Section 1 : Physical Chemistry
Acids and Bases
Brønsted-Lowry Theory of Acids and Bases
The Brønsted-Lowry de nition of acids + bases depends on protons.
• Brønsted-Lowry acid: proton donor
• Brønsted-Lowry base: proton acceptor
Acid-base equilibria involves the transfer of protons.
Strong Acids and Strong Bases
Strong acids + strong bases are completely dissociated in aqueous solution.
• Strong acids inc. HCl, H2SO4 + HNO3. Acidic hydrogen atoms are the H atoms which will produce H+(aq)
when the acid dissolves in water.
- HCl is described as a strong monoprotic acid as 1 mole of HCl produces 1 mole of H+(aq).
• HCl(aq) → H+(aq) + Cl-(aq)
- H2SO4 is described as a strong diprotic acid as 1 mole of H2SO4 produces 2 moles of H+(aq).
• H2SO4(aq) → 2H+(aq) + SO42-(aq)
• Strong bases inc. KOH + CuOH. Strong bases produce OH-(aq) if the base dissolves in water.
- NaOH(aq) → Na+(aq) + OH-(aq)
Weak Acids and Weak Bases
Weak acids + weak bases are slightly dissociated in aqueous solution.
• Weak acids inc. ethanoic acid (CH3COOH). Most carboxylic acids are weak acids. Acidic hydrogen
atoms are the H atoms which will produce H+(aq) when the acid dissolves in water.
- CH3COOH(aq) ⇌ CH3COO-(aq) + H+(aq)
- CH3COOH is described as a weak monoprotic acid as 1 mole of CH3COOH produces 1 mole of H+(aq)
if it dissociated completely.
• Weak bases inc. ammonia. Weak bases produce OH-(aq) if the base dissolves in water.
- NH3(g) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
De nition and Determination of pH
The conc. of H+ in aqueous solution covers a very wide range ∴ a logarithmic scale, the pH scale, is
used as a measure of H+ conc.
pH = –log10[H+] = –log[H+]
[H+] = 10(-pH)
• Neutral solutions have a pH value of 7.00 at 25°C.
• Acidic solutions have a pH of less than 7.00 + alkaline solutions have a pH of greater than 7.00.
- An alkali is a soluble base + when bases dissociate in solution they are called alkalis.
• [H+] = conc. of protons in solution measured in mol dm-3.
- ∴ [H+] = proticity of acid x [acid]
• Monoprotic acids have a proticity of 1 + diprotic acids have a proticity of 2.
Calculating pH of Strong Acids
E.g. calculate the pH of 1 M sulfuric acid.
• For strong diprotic acids, proticity = 2 so [H+] = 2 x 1= 2M
• pH = -log[2] = -0.30 (2 d.p.)
Calculating the Concentration of an Acid from its pH
E.g. determine the conc. of sulfuric acid which has a pH of 1.00
• [H+] = 10(-pH) = 10-1.00 = 0.1 M
• For strong diprotic acids, proticity = 2 so [acid] = [H+] / 2 = 0. = 0.05 M
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, Section 1 : Physical Chemistry
The Ionic Product of Water, Kw
Water slightly dissociates according to the following equilibrium: H2O ⇌ H+ + OH-
• KW is the ionic product of water and KW = [H+][OH-].
• The units of KW are always mol2 dm-6.
• The value of KW varies w/ temp.
- At 25°C, KW = 1.00 x 10-14 mol2 dm-6.
- KW ↑ as temp. ↑, indicating that the dissociation of water is endothermic as the equilibrium is
moving from the LHS to the RHS (more H+) as the temp. ↑.
Calculating the pH of Pure Water
In pure water [H+] = [OH-] ∴ KW = [H+]2 ∴ [H+] = √KW
E.g. calculate the pH of water at 40°C when KW = 2.92 x 10-14 mol2 dm-6.
• [H+] = √(2.92 x 10-14) = 1.709 x 10-7 M
• pH = -log[1.709 x 10-7] = 6.77 (2 d.p.)
Calculating the pH of Strong Bases
When a soluble base dissolves in water, hydroxide ions are present in the solution.
• [OH-] can be converted to [H+] using KW = [H+][OH-] in order to calculate the pH of a strong base.
• [OH-] = no. of moles of OH- produced by 1 mole of the base in solution.
E.g. calculate the pH of 0.250M solution of NaOH [KW = 1.00 x 10-14 mol2 dm-6].
• 1 mole of NaOH contains 1 mole of OH- ions.
• [OH-] = 1 x 0.250 = 0.250M ∴ [H+] = (1.00 x 10-14)/0.250 = 4.00 x 10-14 M
• pH = -log[4.00 x 10-14] = 13.40 (2 d.p.)
Calculating the Concentration of a Base from its pH
E.g. calculate the conc. of KOH w/ a pH of 13.70 [KW = 1.00 x 10-14 mol2 dm-6].
• [H+] = 10(-13.70) = 1.995 x 10-14 M
• [OH-] = (1.00 x 10-14)/(1.995 x 10-14 ) = 0.5013
• KOH has 1 mole of OH- ions per mole of base, so [KOH] = 0.50M (2 d.p.)
Calculating a Value for KW
[H+] may be calculated from the pH, [OH-] may be calculated from the conc.
E.g. at 40°C, a 0.270M solution of NaOH has a pH of 12.98. Calculate the value for Kw at 40°C.
• [H+] = 10-12.98 = 1.047 x 10-13 M
• As NaOH contains 1 mole of OH- per mole of NaOH, [OH-] = [NaOH] = 0.270M
• Kw = (1.047 x 10-13) x 0.270 = 2.83 x 10-14 mol2 dm-6
pKW, pH and pOH
pKW may be used in place of KW to carry out calculations involving strong bases.
pKW = -log(KW)
pKW = pH + pOH
• pOH = -log[OH-]
Ka for weak acids
Ka is the dissociation constant for a weak acid. For a general acid dissociation HA ⇌ H+ + A- …
Ka = [A-][H+]/[HA]
• E.g. for CH3COOH ⇌ CH3COO- + H+, Ka = [CH3COO-][H+]/[CH3COOH]
• Ka always has units mol dm-3.
pKa = -log(Ka)
Ka = 10(-pKa)
Acids and Bases
Brønsted-Lowry Theory of Acids and Bases
The Brønsted-Lowry de nition of acids + bases depends on protons.
• Brønsted-Lowry acid: proton donor
• Brønsted-Lowry base: proton acceptor
Acid-base equilibria involves the transfer of protons.
Strong Acids and Strong Bases
Strong acids + strong bases are completely dissociated in aqueous solution.
• Strong acids inc. HCl, H2SO4 + HNO3. Acidic hydrogen atoms are the H atoms which will produce H+(aq)
when the acid dissolves in water.
- HCl is described as a strong monoprotic acid as 1 mole of HCl produces 1 mole of H+(aq).
• HCl(aq) → H+(aq) + Cl-(aq)
- H2SO4 is described as a strong diprotic acid as 1 mole of H2SO4 produces 2 moles of H+(aq).
• H2SO4(aq) → 2H+(aq) + SO42-(aq)
• Strong bases inc. KOH + CuOH. Strong bases produce OH-(aq) if the base dissolves in water.
- NaOH(aq) → Na+(aq) + OH-(aq)
Weak Acids and Weak Bases
Weak acids + weak bases are slightly dissociated in aqueous solution.
• Weak acids inc. ethanoic acid (CH3COOH). Most carboxylic acids are weak acids. Acidic hydrogen
atoms are the H atoms which will produce H+(aq) when the acid dissolves in water.
- CH3COOH(aq) ⇌ CH3COO-(aq) + H+(aq)
- CH3COOH is described as a weak monoprotic acid as 1 mole of CH3COOH produces 1 mole of H+(aq)
if it dissociated completely.
• Weak bases inc. ammonia. Weak bases produce OH-(aq) if the base dissolves in water.
- NH3(g) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
De nition and Determination of pH
The conc. of H+ in aqueous solution covers a very wide range ∴ a logarithmic scale, the pH scale, is
used as a measure of H+ conc.
pH = –log10[H+] = –log[H+]
[H+] = 10(-pH)
• Neutral solutions have a pH value of 7.00 at 25°C.
• Acidic solutions have a pH of less than 7.00 + alkaline solutions have a pH of greater than 7.00.
- An alkali is a soluble base + when bases dissociate in solution they are called alkalis.
• [H+] = conc. of protons in solution measured in mol dm-3.
- ∴ [H+] = proticity of acid x [acid]
• Monoprotic acids have a proticity of 1 + diprotic acids have a proticity of 2.
Calculating pH of Strong Acids
E.g. calculate the pH of 1 M sulfuric acid.
• For strong diprotic acids, proticity = 2 so [H+] = 2 x 1= 2M
• pH = -log[2] = -0.30 (2 d.p.)
Calculating the Concentration of an Acid from its pH
E.g. determine the conc. of sulfuric acid which has a pH of 1.00
• [H+] = 10(-pH) = 10-1.00 = 0.1 M
• For strong diprotic acids, proticity = 2 so [acid] = [H+] / 2 = 0. = 0.05 M
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, Section 1 : Physical Chemistry
The Ionic Product of Water, Kw
Water slightly dissociates according to the following equilibrium: H2O ⇌ H+ + OH-
• KW is the ionic product of water and KW = [H+][OH-].
• The units of KW are always mol2 dm-6.
• The value of KW varies w/ temp.
- At 25°C, KW = 1.00 x 10-14 mol2 dm-6.
- KW ↑ as temp. ↑, indicating that the dissociation of water is endothermic as the equilibrium is
moving from the LHS to the RHS (more H+) as the temp. ↑.
Calculating the pH of Pure Water
In pure water [H+] = [OH-] ∴ KW = [H+]2 ∴ [H+] = √KW
E.g. calculate the pH of water at 40°C when KW = 2.92 x 10-14 mol2 dm-6.
• [H+] = √(2.92 x 10-14) = 1.709 x 10-7 M
• pH = -log[1.709 x 10-7] = 6.77 (2 d.p.)
Calculating the pH of Strong Bases
When a soluble base dissolves in water, hydroxide ions are present in the solution.
• [OH-] can be converted to [H+] using KW = [H+][OH-] in order to calculate the pH of a strong base.
• [OH-] = no. of moles of OH- produced by 1 mole of the base in solution.
E.g. calculate the pH of 0.250M solution of NaOH [KW = 1.00 x 10-14 mol2 dm-6].
• 1 mole of NaOH contains 1 mole of OH- ions.
• [OH-] = 1 x 0.250 = 0.250M ∴ [H+] = (1.00 x 10-14)/0.250 = 4.00 x 10-14 M
• pH = -log[4.00 x 10-14] = 13.40 (2 d.p.)
Calculating the Concentration of a Base from its pH
E.g. calculate the conc. of KOH w/ a pH of 13.70 [KW = 1.00 x 10-14 mol2 dm-6].
• [H+] = 10(-13.70) = 1.995 x 10-14 M
• [OH-] = (1.00 x 10-14)/(1.995 x 10-14 ) = 0.5013
• KOH has 1 mole of OH- ions per mole of base, so [KOH] = 0.50M (2 d.p.)
Calculating a Value for KW
[H+] may be calculated from the pH, [OH-] may be calculated from the conc.
E.g. at 40°C, a 0.270M solution of NaOH has a pH of 12.98. Calculate the value for Kw at 40°C.
• [H+] = 10-12.98 = 1.047 x 10-13 M
• As NaOH contains 1 mole of OH- per mole of NaOH, [OH-] = [NaOH] = 0.270M
• Kw = (1.047 x 10-13) x 0.270 = 2.83 x 10-14 mol2 dm-6
pKW, pH and pOH
pKW may be used in place of KW to carry out calculations involving strong bases.
pKW = -log(KW)
pKW = pH + pOH
• pOH = -log[OH-]
Ka for weak acids
Ka is the dissociation constant for a weak acid. For a general acid dissociation HA ⇌ H+ + A- …
Ka = [A-][H+]/[HA]
• E.g. for CH3COOH ⇌ CH3COO- + H+, Ka = [CH3COO-][H+]/[CH3COOH]
• Ka always has units mol dm-3.
pKa = -log(Ka)
Ka = 10(-pKa)
