Unit 4: Chemistry at Work Area of Study 1 – Industrial Chemistry Acid-Base Equilibrium VCE Chemistry Unit 4: Chemistry at Work Area of Study 1 – Industrial Chemistry

BRØNSTED-LOWRY THEORY ACIDS AND BASES BRØNSTED-LOWRY THEORY ACID proton donor HCl  H+ (aq) + Cl- (aq) BASE proton acceptor NH3 (aq) + H+ (aq)  NH4+ (aq) Conjugate systems Acids are related to bases ACID PROTON + CONJUGATE BASE Bases are related to acids BASE + PROTON CONJUGATE ACID For an acid to behave as an acid, it must have a base present to accept a proton... HA + B BH+ + A- acid base conjugate conjugate acid base

STRONG ACIDS AND BASES STRONG ACIDS completely dissociate (split up) into ions in aqueous solution e.g. HCl  H+ (aq) + Cl- (aq) MONOPROTIC 1 replaceable H HNO3  H+ (aq) + NO3- (aq) H2SO4  H+ (aq) + HSO4- (aq) HSO4-  H+ (aq) + SO42- (aq) DIPROTIC 2 replaceable H’s BASES completely dissociate into ions in aqueous solution e.g. NaOH  Na+(aq) + OH-(aq)

CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq) WEAK ACIDS Weak acids partially dissociate into ions in aqueous solution When a weak acid dissolves in water an equilibrium is set up e.g. ethanoic acid CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq) The water stabilises the ions The weaker the acid: the less it dissociates the more the equilibrium lies to the left. The relative strengths of acids can be expressed as Ka or pKa values For ethanoic acid: [CH3COO-] [H3O+] Ka = [CH3COOH]

WEAK BASES Partially react with water to give ions in aqueous solution When a weak base dissolves in water an equilibrium is set up e.g. ammonia NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) The weaker the base the less it dissociates the more the equilibrium lies to the left The relative strengths of bases can be expressed as Kb or pKb values. For ammonia: [NH4+] [OH-] Kb = [NH3]

HYDROGEN ION CONCENTRATION [H+] pH Hydrogen ion concentration can be converted to pH pH = - log10 [H+] To convert pH into hydrogen ion concentration [H+] = 10-pH pOH An equivalent calculation for bases converts the hydroxide ion concentration to pOH pOH = - log10 [OH-] In both the above, [ ] represents the concentration in mol L-1 STRONGLY ACIDIC 100 10-1 10-2 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-11 10-12 10-13 10-14 10-0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 pH OH¯ [H+] WEAKLY ACIDIC NEUTRAL STRONGLY ALKALINE WEAKLY ALKALINE

IONISATION CONSTANT OF WATER - Kw Despite being covalent, water conducts electricity to a very small extent. This is due to its self ionisation ... H2O (l) H+ (aq) + OH- (aq) Applying the equilibrium law Kc = [H+] [OH-] [ ] is the equilibrium concentration in mol L-1 [H2O] As the dissociation is small, the water concentration is large (~ 56 M) compared with the dissociated ions and any changes to its value are insignificant; its concentration can be regarded as constant. This “constant” is combined with (Kc) to get a new constant (Kw). Kw = [H+] [OH-] = 10-7 x 10-7 = 1 x 10-14 M2 (at 25°C) Because Kw is based on an equilibrium, it VARIES WITH TEMPERATURE

IONISATION CONSTANT OF WATER - Kw The value of Kw varies with temperature because it is based on an equilibrium. Temperature / °C 0 20 25 30 60 Kw / M2 0.11 0.68 1.0 1.47 5.6 H+ / M 0.33 0.82 1.0 1.27 2.37 pH 7.48 7.08 7 6.92 6.63 What does this tell you about the equation H2O (l) H+ (aq) + OH- (aq) ? • Kw gets larger as the temperature increases • This means the concentration of H+ and OH- ions gets greater • This means the equilibrium has moved to the right • If the concentration of H+ increases then the pH decreases • pH decreases as the temperature increases Because the equation moves to the right as the temperature goes up, it must be an ENDOTHERMIC process

RELATIONSHIP BETWEEN pH AND pOH Because H+ and OH- ions are produced in equal amounts when water dissociates [H+] = [OH-] = 1 x 10-7 M their concentrations will be the same. Kw = [H+] [OH-] = 1 x 10-14 M2 take logs of both sides log [H+] + log [OH-] = -14 multiply by minus - log [H+] - log [OH-] = 14 change to pH and pOH pH + pOH = 14 (at 25°C) N.B. As they are based on the position of equilibrium and that varies with temperature, the above values are only true if the temperature is 25°C Neutral solutions may be regarded as those where [H+] = [OH-]. Therefore a neutral solution is pH 7 only at a temperature of 25°C Kw is constant for any aqueous solution at the stated temperature

BUFFER SOLUTIONS – A BRIEF INTRODUCTION Definition “Solutions which resist changes in pH when small quantities of acid or alkali are added.” Acidic Buffer (pH < 7) made from a weak acid + its sodium salt ethanoic acid sodium ethanoate Alkaline Buffer (pH > 7) made from a weak base + its chloride ammonia ammonium chloride Uses Standardising pH meters Buffering biological systems (e.g. in blood) Maintaining the pH of shampoos

H2CO3 (aq)  HCO3- (aq) + H+ (aq) pH IN THE BODY A number of biochemical reactions involve acid-base chemistry. Without a means of regulating pH, the pH in the body can fluctuate to dangerous levels. For example the blood supply must be maintained within the narrow pH range of 7.35 to 7.45. Illnesses and diseases such as pneumonia, diabetes and emphysema can cause blood pH to fall leading to acidosis while hyperventilation can lead to an increase in blood pH (alkalosis). Natural buffers in the body help maintain the pH within this narrow range. They are made from a mixture of weak acids and their conjugate base. One important buffer is made up of carbonic acid (H2CO3) and the hydrogen carbonate ion (HCO3-): H2CO3 (aq)  HCO3- (aq) + H+ (aq) If H+ ions were added, a net back reaction occurs removing most of these ions. If OH- ions were added, they react with H+ ions. A net forward reaction results producing more H+ ions. The OH- ions have effectively reacted with carbonic acid

Unit 4: Chemistry at Work Area of Study 1 – Industrial Chemistry pH Calculations VCE Chemistry Unit 4: Chemistry at Work Area of Study 1 – Industrial Chemistry

Calculating pH - Strong Acids and Bases Strong acids and alkalis completely dissociate in aqueous solution It is easy to calculate the pH; you only need to know the concentration. Calculate the pH of 0.02 M HCl HCl completely dissociates in aqueous solution HCl  H+ + Cl- One H+ is produced for each HCl dissociating so [H+] = 0.02 M pH = - log [H+] pH = - log [0.02] = 1.7 Calculate the pH of 0.1 M NaOH NaOH completely dissociates in aqueous solution NaOH  Na+ + OH- One OH- is produced for each NaOH dissociating [OH-] = 0.1 M The ionic product of water (at 25°C) Kw = [H+][OH-] Therefore [H+] = Kw / [OH-] pH = - log [H+] = 13

Calculating pH - Weak Acids A weak acid is one which only partially dissociates in aqueous solution A weak acid, HA, dissociates as follows HA H+ + A- (1) Applying the Equilibrium Law Ka = [H+ ] [A-] (2) [HA] The ions are formed in equal amounts, so [H+] = [A- ] therefore Ka = [H+]2 (3) Rearranging (3) gives [H+]2 = [HA] Ka therefore [H+] = [HA] Ka pH = -log [H+] ASSUMPTION HA is a weak acid so it will not have dissociated very much. You can assume that its equilibrium concentration is approximately that of the original concentration.

Calculating pH - Weak Acids Calculate the pH of a weak acid HA of concentration 0.1 M ( Ka = 4 x 10-5 M ) HA dissociates as follows HA H+ + A- Dissociation constant for a weak acid Ka = [H+] [A- ] [HA] Substitute for A- as ions are formed in [H+] = [HA] Ka equal amounts and the rearrange equation ASSUMPTION HA is a weak acid so it will not have dissociated very much. You can assume that its equilibrium concentration is approximately that of the original concentration [H+] = 0.1 x 4 x 10-5 M = 4.00 x 10-6 M = 2.00 x 10-3 M ANSWER pH = - log [H+] = 2.70

pH of Mixtures Strong Acids and Strong Alkalis (Either in EXCESS) Calculate the pH of a mixture of 25 mL of 0.1 M NaOH is added to 20 mL of 0.1 M HCl Original moles of H+ = 0.1 x 20/1000 = 2 x 10-3 mol Original moles of OH- = 0.1 x 25/1000 = 2.5 x 10-3 mol Moles of excess OH- = 5 x 10-4 mol Final volume = 45 mL = 0.045 L [OH-] = 5 x 10-4 / 0.045 = 0.0111 M pOH = 1.95 5. pH = 14 – 1.95 = 12.05 Weak Acid and EXCESS Strong Alkali Calculate the pH of a mixture of 25 mL of 0.1 M NaOH and 22 mL of 0.1 M CH3COOH Original moles of H+ = 0.1 x 22/1000 = 2.2 x 10-3 mol Original moles of OH- = 0.1 x 25/1000 = 2.5 x 10-3 mol Moles of excess OH- = 3 x 10-4 mol Final volume = 47 mL = 0.047 L [OH-] = 3 x 10-4 / 0.047 = 6.38 x 10-3 M pOH = 2.20 5. pH = 14 – 2.20 = 11.80

Calculations Involving Acidity Constants Chloroacetic acid is a weak monoprotic acid with a Ka of 1.3 x 10-3 M. For a 1.0 M solution of chloroacetic acid, calculate: The pH The percentage hydrolysis CH2ClCOOH (aq)  CH2ClCOO- (aq) + H+ (aq) Ka = [CH2ClCOO-] [H+] / [CH3COOH] = 1.3 x 10-3 M From the equation, for every mole of chloroacetic acid that ionises one mole of [CH2ClCOO-] and one mole H+ will be formed. Thus [CH2ClCOO-] = [H+]. [H3O+]2 / [CH3COOH] = 1.3 x 10-3 M Because chloroacetic acid is a weak acid, it only ionises to a small extent. You can assume that its equilibrium concentration is approximately that of its original concentration. [CH3COOH] = 1.0 M.

Calculations Involving Acidity Constants Hence you get: [H3O+]2 / 1.0 M = 1.3 x 10-3 M [H3O+]2 = 1.3 x 10-3 [H3O+] = 0.036 M pH = -log [0.036] = 1.4 The percentage hydrolysis (percentage ionisation) measures the extent of the reaction. It is the fraction of the acid that is ionised (a ratio of the conjugate base and the acid). Percentage hydrolysis = [CH2ClCOO-] / [CH3COOH] x 100 Percentage hydrolysis = 0.036 / 1.0 x 100 Percentage hydrolysis = 3.6%