = 0.1 M and pH = – log10 [H+] = 1.0
Example:
What is the pH of a 0.002 M NaOH solution?
Sodium hydroxide is fully soluble thus totally ionised
Chem 1102 1B 40
Thus [OH– ] = 0.002 M
pOH = – log10 [OH– ] = – log10 (0.002) = 2.7
pH = 14 – 2.7 = 11.3
More Examples
Calculate the pH of:
1) 0.001 M HNO3
2) 0.001 M NaOH
3) 0.001 M Ca(OH)2
Chem 1102 1B 41
Solutions
1) HNO3 is a strong acid thus fully dissociated.
HNO3 + H2O H3O+ + NO3
‐
[H3O+] = 10‐3 M Thus pH = ‐log10 10
‐3 = 3.0
2) Sodium hydroxide is fully soluble
[OH‐] = 10‐3 M and pOH = ‐log1010‐3 = 3.0
pH + pOH = 14 therefore pH = 14‐3.0 = 11
3) Ca(OH)2 gives 2 mole of OH‐ for each mole thus [OH‐] = 0.002M and pOH = ‐log100.002 =
Chem 1102 1B 42
8
Alternate question?
1) What is the pH of a solution formed by mixing 400 mL of 0.05 M HCl with 600 mL of 0.05 M NaOH
Chem 1102 1B 43
Solutions:1) First what is the reaction? Neutralization thus is only
H+ + OH‐ H2OWe are given concentrations and volumes thus when the
solutions are added final volume and concentrations must be calculated.
n= C x V from HCl mol of H+ = 0.05 M x 0.40L = 0.02molFrom NaOH mol of OH‐ = 0.05 M x 0.60L = 0.03molVolume = 1.0 L Excess OH‐ thus [OH‐] = (0.03‐0.02)/1.0L = 0.01MpOH = ‐log100.01 = 2.0 and pH = 12
Chem 1102 1B 44
Alternate question?
2) What is the [H+] of a solution with a pH of 4.5?
Chem 1102 1B 45
Solution:
If pH= 4.5 then [H+] = 10‐4.5 or 3.2 x 10‐5
Harder Example
What is the pH of a 2.0 × 10‐7 M HCl solution?
Chem 1102 1B 46
What are the sources of [H+]?
HCl (aq) → H+ (aq) + Cl‐ (aq) [H+] = 2 × 10‐7 MH2O (l ) ⇌ H+ (aq) + OH‐ (aq) [H+] = 10‐7 M
So is [H+] = 2 × 10‐7 M ??
NO ‐ the water equilibrium will be driven to the left by the acid.
Solution:
0cbxaxwhere
a2
ac4bbx
2
2
Chem 1102 1B 47
Conc (M) H2O ⇌ H+ + OH‐
Initial 2 × 10‐7 0
H2O autoionisation: (2 × 10‐7 + x) x
Kw = [H+][OH‐] = (2 × 10‐7 + x)(x) = 10‐14
Solve quadratic equation x = 4.14 × 10‐8 M,
giving pH = 6.62
Polyprotic Acids
H3PO4 (aq) ⇌ H+ (aq) + H2PO4– (aq)Ka1 = 10
‐2.2
H2PO4– (aq) ⇌ H+ (aq) + HPO4
2– (aq) Ka2 = 10‐7.2
HPO42‐ (aq) ⇌ H+ (aq) + PO4
3– (aq) Ka3 = 10‐12.4
Ka1 > Ka2 >Ka3
Harder to remove +ve charge against increasing ‐vecharge.
consider one equilibrium at a time
Chem 1102 1B 48
9
Polyprotic Acids
Chem 1102 1B 49
Ionisableprotons a long way apart
Ionisableprotons close together
Salts of Weak Acids and Bases
Is a solution of NaNO2 acidic or basic?
Take the ions apart
Na+ is from the strong base NaOH
NO2‐ is from the weak acid and HNO2 .
The base “wins”
pH > 7.
Chem 1102 1B 50
Overall reaction
H2O (aq) + NO2– (aq) ⇌ OH– (aq) + HNO2 (aq)
We can discount the other reaction:
H2O (aq) + Na+ (aq) ⇌ NaOH (aq) + H+ (aq)
Sodium hydroxide will not reform
NaOH is a strong base and cannot coexist with H+
Chem 1102 1B 51
Salts of Weak Acids and Bases
Does a solution of NH4Cl have pH > 7 or < 7?
Separate the salt ions
NH4+ comes from the weak base NH4OH
Cl‐ comes from the strong acid HCl.
The acid “wins”
pH < 7.
Chem 1102 1B 52DEMO: ammonium chloride in water
Overall reaction
H2O (aq) + NH4+ (aq) ⇌ NH3 (aq) + H3O
+ (aq)
We can discount the other reaction:
H2O (aq) + Cl‐ (aq) ⇌ HCl (aq) + OH‐ (aq)
because HCl is a strong acid and cannot coexist
with OH‐
Chem 1102 1B 53
Salts of Weak Acids and Bases
What is the pH of ammonium acetate at 25 °C ?2 possible reactions are:
CH3COO– (aq) + H2O (l) ⇌ CH3COOH (aq) + OH
– (aq)
Kb = 10–9.24
Anions of weak acids hydrolyse→ OH–
NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H3O
+ (aq)
Ka = 10–9.24
Cations of weak bases hydrolyse→ H3O+
In this case Ka = Kb so salt is neutral!
Chem 1102 1B 54
LOOK at the Ka and Kb how different are theyDEMO 5.13 + ammonium acetate in water
10
Salts of Strong Acids and Bases
Does a solution of NaCl have pH > 7 or < 7?We can discount the reaction:
H2O (aq) + Na+ (aq) → NaOH (aq) + H+ (aq)
because NaOH is a STRONG BASE and cannot coexist with H+, and we can discount the reaction:
H2O (aq) + Cl‐ (aq) → HCl (aq) + OH‐ (aq)
because HCl is a STRONG ACID and cannot coexist with OH‐
Salts of strong acids and strong bases are always neutral
Chem 1102 1B 55
pKa of Weak Acid and its Salt
What is the pH of a solution made up as 0.1 M in acetic acid (HAc) and 0.1 M in sodium acetate (pKa of HAc = 4.7)?
Reactions are:
CH3COOH (aq) + H2O (l) ⇌ CH3COO‐ (aq) + H3O
+ (aq)
NaCH3COO (aq) + H2O (l) ⇌ Na+ (aq) + CH3COO‐ (aq)
We have COMMON ION
Chem 1102 1B 56
Solution:
take 0.1 mol of CH3COOH and 0.1 mol NaCH3COO and dilute to 1 L
Can work either from acid or base dissociation constants. Take acid this time:
HAc (aq) ⇌ H+ (aq) + Ac‐ (aq)
Before: 0.1 0 0.1
After: 0.1 – x x 0.1 + x
(xmol HAc dissociates and forms xmol of Ac–)
Chem 1102 1B 57
could solve as quadratic, but make usual assumption that x << 0.1
pH = – log x = 4.7 = pKa
The Common Ion EffectIf you add the salt of an acid to a solution of the same acid then the equilibrium will shift towards neutral.
CH3COOH (aq) + H2O (l) ⇌ CH3COO‐ (aq) + H3O
+ (aq)
Addition of CH3COONa will boost [CH3COO‐]
By Le Châtelier’s principle the first equilibrium will shift to the left to remove CH3COO
‐ and therefore decrease [H3O
+].Same for a baseIMPORTANT FOR BUFFER SOLUTIONS
Chem 1102 1B 58
BuffersA solution containing both:
a weak acid + its salt
OR
a weak base + its salt
withstands pH changes when (limited) amounts of acid or base are added.
Reason: Le Châtelier’s principle.– if we add acid, then reaction HA ⇌ H+ + A– goes to
left to absorb change;
– vice‐versa if we add base
Chem 1102 1B 59
Sodium bicarbonate is used as a buffer in swimming pools
DEMO 5.11
Buffers
Consider the change in pH of pure water (pH = 7) if we add an equal amount of 10–3 M HCl:
Take 500 mL of water pH = 7.0 ([H+] = 10‐7M
Add 500 mL of 10‐3 M HCl
[H+] = 5 × 10–4 M (can neglect amount already present in water), so pH goes from 7 to 3.3!
Huge change!
What about buffers?
Chem 1102 1B 60
11
ExampleConsider a buffer solution with 0.1 M each of sodium acetate (NaAc) & acetic acid (HAc):
(found earlier that pH = 4.7)
What is the pH when 10–3 M HCl is added?
Chem 1102 1B 61
HAc ⇌ H+ + Ac–
initially: 0.1 10–3 0.1eqm: 0.1+10–3–x x 0.1–10–3 + x
i.e., suppose all but x of the added H+ forms HAc, but x will be very small even in comparison to 10‐3.
Chem 1102 1B 62
x = 1.02 × Ka = 0.000020 << 0.001
pH = – log x = 4.69
the pH hardly changes from 4.7!
Solution is buffered against pH change
Example (continued)
Chem 1102 1B 63
Henderson ‐ Hasselbalch Equation
For a buffer solution, which contains similar concentrations of a conjugate acid/base pair of a weak acid:
Chem 1102 1B 64
Since the dissociation of HA or
protonation of A‐ doesn’t lead to a significant change in the concentrations of these species.
Chem 1102 1B 65
Buffer Preparation
If the pH of a required buffer is the same as pKa of an available acid then use equimolaramounts of acid and conjugate base.
If the required pH differs from the pKa then use the Henderson‐Hasselbalch equation.
Chem 1102 1B 66
12
Buffer Preparation
]HA[
]A[logpKpH 10a
Chem 1102 1B 67
1) Choose conjugate acid‐base pair –chosen mostly by the desired pH. The buffer is most effective when the ratio of the acid/base is close to 1 pH close to pKa.
2) Calculate the ratio of the buffer component concentrations. Find the ratio of [A‐]/[HA] that gives the desired pH. Use the Henderson‐Hasselbalch equation
Example 1In the H3PO4 / NaH2PO4 / Na2HPO4 / Na3PO4
system, how could you make up a buffer with a pH of 7.40?
DATA: Ka1 = 7.2×10‐3, Ka2 = 6.3×10
‐8, Ka3 = 4.2×10‐13
Chem 1102 1B 68
H3PO4 ⇌ H2PO4‐⇌ HPO4
2‐ ⇌ PO43‐
38.12]10x2.4[logp
20.7]10x3.6[logp
14.2]10x2.7[logp
1310
810
310
a3
a2
a1
K
K
K
58.110
log20.740.7
logpKpH
20.0
10
10a
]PO[H
][HPO
]PO[H
][HPO
]PO[H
][HPO
42
24
42
24
42
24
Chem 1102 1B 69
pKa1 = 2.14 pKa2 = 7.20 pKa3 = 12.38
Must use mixture of H2PO4‐ and HPO4
‐. Could go
through whole procedure, or simply use
Henderson‐Hasselbalch equation
so the required ratio of Na2HPO4 to NaH2PO4
1.58:1
Example 2 ]HA[
]A[logpKpH 10a
Chem 1102 1B 70
Prepare a buffer solution with a pH = 3.90.
4.110]HCOOH[
]HCOO[
16.0]HCOOH[
]HCOO[log
]HCOOH[
]HCOO[log74.390.3
16.0
For every 1.0 mol of HCOOH we need 1.4 mol of HCOONa
The pKa of formic acid is 3.74 or Ka = 1.8 x 10‐4. The buffer
components can be formic acid HCOOH and the formateion HCOO‐ supplied by a soluble salt such as sodium
formate, HCOONa.
To calculate the component ratios use the equation
Buffer Capacity
Buffer capacity is related to the amount of strong acid or base that can be added without causing significant pH change.
Depends on amount of acid & conjugate base in solution:
‐ highest when [HA] and [A–] are large‐ highest when [HA] ≈ [A–] (most effective buffers have acid/base ratio less than 10 and more than 0.1)
pH range is ±1
Chem 1102 1B 71
Buffers in Natural SystemsBiological systems, e.g. blood, contain buffers: pH control essential because biochemical reactions are very sensitive to pH
Human blood is slightly basic, pH ≈ 7.39 – 7.45
In a healthy person, blood pH is never more than 0.2 pH units from its average value
pH < 7.2, “acidosis”; pH > 7.6, “alkalosis”
Death if pH < 6.8 or > 7.8
Chem 1102 1B 72
13
Buffers in Natural SystemsTo hold the pH of the blood close to 7.4 the body uses 3 buffer systems:
• Carbonate the most important
• Phosphate
• proteins
Chem 1102 1B 73
The acid is carbonic acid, H2CO3
the base is the carbonate HCO3‐.
The pKa of H2CO3 is 6.37.
Since the pH of an equimolar mixture of acid and conjugate base is equal to its pKa, a buffer made of equal
concentrations of H2CO3 and HCO3‐ has a pH of 6.37
Blood Buffering
10]CO[H
][HCO
32
3
Chem 1102 1B 74
• Blood however has a pH of 7.4. The carbonate buffer can maintain this pH only if ‐
[H2CO3] is not equal to [HCO3‐]
In Fact
The normal concentrations in blood are about 0.0025 M H2CO3
BLOOD 10]CO[H
][HCO
32
3
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This means that it is a better buffer system for acids which lowers the ratio than for bases which increases the ratio.
This is in harmony with the functioning of the body as more acidic than basic substances enter the blood.
This ratio can be easily maintained as the body can readily alter the amount of CO2
entering the blood
Buffer System in Blood
“Extracellular” buffer (outside cell)
H+ (aq) + HCO3– (aq) ⇌ H2CO3 (aq)
H2CO3 (aq) ⇌ H2O (l) + CO2 (g)
Removal of CO2 shifts equilibria to right, reducing [H+], i.e., raising the pH
The pH can be reduced (made acidic) by:
H2CO3 (aq) + OH– (aq) ⇌ HCO3
– (aq) + H2O (l)
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How much acid or base is in a given amount of solution?
• If solution is ACIDIC – add base until the acid is neutralised
• If solution is BASIC – add acid until base is neutralised
1. The unknown solution goes in the conical flask 2. The solution of known CONCENTRATION goes in the burette
3. The INDICATOR is chosen to change colour at the appropriate pH.
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TitrationsEquivalence Point:
•When number of moles of
added base = original no. of moles of acid
– Strong acid/strong base pH = 7
– Weak acid/strong base pH > 7
– Strong acid/weak base pH < 7
•End Point:
•When a colour change in the indicator is observed
•⇒ Choose an indicator that changes colourclose to the equivalence point
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Indicatorsweak acid ⇌ weak base
•Each form has a different colour.
•The pH at which acid → base depends on the pKa of the indicator.
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Salts of Acids and Bases
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Salt acidic/basic/neutral
NaOH BASIC – STRONGLY
NaHCO3 SLIGHTLY BASIC
Na2CO3 BASIC
NaCl NEUTRAL
NH4Cl ACIDIC
CH3COONH4 NEUTRAL
Al2(SO4)3 ACIDIC
KHSO4 ACIDIC
CH3COOH ACIDIC
HCl ACIDIC –STRONGLY
Salts of Acids and Bases
ReasoningNaHCO3 Na+ +HCO3
‐ HCO3‐+ H2O ?
Na2CO3 Na+ +CO32‐ CO3
2‐+ H2O HCO3‐+ OH‐
NH4Cl NH4+ + Cl‐ NH4
+ + H2O NH3 + H3O+
Al2(SO4)3 SO42‐ + Al(H2O)6
3+ + H2O Al(H2O)5(OH‐)2+ + H3O
+
KHSO4 K+ + HSO4‐ + H2O K+ + H3O
+ + SO42‐
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Titrations: Strong Acid / Strong Base
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25.0 mL of HCl solution is titrated with NaOH
Titrations: Weak Acid / Strong Base
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Equivalence point pH > 7 (value depends on starting concentrations)
Change is more gradual
Titrations: Weak Base / Strong Acid
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Equivalence point pH < 7 (value depends on starting concentrations)
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5. Acids and Bases: Summary
•You should understand:
– The Brønsted‐Lowry concept of an acid as a proton donor and a base as a proton acceptor.
– The difference between strong acid (or base) and a weak acid (or base); and the role of the conjugate base (or acid) in equilibrium.
– The temperature dependence of Kw ∝ exp(‐∆H0/RT) and therefore pH.
– Trends in the relative strengths of acids and bases due to bond strength and electronegativity.
– How a buffer solution works due to the presence of a weak acid (or base) and its salt.
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• You should know:– The definition of the acid dissociation constant
Ka = [H+][A‐]/[HA].– The “p” convention, e.g.,
pH = ‐log10[H+], pOH = ‐log10[OH‐], pKa = ‐log10[Ka].– That pH + pOH = pKw = 14, and that pKa + pKb = pKw = 14.
You should be able to determine– The conjugate base of a given acid, and the conjugate acid of a
given base.– The pH of a strong or weak acid or base in solution.– The extent of ionisation of a weak acid or base in solution.– The pH of a buffer using Henderson‐Hasselbalch pH ≈ pKa +
log([added base]/[added acid]).– The equivalence point of a titration.
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