Review
of
Acids
HCl Strong Acid HCl → H+ + Cl-
0.10 M 0.10 M
pH = -Log[H+] = 1.0 No ICE
HF Weak Acid HF ⇌ H+ + F-
I 0.10 M 0 0
C x x x
E 0.10 - x x x
small Ka
x2 = 3.5 x 10-4 x = 0.005916 M
0.10
pH = -Log[0.005916] = 2.23
NaOH Strong Base NaOH → Na+ + OH-
0.20 M 0.20 M
pOH = -Log[OH-] = 0.70 No ICE
NH3 Weak Base NH3 + H2O ⇌ NH4+ + OH-
I 0.20 M 0 0
C x x x
E 0.20 - x x x
small Kb
x2 = Kb = Kw = 1.0 x 10-14 = 1.786 x 10-5
0.20 Ka 5.6 x 10-10
x = 0.001890 M
pOH = -Log[0.001890] = 2.73
pH = 11.27
CaO Basic Anhydride CaO + H2O → Ca(OH)2
SO2 Acid Anhydride SO2 + H2O → H2SO3
Salts
NaCl neutral salt NaCl → Na+ + Cl-
KCN basic salt
CN- + H2O ⇌ HCN + OH-
hydrolysis
NH4Cl acidic salt
NH4+ + H2O ⇌ H3O+ + NH3
hydrolysis
FeCl3 acidic salt
Fe(H2O)63+ ⇌ H+ + Fe(H2O)5(OH)2+
hydrolysis- copy off chart!
Know the pH scale
0 7 14
HCl HF NaCl NH3 NaOH
Assuming 1M
Buffers Maintain the pH
Weak Acid + Salt ( conjugate)
HF KF
HF ⇌ H+ + F-
High Low High
Indicator Change colour with pH
HInd ⇌ H+ + Ind-
Acid Low Basic
Indicators Change colour with pH
methyl red
HInd ⇌ H+ + Ind-
red yellow page 7
@ trans pt [HInd] = [Ind-] colour is orange
and Ka = [H+]
pH = 4.8 + 6.0 = 5.4
2
Ka = [H+] = 10-5.4 = 4 x 10-6
Acid Rain SO2 NO2 pH ≤ 5
Normal Rain CO2 pH = 6
Amphriprotic H2C2O4 acid
HC2O4- ampriprotic
C2O42- base
Is HC2O4- an acid or base?
Ka (HC2O4-) = 6.4 x 10-5
Kb (HC2O4-) = 1.0 x 10-14 = 1.7 x 10-13
5.9 x 10-2 acid
Is NH4CH3COO an acid or base?
Ka (NH4+) = 5.6 x 10-10
Kb (CH3COO-) = 1.0 x 10-14 = 5.6 x 10-10
1.8 x 10-5 neutral
100.0 mL 0.100 M H2SO4 is mixed with 220.0 mL of 0.100 M NaOH. What is the pH of the resulting solution?
H2SO4 + 2NaOH
0.1000 L x 0.100 mole 0.2200 L x 0.100 mole
1 L 1 L
I 0.0100 mole 0.0220 mole
C 0.0100 mole 0.0200 mole
E 0 mole 0.0020 mole
[NaOH] = 0.0020 mole = 0.00625 M
0.320 L pOH = 2.20
pH = 11.80
Complete the reaction and state if the reactants or products are favoured.
HCO3- + HSO3-
Base Acid
HCO3- + HSO3- ⇌ H2CO3 + SO32-
weaker weaker stronger stronger base acid acid base
Reactants are favoured
Know the Equations
E + H2O ⇌ H+ + OH-
Kw = [H+][OH-] = 1.0 x 10-14
pH = -Log[H+] pOH = -Log[OH-]
[H+] = 10-pH [OH-] = 10-pOH
pH + pOH = pKw = 14 @ 25 oC
@ other temperatures pure water is always neutral
& [H+] = [OH-] & pH = pOH ≠ 7