Acids and Bases
Acids and Bases • Electronegativity: the electron-attracting power of an atom in a molecule • Fluorine (the most electronegative element) is assigned a value of 4.0, and the values range down to caesium and francium which are the least electronegative at 0.7 • Electronegativity increases across the table and decreases down a group • A large electronegativity difference leads to an ionic bond • A small electronegativity difference leads to a polar covalent bond • Uses of common acids
• • • • •
Substance Acetic acid (vinegar) Citric acid Phosphoric acid Boric acid Aluminium salts
Formula CH3COOH H3C6H5O7 H3PO4 B(OH)3 H3BO3 NaAl(SO4)2•12H2O
Hydrochloric acid
HCl
Substance Sodium hydroxide
Formula NaOH
Ammonia Sodium carbonate Sodium hydrogen carbonate
NH3 Na2CO3 NaHCO3
Trisodium Phosphate
Na3PO4
Use Flavouring, preservative Flavouring Rust remover Mild antiseptic; insecticide In baking power, with sodium hydrogen carbonate • Brick and ceramic tile cleaner
• • • • •
Use • Oven cleaner, unblocking plumbing • Household cleaner • Water softener, grease remover • Fire extinguisher, rising agent in cake mixes (baking soda), mild antacid • Cleaner for surfaces before painting or wallpapering
Organic acids have a carboxyl group in them Many acids are oxy acids where the proton is attached to an oxygen atom Hydrohalic acids: HF, HCl, HBr and HI Organic bases have a nitrogen atom with a lone pair Hydroxides: LiOH, NaOH, KOH
Arrhenius 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 Arrhenius was the first to recognise the nature of acids and bases He postulated o Acids produce hydrogen ions in aqueous solutions – H+ o Bases produce hydroxide ions – OHo Acids and bases will neutralise each other when mixed they produce water and an ionic salt, neither of which are acidic or basic The more general definition is the Brønsted-Lowry definition They considered that acid-base reactions involved proton (H+) transfer between an acid and a bases, which they defined as follows o An acid is a proton donor o Base is a proton acceptor Acids that can donate only one proton are monoprotic, and acids that can donate more than one proton are polyprotic Gilbert Newton Lewis: o Acid = lone pair acceptor o Base = line pair donor
• In water: an acid ionises to produce H+(aq) – actually H3O+(aq) but we usually just write H+(aq)
Acid, Bases and Equilibrium
HA(aq) + H2O(l) H3O+(aq) + A-(aq) • A strong acid has equilibrium to the right (HA dissociated completely) • The majority of acids exist as weak acids, an acid which dissociates only partially • An acid-base dissociated will have a particular equilibrium constant which will determine the extent of the reaction (whether it lies to the left or right of the equation) • The equilibrium equation
[𝐻3 𝑂+ ][𝐴− ] 𝐾= [𝐻𝐴] • Ka is the acid dissociation constant
Conjugate Acid-Base Pairs • • • •
A conjugate base has one less proton that its conjugate acid NH4+ is the conjugate acid of NH3 NH3 is the conjugate base of NH4+ HSO4o Conjugate base is SO42o Conjugate acid is H2SO4 • H2SO4 is a dibasic or diprotic acid o H2SO4(aq) + H2O(l) H3O+(aq) + HSO4-(aq) o HSO4-(aq) + H2O(l) H3O+(aq) + SO42-(aq) Acid H3O+ H2SO4 HClO4 CH3COOH HPO42-
Conjugate base H2O HSO4ClO4CH3COOPO43-
Base OHH2O CNNH3 HPO42-
Conjugate acid H2O H3O+ HCN NH4+ H2PO4-
Acid-Base Reactions • • • •
HF(aq) + H2O(l) F-(aq) + H3O+(aq) HCOOH(aq) + CN-(aq) HCOO-(aq) + HCN(aq) H2PO4-(aq) + OH(aq) HPO42-(aq) + H2O(l) NH4+(aq) + CO32- NH3(aq) +HCO3-(aq)
Autoionisation of Water • Water can react with itself to generate hydronium and hydroxide ions according to the equation
H2O(l) H+(aq) + OH-(aq) • This reaction, in which a proton is transferred between two identical molecules, is called autoionisation • Equilibrium constant given special symbol Kw=[H+][OH-] • At 25°C, Kw = 1.0 × 10-14 • It is a very small value, which shows that the autoionisation of water proceeds only to a very small extent and that the equilibrium concentrations of [H3O+] and [OH-] in pure water are small • Neutral solution: [H+]=[OH-] = 1.0 × 10-7 mol L-1 • Acidic solution [H+] > 1.0 × 10-7 M
• Basic solution [H+] < 1.0 × 10-7 M
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+] • Example o If [H+] = 1.0 × 10-6 M then pH = -log10[10-6] = -(-6) = 6.0 o If [H+] = 1.6 × 10-5 M then pH = -log10[10-4.8] = -(-4.8) = 4.8 • pH = -log10[H+] • pOH = -log10[OH-] • pKw = -log10[Kw] = 14 at 25°C • Acid: pH < 7 • Neutral: pH = 7 • Basic pH > 7 • Since Kw = [H+][OH-] o log10Kw = log10[H+] + log10[OH-] o -log10[H+] – log[OH-] = -log10Kw • pH + pOH = 14 pOH = 14 – pH
Temperature Dependence of pH • The formation of hydrogen ions and hydroxide ions from water is an endothermic process
H2O(l) H+(aq) + OH-(aq) • • • • • • • • •
The forward reaction absorbs heat If you increase the temperature of the water, the equilibrium moves to lower the temperature again It will do that by absorbing the extra heat That means that the forward reaction will be favoured, and more hydrogen ions and hydroxide ions will be formed the effect of that is to increase the value of Kw as temperature increases Kw = 1.0 × 10-14 only at 25°C For T > 25°C, Kw > 10-14 For T < 25°C, Kw < 10-14 pH + pOH ≠ 14 if T≠ 25°C Neutral pH ≠ 7 if T ≠ 25°C
Weak Acids • Most acids or bases are weak – they do not completely ionise in water
HA(aq) H+(aq) + A-(aq) • Acid dissociation constant
𝐾𝑎 =
[𝐻+ ][𝐴− ] [𝐻𝐴]
Relationship between the Ka and pKa • The more positive the pKa the weaker the acid (and the stronger the conjugate base) • The larger the value of the Ka the stronger the acid and the lower the value of pK a
pKa = -log10Ka
Example • Find the pH of 0.1 M acetic acid (CH3COOH (HAc)) • Data pKa = 4.7 Ka= 10-4.7
Before Equilibrium After Equilibrium Equilibrium
HA 0.1 0.1 - x 0.1 - x
H2O -
A0 +x x
H3O+ 0 +x x
F0 +x x
H3O+ 0 +x x
𝑥2
Ka = 10-4.7 = 0.1−𝑥 • • • • • • •
Since the equilibrium constant is very small we assume x < 0.1 i.e. (0.1 – x) = 0.1 10-4.7 ≈ x2/0.1 x2 ≈ 0.1 × 10-4.7 = 10-5.7 x = √10−5.7 = 0.00141253754 pH = -log10[H+] = -log10[0.00141253754] = 2.9 Check that x=0.00141253754 < 0.1, (0.1 – x) = 0.0986 M
Example • Find % ionisation of 0.50 M HF (pKa = 3.1)
Before Equilibrium After Equilibrium Equilibrium
HF 0.50 0.50 - x 0.50 - x
H2O 𝑥2
Ka = 10-3.1 = 0.50−𝑥 • • • •
If x < 0.50, 10-3.1 ≈ x2/0.50 x2 = 0.50 × 10-3.1 x = 2 × 10-2 M % ionisation = x/0.50 ×100 = 4% (indeed a weak acid)
Weak Bases
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) • Equilibrium constant is called base ionisation constant, Kb
[𝑁𝐻4 + ][𝑂𝐻− ] 𝐾𝑏 = [𝑁𝐻3 ] • Can calculate pOH and hence pH, given Kb
• • • • •
Substance Acetic acid (vinegar) Citric acid Phosphoric acid Boric acid Aluminium salts
Formula CH3COOH H3C6H5O7 H3PO4 B(OH)3 H3BO3 NaAl(SO4)2•12H2O
Hydrochloric acid
HCl
Substance Sodium hydroxide
Formula NaOH
Ammonia Sodium carbonate Sodium hydrogen carbonate
NH3 Na2CO3 NaHCO3
Trisodium Phosphate
Na3PO4
Use Flavouring, preservative Flavouring Rust remover Mild antiseptic; insecticide In baking power, with sodium hydrogen carbonate • Brick and ceramic tile cleaner
• • • • •
Use • Oven cleaner, unblocking plumbing • Household cleaner • Water softener, grease remover • Fire extinguisher, rising agent in cake mixes (baking soda), mild antacid • Cleaner for surfaces before painting or wallpapering
Organic acids have a carboxyl group in them Many acids are oxy acids where the proton is attached to an oxygen atom Hydrohalic acids: HF, HCl, HBr and HI Organic bases have a nitrogen atom with a lone pair Hydroxides: LiOH, NaOH, KOH
Arrhenius 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 Arrhenius was the first to recognise the nature of acids and bases He postulated o Acids produce hydrogen ions in aqueous solutions – H+ o Bases produce hydroxide ions – OHo Acids and bases will neutralise each other when mixed they produce water and an ionic salt, neither of which are acidic or basic The more general definition is the Brønsted-Lowry definition They considered that acid-base reactions involved proton (H+) transfer between an acid and a bases, which they defined as follows o An acid is a proton donor o Base is a proton acceptor Acids that can donate only one proton are monoprotic, and acids that can donate more than one proton are polyprotic Gilbert Newton Lewis: o Acid = lone pair acceptor o Base = line pair donor
• In water: an acid ionises to produce H+(aq) – actually H3O+(aq) but we usually just write H+(aq)
Acid, Bases and Equilibrium
HA(aq) + H2O(l) H3O+(aq) + A-(aq) • A strong acid has equilibrium to the right (HA dissociated completely) • The majority of acids exist as weak acids, an acid which dissociates only partially • An acid-base dissociated will have a particular equilibrium constant which will determine the extent of the reaction (whether it lies to the left or right of the equation) • The equilibrium equation
[𝐻3 𝑂+ ][𝐴− ] 𝐾= [𝐻𝐴] • Ka is the acid dissociation constant
Conjugate Acid-Base Pairs • • • •
A conjugate base has one less proton that its conjugate acid NH4+ is the conjugate acid of NH3 NH3 is the conjugate base of NH4+ HSO4o Conjugate base is SO42o Conjugate acid is H2SO4 • H2SO4 is a dibasic or diprotic acid o H2SO4(aq) + H2O(l) H3O+(aq) + HSO4-(aq) o HSO4-(aq) + H2O(l) H3O+(aq) + SO42-(aq) Acid H3O+ H2SO4 HClO4 CH3COOH HPO42-
Conjugate base H2O HSO4ClO4CH3COOPO43-
Base OHH2O CNNH3 HPO42-
Conjugate acid H2O H3O+ HCN NH4+ H2PO4-
Acid-Base Reactions • • • •
HF(aq) + H2O(l) F-(aq) + H3O+(aq) HCOOH(aq) + CN-(aq) HCOO-(aq) + HCN(aq) H2PO4-(aq) + OH(aq) HPO42-(aq) + H2O(l) NH4+(aq) + CO32- NH3(aq) +HCO3-(aq)
Autoionisation of Water • Water can react with itself to generate hydronium and hydroxide ions according to the equation
H2O(l) H+(aq) + OH-(aq) • This reaction, in which a proton is transferred between two identical molecules, is called autoionisation • Equilibrium constant given special symbol Kw=[H+][OH-] • At 25°C, Kw = 1.0 × 10-14 • It is a very small value, which shows that the autoionisation of water proceeds only to a very small extent and that the equilibrium concentrations of [H3O+] and [OH-] in pure water are small • Neutral solution: [H+]=[OH-] = 1.0 × 10-7 mol L-1 • Acidic solution [H+] > 1.0 × 10-7 M
• Basic solution [H+] < 1.0 × 10-7 M
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+] • Example o If [H+] = 1.0 × 10-6 M then pH = -log10[10-6] = -(-6) = 6.0 o If [H+] = 1.6 × 10-5 M then pH = -log10[10-4.8] = -(-4.8) = 4.8 • pH = -log10[H+] • pOH = -log10[OH-] • pKw = -log10[Kw] = 14 at 25°C • Acid: pH < 7 • Neutral: pH = 7 • Basic pH > 7 • Since Kw = [H+][OH-] o log10Kw = log10[H+] + log10[OH-] o -log10[H+] – log[OH-] = -log10Kw • pH + pOH = 14 pOH = 14 – pH
Temperature Dependence of pH • The formation of hydrogen ions and hydroxide ions from water is an endothermic process
H2O(l) H+(aq) + OH-(aq) • • • • • • • • •
The forward reaction absorbs heat If you increase the temperature of the water, the equilibrium moves to lower the temperature again It will do that by absorbing the extra heat That means that the forward reaction will be favoured, and more hydrogen ions and hydroxide ions will be formed the effect of that is to increase the value of Kw as temperature increases Kw = 1.0 × 10-14 only at 25°C For T > 25°C, Kw > 10-14 For T < 25°C, Kw < 10-14 pH + pOH ≠ 14 if T≠ 25°C Neutral pH ≠ 7 if T ≠ 25°C
Weak Acids • Most acids or bases are weak – they do not completely ionise in water
HA(aq) H+(aq) + A-(aq) • Acid dissociation constant
𝐾𝑎 =
[𝐻+ ][𝐴− ] [𝐻𝐴]
Relationship between the Ka and pKa • The more positive the pKa the weaker the acid (and the stronger the conjugate base) • The larger the value of the Ka the stronger the acid and the lower the value of pK a
pKa = -log10Ka
Example • Find the pH of 0.1 M acetic acid (CH3COOH (HAc)) • Data pKa = 4.7 Ka= 10-4.7
Before Equilibrium After Equilibrium Equilibrium
HA 0.1 0.1 - x 0.1 - x
H2O -
A0 +x x
H3O+ 0 +x x
F0 +x x
H3O+ 0 +x x
𝑥2
Ka = 10-4.7 = 0.1−𝑥 • • • • • • •
Since the equilibrium constant is very small we assume x < 0.1 i.e. (0.1 – x) = 0.1 10-4.7 ≈ x2/0.1 x2 ≈ 0.1 × 10-4.7 = 10-5.7 x = √10−5.7 = 0.00141253754 pH = -log10[H+] = -log10[0.00141253754] = 2.9 Check that x=0.00141253754 < 0.1, (0.1 – x) = 0.0986 M
Example • Find % ionisation of 0.50 M HF (pKa = 3.1)
Before Equilibrium After Equilibrium Equilibrium
HF 0.50 0.50 - x 0.50 - x
H2O 𝑥2
Ka = 10-3.1 = 0.50−𝑥 • • • •
If x < 0.50, 10-3.1 ≈ x2/0.50 x2 = 0.50 × 10-3.1 x = 2 × 10-2 M % ionisation = x/0.50 ×100 = 4% (indeed a weak acid)
Weak Bases
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) • Equilibrium constant is called base ionisation constant, Kb
[𝑁𝐻4 + ][𝑂𝐻− ] 𝐾𝑏 = [𝑁𝐻3 ] • Can calculate pOH and hence pH, given Kb
