1. Monoprotic Acid- Base Equilibria K w = [ H + ] [ HO - ] = 1.0 x 10 -14 -log K w = pH + pOH = 14.00 at 25 o C So what is the pH of 1.0 x 10 -8 M KOH? [H + ] = K w / [ HO - ] =(1.0 x 10 -14 )(1.0 x 10 -8 ) =1.0 x 10 -6 pH= 6.00 Incorrect !! KOH is a base so pH >7. What ’ s wrong ? Water is an acid. It ionizes to form H + and HO - In pure water: [HO - ] = 1.0 x 10 -7 M Hence the reactions in the solution are: KOH K + + HO - H 2 O H + + HO -

2. The chemical species in the solution are K +, HO - and H + The charge balance is : [ K + ] + [ H + ] = [ HO - ](1) The mass balance is : [ K + ] = 1.0 x 10 -8 M (2) The equilibrium constant expression : K w = [ H + ] [ HO - ](3) From (1) : [ HO - ]= [ K + ] + [ H + ] = 1.0 x 10 -8 + [ H + ] (4) Putting (4) into (3) : [ H + ](1.0 x 10 -8 + [ H + ]) = 1.0 x 10 -14 [ H + ] = 9.6 x 10 -8 M or – 1.1 x 10 -7 M

3. Rejecting the negative value, we have : When concentration is low (  10 -8 M), the pH is 7.00. There is insufficient acid or base to significantly affect the pH of water itself At intermediate concentration (~ 10 -6 to 10 -8 M), the effects of water ionization and the added acid or base is comparable [H + ] = 9.6 x 10 -8 M pH = -log [H + ] = 7.02 When concentration is high (  10 -6 M), the pH has the value we would calculate by just considering the added H + or HO -

4. Weak Acids and Bases A weak acid is one that not completely dissociated Examples of weak acids : carboxylic acids, alcohols HA + H 2 O H 3 O + + A - Acid-dissociation constant, K a = Similarly, a weak base is one whose dissociation does not go to completion Examples of weak base: primary and secondary amines B + H 2 O BH + + HO - Base-dissociation constant, K b = [H + ][A - ] [HA] [BH + ][HO - ] [B]

5. Salts of Weak Acids and Bases Recall : In the Brønsted and Lowry classification: the products of a reaction between an acid and a base are also classified as acids and bases. Example : acetic acid methylamineacetate ion methyl- ammonium ion acid conjugat e base base conjugat e acid Hence the anion of a weak acid, eg - OAc, is a Brønsted base, which will accept protons : - OAc + H 2 O HOAc + - OH The stronger the conjugate base the more strongly the salt will combine with a proton, as from water, to shift the ionization to the right.

6. Equilibrium constant: K H = K b = [HOAc][HO - ] [AcO - ] hydrolysis constant Multiply numerator and denominator by [H + ] K b = = = K w = K a K b [HOAc][HO - ] [AcO - ] [H+][H+] [H+][H+] [HOAc] K w [AcO - ][H + ] KwKw KaKa Similar equation can be derived for the cations of salts of weak bases, ie : BH + + H 2 O B + H 3 O + K a = = [B][H 3 O + ] [BH + ] KwKw KbKb

7. Example : Consider a weak acid, HA with K a =10 -4 The conjugate base, A - has K b = K w /K a = 10 -10 the conjugate base of a weak acid is a weak base If K a was 10 -5, then K b = K w /K a = 10 -9 as HA becomes a weaker acid, A - becomes a stronger base (but never a strong base) Weak Is Conjugate to Weak

8. Weak Acid Equilibria How does one find the pH of a solution of a 0.0500 M weak acid, HA ? (Given: K a = 1.07 x 10 -3 ) HA + H 2 O H 3 O + + A - - A + H 2 O HA + - OH 2H 2 O H 3 O + + - OH Charge balance: [H 3 O + ] = [A - ] + [ - OH] (1) Mass balance: C HA = [A - ] + [HA] (2) = 0.0500 M Equilibria: K a = (3) [H 3 O + ][A - ] [HA]

9. In weak acid, the concentration of H 3 O + due to acid dissociation >>the H 3 O + from water dissociation. Thus, [A - ] >> [HO - ] Thus (1) becomes: [H 3 O + ]  [A - ] From (3) : 1.07 x 10 -3 = [H 3 O + ] = 6.80 x 10 -3 M = [A - ] pH = -log (6.80 x 10 -3 )= 2.17 [H 3 O + ][H 3 O + ] 0.0500 - [H 3 O + ] K w = [H 3 O + ][ - OH] (4) There are 4 chemical species ([A - ], [ - OH], [HA] and [H 3 O + ]) and 4 quations

10. [HA] = 0.0500 - [H 3 O + ] = 0.0432 M [ - OH] = K w /[H 3 O + ] = 1.47 x 10 -12 M [H 3 O + ] from H 2 O dissociation = 1.47 x 10 -12 M [H 3 O + ] from HA dissociation = 6.80 x 10 -3 M The assumption that [H 3 O + ] is derived mainly from HA is acceptable

11. Fraction of Dissociation The fraction of dissociation, , is defined as the fraction of the acid HA that forms A -  = = = For a solution of a 0.0500 M weak acid, HA, with K a = 1.07 x 10 -3, the fraction of dissociation is :  = = 0.136 = 13.6% [A - ] [A - ] + [HA] [A - ] [A - ] + (C HA - [A - ]) [A - ] C HA (6.80 x 10 -3 ) 0.0500

12. Weak-Base Equilibria The treatment of weak bases is almost the same as that of weak acids B + H 2 O BH + + HO - 2H 2 O H 3 O + + - OH K b = Charge balance : [BH + ] + [H 3 O + ] = [HO - ] Mass balance : C B = [BH + ] + [B] Assuming that nearly all of the HO - comes from the reaction of B + H 2 O [BH + ]  [HO - ] K b = If C B = 0.0372 M and K b =2.6 x 10 -6, then [BH + ][HO - ] [B] [BH + ] C B – [BH + ]

13. [BH + ] = 3.1 x 10 -4 M = [HO - ] from base hydrolysis From : C B = [BH + ] + [B] [B] = 0.0372 – (3.1 x 10 -4 ) = 0.03689 M From : [H 3 O + ] = K w /[HO - ] = (1.0 x 10 -14 )/(3.1 x 10 -4 ) = 3.2 x 10 -11 M = [HO - ] from water hydrolysis pH = -log (3.2 x 10 -11 ) = 10.49 Fraction of association  = = 0.0083 =0.83% [BH + ] [BH + ] + [B]

14. Buffers A buffer : -resists changes in pH when acids or bases are added or when dilution occurs -is a composed of a weak-acid and its salt (conjugate base) or a weak-base and its salt (conjugate acid) If A moles of a weak acid is mixed with B moles of its conjugate base, the moles of acid remain close to A and the moles of base remain close to B. Why ? Consider an acid, 0.10 M solution of HA, with pK a = 4.00 and its conjugate base, 0.10 mol A -, with pK b = 10.00 HA H + + A - pK a = 4.00 Let x be the concentration of acid that dissociates

15. [H + ][A - ] [HA] x2x2 C HA - x K a = = = = 1.0 x 10 -4 x =3.1 x 10 -3 Fraction of dissociation,  == 0.031 This means that only 3.1% of the acid has dissociated HA dissociates very little. Adding extra A - to the solution will make HA dissociate even less. What about the reaction of A - with water ? x2x2 0.10 - x x C HA A - + H 2 O HA + HO - pK b = 10.00 Let y be the concentration of A - that reacts with water

16. K b = = = 1.0 x 10 -10 y = 3.2 x 10 -6 Fraction of association  = = 3.2 x 10 -5 The degree of association is 3.2 x 10 -3 % A - does not react very much with water and the presence of HA makes A - react even less with water. Hence if 0.10 mol A - and 0.10mol HA are added to water, there will be close to 0.10 mol A - and close to 0.10mol HA in the solution in equilibrium. This approximation however breaks down for dilute solutions or at extremes of pH y CACA - [HA][HO - ] [A - ] y2y2 0.10 - y

17. Henderson-Hasselbalch Equation The Henderson-Hasselbalch equation is : -a central equation for buffers -basically a rearranged form of the K a equilibrium expression K a = log K a = log = log[H + ] + log -log [H + ] = -log K a + log pH = pK a + log [H + ][A - ] [HA] [H + ][A - ] [HA] [A - ] [HA] [A - ] [HA] [A - ] [HA] The Henderson-Hasselbalch equation tells us the pH of a solution provided we know the ratio of the concentrations of conjugate and base and the pK a for the acid

18. If a solution is prepared from a weak base B and its conjugate acid, the equation becomes pH = pK a + log where pK a = acid dissociation constant of the weak acid BH + [B] [BH + ] What does the equation tell us: pH = pK a + log HA H + + A - - when pH = pK a, [A - ] = [HA] - for every power-of-10 change in the ratio [A - ] /[HA], the pH changes by 1 unit - as the base concentration ([A - ] or [B]) increases, the pH goes up. Similarly, as the acid concentration increases, the pH comes down [A - ] [HA]

19. Example : What is the ratio of [H 2 CO 3 ]/[HCO 3 - ] in the blood buffered to a pH of 7.40 ? Given: K a = 4.4 x 10 -7 M H 2 CO 3 HCO 3 - + H + K a = = 4.4 x 10 -7 M = pH = -log[H + ] = 7.40 [H + ] = 10 -7.40 = = 9.0 x 10 -2 [H + ][HCO 3 - ] [H 2 CO 3 ] [HCO 3 - ]KaKa [H + ] [H 2 CO 3 ] [HCO 3 - ] 10 -7.40 4.4 x 10 -7

20. Preparing a Buffer What should you do if you would wish to prepare 1.00 l of buffer containing 0.100 M tris at pH 7.60 from solid tris hydrochloride and ~1M NaOH? Step 1: Weigh out 0.100 mol of tris hydrochloride and dissolve it in a beaker containing about 800 ml of water Step 2: Place a pH electrode in the solution and monitor he pH Step 3: Add NaOH solution until the pH is exactly 7.60 Step 4: Transfer the solution to a volumetric flask and wash the beaker a few times. Add the washings to the volumetric flask Step 5: Dilute to the mark and mix thoroughly Reason for adding about 800 ml of water is to make the volume reasonably close to the final volume during pH adjustment. Addition of large quantities of solvent will affect the ionic strength affect the pH

21. Addition of Strong Acid/Base to a Buffer HA H + + A - pH = pK a + log What happens when a small amount of strong acid/base is added to the buffered solution ? If a small amount of a strong acid is added : - it will combine with an equal amount of the A - to convert it to HA. - the change in the ratio [A - ]/[HA] is small and hence the change in pH is small If a small amount of a strong base is added : - it will combine with part of the HA to form an equivalent amount of A - - the change in the ratio [A - ]/[HA] is small and hence the change in pH is small [A - ] [HA]

22. Buffer Capacity The buffer capacity, , is a measure of how well a solution resists changes in pH when strong acid or base is added.  where C a and C b are the number of moles of strong acid and strong base per liter needed to produce a unit change in pH. The larger the value of  the more resistant the solution is to pH change  is determined by the concentrations of HA and A -. The higher the HA and A - concentrations, the more acid or base the solution can tolerate dC b dpH dC a dpH

23. A buffer is most effective in resisting changes in pH when pH = pK a maximum buffer capacity