Chemistry 130 Acid and Base equilibria Dr. John F. C. Turner 409 Buehler Hall [email protected] Chemistry 130 Acids and bases The Brùnsted-Lowry definition of an acid states that any material that produces the hydronium ion is an acid. A Brùnsted-Lowry acid is a proton donor: + - HA H2 Ol H3 Oaq Aaq + + The hydronium ion, H 3 O a q or H a q has the same structure as ammonia. It is pyramidal and has one lone pair on O. Chemistry 130 Acids and bases The Brùnsted-Lowry definition of a base states that any material that accepts a proton is a base. A Brùnsted-Lowry base is a proton acceptor: - - + Baq H2 Ol HOaq BHaq In aqueous solution, a base forms the hydroxide ion: Chemistry 130 Conjugate acids and bases For any acid-base reaction, the original acid and base are complemented by the conjugate acid and conjugate base: - + NH3aq H2Ol HOaq NH4aq On the RHS, On the LHS, Water is the proton donor Hydroxide ion is the proton acceptor Ammonia is the proton acceptor Ammonium ion is the proton donor Water and hydroxide ion are conjugate acid and base Ammonia and ammonium ion are also conjugate acid and base Chemistry 130 Conjugate acids and bases The anion of every acid is the conjugate base of that acid The cation of every base is the conjugate acid of that base Acid-base conjugates exist due to the dynamic equilibrium that exists in solution. - + NH3aq H2Ol HOaq NH4aq - + NH3aq H2Ol HOaq NH4aq Chemistry 130 Acid and base constants We use equilibrium constants to define the position of equilibrium and to reflect the dynamic nature of the system. For an acid, we define + - HClaq H2 Ol H3 Oaq Claq + - [Products] [H3 Oaq][Claq] K A = = [Reactants] [HClaq] Water is not included as the change in concentration of water is negligible (water is ~55 M when pure) for a dilute solution. Chemistry 130 Acid and base constants We use equilibrium constants to define the position of equilibrium and to reflect the dynamic nature of the system. For a base, we define + - NaOHaq H2 Ol Naaq OHaq + - [Products] [Naaq][OHaq] K B = = [Reactants] [NaOHaq] Again, water is not included as the change in concentration of water is negligible (water is ~55 M when pure) for a dilute solution. Chemistry 130 Acid and base constants We also use the logarithm of the acid or base constant as an indicator of acid or base strength: + - + - [H3 Oaq][Claq] [Naaq][OHaq] K A = K B = [HClaq] [NaOHaq] pK A = −lg K A pK B = −lg K B Note that the logarithm used here is to base 10, not the natural log lg = log10 ln = loge Chemistry 130 Conjugate acid and base strengths The dynamic nature of the equilibrium between the conjugate acid-base pair means that a strong acid will have a weak conjugate base A strong base will have a weak conjugate acid. In both these cases, ©strong© and ©weak© are defined qualitatively the position of the acid or base equilibrium. A strong acid forms almost exclusively the hydronium ion and the concentration of the undissociated acid is negligible; the conjugate base must be weak. Chemistry 130 Conjugate acid and base strengths There are a variety of strong acids ± any material that Acid pK A has a larger pKA than the hydronium ion will form the hydronium ion in solution. HI −9 Any material with an acid constant smaller than the HBr −8 hydronium ion will establish a measurable equilibrium HCl −6 between the acid and the dissociated hydronium ion and H2 SO4 −3 associated conjugate base ± the anion. + H3 Oaq −1.7 Water therefore acts as a leveling solvent and restricts HNO3 −1.3 the degree of acidity possible in aqueous solution. Chemistry 130 Autoprotolysis of water Water can act as both an acid and a base ± it can form the hydronium ion as well as the hydroxide ion in solution. Pure water also undergoes a ©self-equilibrium©: + - H2 Ol H2 Ol H3 Oaq OHaq + - [Products] [H3 Oaq][OHaq] K = = W Reactants 2 [ ] [H2Ol] + - −14 K W = [H3Oaq][OHaq] = 1×10 pK W = 14 This is autoprotolysis or autionization. Chemistry 130 pKW , pH and pOH The autoprotolysis of water relates the hydronium ion concentration and the hydroxide ion concentration to the autoprotolysis constant of water: + - H2 Ol H2 Ol H3 Oaq OHaq + - −14 K W = [H3Oaq][OHaq] = 1×10 pK lg H O+ OH- lg H O+ lg OH- 14 w = − 10 [ 3 aq][ aq] = − 10 [ 3 aq]− 10 [ aq] = pK lg H O+ lg OH- p H pOH 14 w = − 10 [ 3 aq]− 10 [ aq] = = pK w = pH pOH = 14 This relationship dictates the concentrations of hydroxide and hydronium ion in all aqueous solutions Chemistry 130 Simple pH and pOH calculations Q: What is the pH of a solution of 0.0002 M solution of HI? Molarity of solution = 0.0002 mol L−1 The ionization equation of HI in water is + - HIaq H2 Ol H3 Oaq Iaq The K A of HI is very large and we know that HIaq is a strong acid + −1 Therefore, [H3 Oaq] = 0.0002 mol L lg H O+ 3.699 10 [ 3 aq ] = − pH lg H O+ 3.699 = − 10 [ 3 aq] = Chemistry 130 Simple pH and pOH calculations Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.0002 M solution of HI? We have already calculated the p H of the solution p H lg H O+ 3.699 = − 10 [ 3 aq] = The relationship between p H, pOH and pK W is pK w = p H pOH = 14 3.699 pOH = 14 pOH = 14 − 3.699 = 10.301 pOH lg OH- 10.301 = − 10 [ aq ] = - −10.301 −11 −1 [OHaq] = 10 = 5 × 10 mol L Chemistry 130 Simple pH and pOH calculations Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.00075 M solution of KOH? pOH calculation OH- 0.00075 mol L−1 lg OH- 3.125 [ aq] = [ aq] = − pOH lg OH- 3.125 = − 10 [ aq ] = As pK w = p H pOH = 14 p H 3.125 = 14 p H = 14 − 3.125 = 10.875 As p H lg H O+ = − 10 [ 3 aq] lg H O+ 10.875 10 [ 3 aq] = − + −10.875 −11 −1 [H3Oaq] = 10 = 1.33 × 10 mol L Chemistry 130 Weak acids and bases These calculations are straightforward as HI is a strong acid and KOH is a strong base. For weak acids and bases, which do not fully ionize in solution, we have to use the rules for equilibria with which we are already familiar. We also use pK A = −lg K A pK B = −lg K B which we have defined for the generic acid reaction and base reaction in water. Chemistry 130 Weak acid/base calculations Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.5 M solution of acetic acid? + - −5 CH3 CO2 Haq H2 Ol H3 Oaq CH3 CO2aq K A = 1.8 × 10 Acetic acid Acetate ion + - CH3 CO2 Haq H3 Oaq CH3 CO2aq Initial concentrations 0.5 ≈0 0 Change −x x x Equilibrium concentrations 0.5−x x x At equilibrium, the new concentrations are: [CH3 CO2 Haq] = 0.5−x - [CH3 CO2aq] = x + [H3 Oaq] = x Chemistry 130 Weak acid/base calculations −5 The acid constant for acetic acid is K A = 1.8 × 10 −14 whereas the autoprotolysis constant for water is K W = 1 × 10 The contribution of the self-ionization of water is of the order of 10-7 to the concentration of hydronium ion. In general, if the acid constant for the weak acid is sufficiently large in comparison to the autoprotolysis constant for water, we can ignore the autoprotolysis of water with respect to the hydrogen ion concentration Chemistry 130 Weak acid/base calculations Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.5 M solution of acetic acid? + - −5 CH3CO2 Haq H2 Ol H3Oaq CH3 CO2aq K A = 1.8 × 10 Acetic acid Acetate ion We now set up the equilibrium constant for aqueous acetic acid: - + [CH3 CO2aq][H3 Oaq] K A = [CH3 CO2 Haq] and use the concentrations that we have already calculated: [CH3 CO2 Haq] = 0.5−x - + - [CH3 CO2aq][H3 Oaq] x x [CH3 CO2aq] = x K A = = + [CH3 CO2 Haq] 0.5−x [H3 Oaq] = x Chemistry 130 Weak acid/base calculations Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.5 M solution of acetic acid? + - −5 CH3 CO2 Haq H2 Ol H3 Oaq CH3 CO2aq K A = 1.8 × 10 Acetic acid Acetate ion We now solve the equilibrium equation that we have set up: x x −5 K = = 1.8 × 10 A 0.5−x x2 = 1.8 × 10−5 0.5−x There are two ways of solving this equation 1. use the quadratic formula 2. make the assumption that x≪0.5 Chemistry 130 Weak acid/base calculations Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.5 M solution of acetic acid? + - −5 CH3 CO2 Haq H2 Ol H3 Oaq CH3 CO2aq K A = 1.8 × 10 Acetic acid Acetate ion Using assumption 2, that x≪0.5 x2 x2 K = ≈ = 1.8 × 10−5 A 0.5−x 0.5 x2 ≈ 0.5 × 1.8 × 10−5 ≈ 9 × 10−6 1 2 x ≈ 9 × 10−6 ≈ 0.003 + −1 + [H3 Oaq ] = 0.003 mol L pH = −lg [H3 Oaq] = 2.52 - −12 −1 pOH = 14−pH = 11.48 [OHaq] = 3.33 × 10 mol L Chemistry 130 Weak acid/base calculations Q: Calculate pOH, pH and the concentrations of both hydroxide ion and hydronium in a solution of 1.5 M solution of hydroxylamine? - + −9 NH2 OHaq H2Ol OHaq NH3 OHaq KB = 9.1 × 10 Hydroxylamine Hydroxylammonium - + NH2OHaq OHaq NH3 OHaq Initial concentrations 1.5 ≈0 0 Change −x x x Equilibrium concentrations 1.5−x x x At equilibrium, the new concentrations are: [NH2 OHaq] = 1.5−x + [NH3 OHaq] = x - [OHaq] = x Chemistry 130 Weak acid/base calculations Q: Calculate pOH, pH and the concentrations of both hydroxide ion and hydronium in a solution of 1.5 M solution of hydroxylamine? - + −9 NH2 OHaq H2Ol OHaq NH3OHaq K B = 9.1 × 10 Hydroxylamine Hydroxylammonium We now set up the base equilibrium constant for aqueous hydroxylamine: + - [NH3 OHaq ][OHaq] K A = [NH2OHaq] and use the concentrations that we have already calculated: [NH OH ] = 1.5−x 2 aq + - 2 + [NH3 OHaq ][OHaq] x [NH3OHaq] = x K B = = - [NH2OHaq] 1.5−x [OHaq] = x Chemistry 130 Weak acid/base calculations Q: Calculate pOH, pH and the concentrations of both hydroxide ion and hydronium in