JF Basic Chemistry Tutorial : Acids & Bases
Shane Plunkett
plunkes@tcd.ie
Acids and Bases
• Three Theories
• pH and pOH
• Titrations and Buffers
Recommended reading
• M.S. Silberberg, Chemistry, The Molecular Nature of Matter and Change
Acids and Bases
• Dealing with ions
• Arrhenius theory – an acid which contains hydrogen and can
dissociate in water to produce positive hydrogen ions
e.g.
HX +H2O ↔ H+ (H3O+) + X-
– A base reacts with a protonic acid to give water (and a salt)
e.g.
HCl + NaOH  NaCl + H2O
• BrØnsted-Lowry Theory – acids are proton donors; bases are proton
acceptors
e.g.
HCN + H2O ↔ H3O+ + CN-
HCN is an acid, in that it donates a proton to water. Water is acting as a
base, as it accepts that proton
• Lewis Theory – an acid accepts a pair of electrons; a base donates a
pair of electrons
e.g.
HCl + NaOH  NaCl + H2O
• The strength of an acid depends on the extent to which is dissociates
and is measured by its dissociation constant
Strong and weak acids and bases
• Strong acid – fully dissociates in water, i.e. almost every molecule
breaks up to form H+ ions
• Some strong acids are…HCl, H2SO4, HNO3
• Weak acid – partially dissociates in water
• Some weak acids are…carboxylic acids such as CH3COOH,
C2H5COOH
• Strong base – fully dissociates in water, i.e. almost every molecule
breaks up to form OH- ions
• Some strong bases are….NaOH, compounds which contain OH- ions
or O2- ions
• Weak base – partially dissociates in water
• Some weak bases…nitrogen-containing compounds, such as NH3
• Strengths can be determined by the acid or base dissociation
constant
Acids
•
•
•
•
Act as proton donors
Electron pair acceptors
Strong acids dissociate fully in water.
Weak acids partially dissociate.
• Ka: acid dissociation constant
HA + H2O  H3O+ + AKa = [H3O+][A-]
[HA]
• Higher Ka values mean stronger acids
Bases
•
•
•
•
•
Act as proton acceptors
Electron pair donors
Strong bases dissociate fully in water
Weak bases partially dissociate
Kb: base dissociation constant
pH and pOH
[H3O+] can vary greatly  logarithmic scale used
pH = -log [H3O+]
pOH = -log [OH-]
pH > 7 basic
pH = 7 neutral
pH < 7 acidic
Can also express dissociation constants in terms of logs:
pKa = -log Ka
•  the higher the Ka the lower the pKa
• Similarly for bases
•
•
•
•
•
•
•
Ionic product of water
2H2O  H3O+ + OHKw is the ion-product constant. Kw is the product of the molar
concentrations of H3O+ and OH- ions at a particular temperature
+][OH-]
Kw = [H
O
3
________
[H2O] is constant
[H2O]2
Kw = [H3O+][OH-] = 1.0 × 10-14 at
25°C
[H3O+] = [OH-] Solution is neutral
pKw = -log Kw
[H3O+] > [OH-] Solution is acidic
[H3O+] < [OH-] Solution is basic
• Can incorporate pH and pOH
Kw = [H3O+][OH-]
-log Kw = -log [H3O+][OH-]
pKw = pH + pOH = 14 (at 25°C)
Titrations
• Acid-base stoichiometric point can be determined by using an
indicator
• Indicators are dyes whose colour depends on the pH:
HIn is protonated form and In- is its conjugate base
HIn + H2O ↔ H3O+ + InKIn = [H3O+][In-]
[HIn]
• If pH goes from low to high, the indicator goes from HIn to In- and a
colour change is noted
14
pH
14
7
pH
1
14
7
pH
1
Vol of acid added to base
Strong acid-base curve
7
1
Vol of acid added to base
Strong acid-weak base
Vol of acid added to base
Strong base-weak acid
Buffers
• Allow pH to be maintained over
small additions of acid or base


A
H
O
3
• Made up of a weak acid and its K 
a
conjugate base, e.g.
HA
HA + H2O  A- + H3O+
 HA  

H 3O  K a   
acid
conjugate base
 A 
• The equilibrium will shift to the
right on addition of a small  log H O   logK  log  HA  
3
a
 A 
amount to base and shifts to the


left on addition of small amounts

A
of acid
pH  pK a  log
HA 
• Henderson-Hasselbalch equation
allows determination of pH in
buffer systems:
 



 


 
 
Questions
What is the pH of a 0.14M aqueous solution of HCN (pKa = 9.31)?
Ka = 4.9 × 10-10
pKa = -log Ka
HCN + H2O  H3O+ + CNKa = [H3O+][CN-]
[HCN]
Initial conc
HCN
H3O+
CN-
0.14
0
0
x
x
Equil. Conc 0.14 – x
Ka = [x][x]
[0.14-x]
= 4.9 ×10-10
x2
0.14-x
.
= 4.9 × 10-10
x2 = 4.9 × 10-10 (0.14 – x)
2
-10
-11
x2 = 6.86 × 10-11 – 4.9 ×10-10x x + 4.9×10 x – 6.86×10 = 0
Solve for x
x = 8.25×10-6  [H3O+] = 8.25×10-6
pH = -log [H3O+]
Question
pH = -log (8.25×10-6)
pH = 5.08
The pH of 0.1M CH3COOH is 2.87. What is the value of the acid
dissociation constant, Ka?
pH = -log [H3O+]
 [H3O+] = 1.35×10-3
= 2.87
CH3COOH + H2O  H3O+ + CH3COOKa = [H3O+][CH3COO-]
[CH3COOH]
CH3COOH
H3O+
CH3COO-
Initial conc
0.1
0
0
Equil. Conc
0.1 – x
x
x
Ka = [1.35×10-3][1.35×103]
[0.1]
Ka = 1.8 × 10-5
Question
Estimate the pH of 10-7M HCl(aq)
pH = -log [H3O+]
Two contributions to H3O+ concentration: HCl and H2O (aqueous soln)
pH of pure water = 7
pH = -log [H3O+]
If -log [H3O+] = 7
 [H3O+] = 1 ×
10-7
HCl + H2O  H3O+ + Cl-
HCl is a strong acid.  dissociates completely in water  [H3O+] = 10-7
Total [H3O+] = 1.0 × 10-7 + 10-7
= 2.0 × 10-7
pH = -log (2.0 × 10-7)
pH = 6.7
Question
Calculate the concentration of OH- for an aqueous solution with a pH
of 9.60
Two contributions to OH- concentration.
pH of pure water = 7.0
 [OH-] = 1.0 ×10-7
pH of solution = 9.60
pH + pOH = 14
 pOH = 14 – 9.60 = 4.4
pOH = -log [OH-] = 4.4
[OH-] = 3.98 × 10-5
 Total [OH-] = 3.99 × 10-5  4.0 × 10-5
2001 Paper 1 Question 8 (b)
A buffer solution is made by adding 0.3 mol of acetic acid and 0.3 mol of
sodium acetate to enough water to make 1L of solution. Determine the
pH of the buffer solution. The Ka for acetic acid is 1.8 × 10-5
pH = pKa + log [A-] / [HA]
A- = sodium acetate
HA = acetic acid
pH = -log (1.8 × 10-5) + log (0.3) / (0.3)
pH = 4.74
Calculate the pH of the buffer solution after 0.02 moles of sodium
hydroxide is added
1 litre of buffer contains 0.3 moles of sodium acetate and 0.3 moles of
acetic acid. Sodium hydroxide is a strong base. The acetic acid will
react with the base added to try to maintain the pH. What the acid
loses in concentration, the salt (sodium acetate) will gain.
CH3COOH + OH-  CH3COO- + H2O
CH3COOH
OH-
CH3COO-
Initial conc
0.3
0
0.3
Change
0.3 – 0.02
0.02
0.3 + 0.02
Equil. Conc.
0.28
0.02
0.32
pH = pKa + log [A-] / [HA]
pH = - log (1.8 × 10-5) + log (0.32) / (0.28)
pH = 4.74 + 0.058
pH = 4.8