Chemistry 130. Acid and Base equilibria

Transcription

1 Acid and Base equilibria Dr. John F. C. Turner 409 Buehler Hall

2 Acids and bases The BrønstedLowry definition of an acid states that any material that produces the hydronium ion is an acid. A BrønstedLowry acid is a proton donor: HA H 3 A aq The hydronium ion, H 3 O aq or H aq has the same structure as ammonia. It is pyramidal and has one lone pair on O.

3 Acids and bases The BrønstedLowry definition of a base states that any material that accepts a proton is a base. A BrønstedLowry base is a proton acceptor: B aq H BH aq In aqueous solution, a base forms the hydroxide ion:

4 Conjugate acids and bases For any acidbase reaction, the original acid and base are complemented by the conjugate acid and conjugate base: NH 3 aq H NH 4 aq On the RHS, Water is the proton donor Ammonia is the proton acceptor On the LHS, Hydroxide ion 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

5 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 Acidbase conjugates exist due to the dynamic equilibrium that exists in solution. NH 3 aq H NH 4 aq NH 3 aq H NH 4 aq

6 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 HCl aq H 3 Cl aq K A = [Products [Reactants [Cl aq = [H O 3 aq [HCl aq Water is not included as the change in concentration of water is negligible (water is ~55 M when pure) for a dilute solution.

7 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 NaOH aq Na aq OH aq K B = [Products [Reactants [OH aq = [Na aq [NaOH aq Again, water is not included as the change in concentration of water is negligible (water is ~55 M when pure) for a dilute solution.

8 Acid and base constants We also use the logarithm of the acid or base constant as an indicator of acid or base strength: K A [Cl aq = [H O 3 aq [HCl aq K B [OH aq = [Na aq [NaOH aq 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 = log 10 ln = log e

9 Conjugate acid and base strengths The dynamic nature of the equilibrium between the conjugate acidbase 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.

10 Conjugate acid and base strengths There are a variety of strong acids any material that has a larger pk A than the hydronium ion will form the hydronium ion in solution. Any material with an acid constant smaller than the hydronium ion will establish a measurable equilibrium between the acid and the dissociated hydronium ion and associated conjugate base the anion. Water therefore acts as a leveling solvent and restricts the degree of acidity possible in aqueous solution. Acid pk A HI 9 HBr 8 HCl 6 H 2 SO 4 3 H HNO 3 1.3

11 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 'selfequilibrium': H 2 H 3 OH aq K W = [Products [Reactants K W = [H 3 pk W = 14 [OH aq [OH aq = [H O 3 aq [H 2 2 = This is autoprotolysis or autionization.

12 pk W, ph and poh The autoprotolysis of water relates the hydronium ion concentration and the hydroxide ion concentration to the autoprotolysis constant of water: H 2 H 3 OH aq K W pk w pk w = [H 3 O aq [OH aq = = lg 10 [H 3 O aq [OH aq = lg 10 [H 3 O aq lg 10 [OH aq = 14 = lg 10 [H 3 O aq lg 10 [OH aq = p H poh = 14 pk w = ph poh = 14 This relationship dictates the concentrations of hydroxide and hydronium ion in all aqueous solutions

13 Simple ph and poh calculations Q: What is the ph of a solution of M solution of HI? Molarity of solution = mol L 1 The ionization equation of HI in water is HI aq H 3 I aq The K A of HI is very large and we know that HI aq is a strong acid Therefore, [H 3 O aq = mol L 1 lg 10 [H 3 O aq = ph = lg 10 [H 3 O aq = 3.699

14 Simple ph and poh calculations Q: Calculate poh, the ph and the concentration of hydroxide ion in a solution of M solution of HI? We have already calculated the p H of the solution p H = lg 10 [H 3 O aq = The relationship between p H, poh and pk W is pk w = p H poh = poh = 14 poh = = poh = lg 10 [OH aq = [OH aq = = mol L 1

15 Simple ph and poh calculations Q: Calculate poh, the ph and the concentration of hydroxide ion in a solution of M solution of KOH? poh calculation [OH aq = mol L 1 lg [OH aq = poh = lg 10 [OH aq As pk w = p H poh = 14 p H = 14 p H = = As p H = lg 10 [H 3 lg 10 [H 3 O aq = [H 3 O aq = = mol L 1 = 3.125

16 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.

17 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? CH 3 CO 2 H aq H 3 Acetic acid CH 3 CO 2 aq Acetate ion K A = H 3 CH 3 CO 2 aq CH 3 CO 2 H aq Initial concentrations Change x x x Equilibrium concentrations 0.5 x x x At equilibrium, the new concentrations are: [CH 3 CO 2 H aq = 0.5 x [CH 3 CO 2 aq = x [H 3 O aq = x

18 Weak acid/base calculations The acid constant for acetic acid is K A = whereas the autoprotolysis constant for water is K W = The contribution of the selfionization 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

19 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? CH 3 CO 2 H aq H 3 Acetic acid CH 3 CO 2 aq Acetate ion K A = We now set up the equilibrium constant for aqueous acetic acid: K A = [CH CO 3 2 aq [H 3 O aq [CH 3 CO 2 H aq and use the concentrations that we have already calculated: [CH 3 CO 2 H aq = 0.5 x [CH 3 CO 2 aq = x K A = [CH CO 3 2 aq [H 3 O aq [CH [H 3 O 3 CO 2 H aq aq = x = x x 0.5 x

20 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? CH 3 CO 2 H aq H 3 Acetic acid CH 3 CO 2 aq Acetate ion K A = We now solve the equilibrium equation that we have set up: x x K A = = x x 2 = x There are two ways of solving this equation 1. use the quadratic formula 2. make the assumption that x 0.5

21 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? CH 3 CO 2 H aq H 3 Acetic acid CH 3 CO 2 aq Acetate ion K A = Using assumption 2, that x 0.5 x 2 K A = 0.5 x = x x2 x [H 3 O aq = mol L 1 ph = lg [H 3 O aq = 2.52 poh = 14 ph = [OH aq = mol L 1

22 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? NH 2 OH aq OH aq Hydroxylamine NH 3 OH aq Hydroxylammonium K B = OH aq NH 3 OH aq NH 2 OH aq Initial concentrations Change x x x Equilibrium concentrations 1.5 x x x At equilibrium, the new concentrations are: [NH 2 OH aq = 1.5 x [NH 3 OH aq = x [OH aq = x

23 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? NH 2 OH aq OH aq Hydroxylamine NH 3 OH aq Hydroxylammonium K B = We now set up the base equilibrium constant for aqueous hydroxylamine: K A = [NH OH [OH 3 aq aq [NH 2 OH aq and use the concentrations that we have already calculated: [NH 2 OH aq = 1.5 x [NH 3 OH aq = x [OH aq = x K B [OH aq = [NH OH 3 aq [NH 2 OH aq = x x

24 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? NH 2 OH aq OH aq Hydroxylamine NH 3 OH aq Hydroxylammonium K B = The equation that we need to solve is: K B = [NH OH 3 aq [OH aq x 2 = [NH 2 OH aq 1.5 x Using the quadratic method, x 2 = = x so x 2 = 1.5 x = x x x = 0

25 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? NH 2 OH aq OH aq Hydroxylamine NH 3 OH aq Hydroxylammonium K B = For a quadratic of the form ax 2 bx c = 0, x = b± b2 4ac 1 2 In this case, a = 1 b = c = and so x = ± x = ± = mol L 1 2 2a

26 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? NH 2 OH aq OH aq Hydroxylamine NH 3 OH aq Hydroxylammonium K B = The equation that we need to solve is: K B = [NH OH [OH 3 aq aq x 2 = [NH 2 OH aq 1.5 x Using the assumption that x 1.5 x 2 x2 1.5 x = so x 2 = = = x mol L 1

27 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? NH 2 OH aq OH aq Hydroxylamine NH 3 OH aq Hydroxylammonium K B = In this case there is no difference between the methods to the third place of decimal and so x = [OH aq = mol L 1 poh = lg 10 [OH aq = lg = 3.93 As poh p H = 14 p H = 14 p OH = = and so as ph = lg 10 [H 3 [H 3 O aq = = mol L 1

28 Weak acid/base calculations Given that ph and poh are related via the relationship K W = [H 3 O aq [OH aq = pk w = ph poh = 14 then in aqueous solution, only one of these quantities is required for all to be calculable. Similarly, if K A is known for an acid or K B is known for a base, then using the change on the establishment of equilibrium will give the hydronium ion concentration or the hydroxide concentration and so all is known. If we know the ph or poh, we can also calculate K A or K B for the system, using similar methods.

29 Weak acid/base calculations Q: Calculate pk B for dimethylamine given that a M solution has a ph of Me 2 NH aq OH aq Dimethylamine Me 2 NH 2 aq Dimethylammonium When [Me 2 NH aq = mol L 1, p H = p H p OH = 14 so poh = 14 so p OH = = 2.02 Given that poh = lg 10 [OH aq [OH aq = = mol L 1

30 Weak acid/base calculations Q: Calculate pk B for dimethylamine given that a M solution has a ph of Me 2 NH aq OH aq Dimethylamine Me 2 NH 2 aq Dimethylammonium From the p H calculation, [OH aq = mol L 1 OH aq Me 2 NH 2 aq Me 2 NH aq Initial concentrations Change Equilibrium concentrations

31 Weak acid/base calculations Q: Calculate pk B for dimethylamine given that a M solution has a ph of Me 2 NH aq OH aq Dimethylamine Me 2 NH 2 aq Dimethylammonium We now know all the concentrations requred to calculate K B [Me 2 NH aq = [OH aq = mol L 1 [Me 2 NH 2 aq = mol L 1 K B = [Me NH 2 2 aq [Me 2 NH aq [OH aq pk B = lg 10 K B = 3.23 = mol L 1 = =

32 Polyprotic acids A polyprotic acid is one that can ionize more than once. Common examples include Sulfuric acid Phosphoric acid Carbonic acid H 2 SO 4 H 3 PO 4 H 2 CO 3 Note that acids such as acetic acid are not polyprotic CH 3 CO 2 H aq H 3 CH 3 CO 2 aq CH 3 CO 2 aq O l H 3 2 CH 2 CO 2 aq

33 Polyprotic acids Each ionization of a polyprotic acid has an associated acid constant. For phosphoric acid H 3 PO 4 H 3 PO 4 aq H 3 PO 4 aq K A 1 = H 2 PO 4 aq H 3 2 HPO 4 aq K A 2 = HPO 4 aq O l H 3 3 PO 4 aq K A 3 = Note that each acid constant differs from the one before and, although the first ionization, in this case, is strong, the others are not. H 3 PO 4 aq K A 1 = Strong acid H 2 PO 4 aq 2 HPO 4 aq K A 2 = Weak acid K A 3 = Weak acid

34 Polyprotic acids If the first ionization of a polyprotic acid is described by a large acid constant, then equal concentrations of the hydronium ion and dihydrogenphosphate ons are produced. The equilibrium for the second ionization is H 2 PO 4 aq H 3 2 HPO 4 aq K A 2 = and the equilibrium constant is K A 2 K A 2 = [HPO 2 4 aq [H 3 O aq = [HPO 2 4 aq [H 2 PO 4 aq 2 = [HPO 4 aq = [H 2 PO 4 aq [H 2 PO 4 aq = and so the acidity of the second ionization is independent of the initial concentration of phosphoric acid

35 Acid and Base equilibria Dr. John F. C. Turner 409 Buehler Hall

36 Salts of strong and weak acids When a salt is dissolved, the equilibria for the conjugate acid and base are established. Dissolving the salt introduces the conjugate acid or the conjugate base into the solution and the normal equilibria occur Salts of strong acids and strong bases form neutral solutions Salts of strong acids and weak bases form acidic solutions Salts of weak acids and strong bases form basic solutions The ph of solutions of salts of weak acids and weak bases depend on the acid constant of the acid and the base constant of the base

37 Salts of strong and weak acids The reaction that occurs when an anion associated with a weak acid is dissolved in water changes the ph of the solution. This happens because the anion is the conjugate base of the associated acid and the acidbase equilibrium for that acid is established. For nitrous acid, the associated anion is nitrite, NO 2 NO 2 aq OH aq HNO 2 aq K B = [OH aq We can multiply K B by [H O 3 aq to give K [H 3 O B aq K B = [OH [HNO aq 2 aq NO [H O 3 aq 2 aq [H 3 O aq For nitrous acid, K A,HNO2 = = [OH aq [HNO 2 aq NO 2 aq [HNO 2 aq [H O 3 aq [H 3 NO 2 aq

38 Salts of strong and weak acids For nitrous acid, the associated anion is nitrite, NO 2 aq OH aq HNO 2 aq K B = [OH aq [HNO 2 aq NO 2 aq We can multiply K B by [H O 3 aq to give K [H 3 O B aq K B = [OH [HNO aq 2 aq NO [H O 3 aq = K W 2 aq [H 3 O aq K A,HNO2 For nitrous acid, K A,HNO2 = and so K B = = = [OH aq [HNO 2 aq [H O 3 aq [H 3 NO 2 aq

39 Salts of strong and weak acids For nitrous acid, the associated anion is nitrite, NO 2 aq OH aq HNO 2 aq K B = [OH aq [HNO 2 aq NO 2 aq We can multiply K B by [H O 3 aq to give K [H 3 O B aq K B = [OH [HNO aq 2 aq NO [H O 3 aq = K W 2 aq [H 3 O aq K A,HNO2 For nitrous acid, K A,HNO2 = and so K B = = = [OH aq [HNO 2 aq [H O 3 aq [H 3 NO 2 aq

40 Strength of conjugate acids and bases The acid and base strength of a conjugate acidbase pair, such as nitritenitrous acid NO 2 HNO 2 acetic acidacetate CH 3 CO 2 H CH 3 CO 2 ammoniaammonium NH 3 NH 4 are related by the relationship K A K B = K W pk A pk B = pk W = 14

41 Strength of conjugate acids and bases This situation occurs because the conjugate base of a weak acid and the conjugate acid of a weak base are both appreciably strong. The acid or base strength of a conjugate acid or base of a strong base or acid is extremely weak and is negligible in most applications. The appreciable strength of a conjugate acid or base and the presence of an equilibrium, because the base or acid is weak means that additions of acid or base to a solution that contains the acidbase conjugate pair will not effect the ph of the solution greatly. These solutions are termed 'buffers'.

42 Buffer solutions A buffer solution is one that contains a conjugate acidbase pair and is used to provide a relatively constant ph in chemical reactions, biological and medical systems and in industrial settings. Because of the presence of the conjugate acidbase, we can write HA aq A aq Weak acid Weak acid H 3 K A = [A aq [H 3 O aq [HA aq [HA aq and so K A [ A aq = [H 3 O aq from which we can calculate the ph if we know the concentrations of the acid and conjugate base and the acid constant for the acid.

43 Buffer solutions Given that HA aq A aq Weak acid Weak acid H 3 [HA [H 3 O aq aq = K A [ A aq p H = lg 10 [H 3 O aq = lg 10 K A p H = lg 10 K A lg 10 [A aq [HA aq [HA aq [ A aq = lg 10 K A lg 10 [HA aq [ A aq

44 Buffer solutions For a buffer, the ph is given by ph = p K A lg 10 [ A aq [HA aq which is the HendersonHasselbalch equation.

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