IV. ACID-BASE EQUILIBRIA (Chapters 8 10)

IV. ACID-BASE EQUILIBRIA (Chapters 8 10) MONOPROTIC ACID-BASE EQUILIBRIA (Chapter 8) Strong Acids and Bases HCl, HBr, HI, HClO 4, HNO 3, and H 2 SO 4 (1st H + only) Completely ionized in dilute solutions: HX + H 2 O H 3 O + + X - (pk a s for these equilibria are negative) These strong acids are leveled in H 2 O: 0.1M HCl, HBr, HI solutions all have a ph of 1 Likewise, strong bases such as LiOH, NaOH, KOH, alkoxides (e.g. CH 3 O - ), and quaternary ammonium hydroxides (e.g. (CH 3 ) 4 N-OH) are also completely ionized in dilute aqueous solution. Strong Acids: ph -log [H + ] (in reality, ph -log A(H + )) Strong Bases: ph -log [H + ], where [H + ] is found from: [H + ] or: ph 14 - poh K w [OH - ] 1 CAUTION: What is the ph of 10-8 M KOH? ph 14.00 (-log 10-8 ) 6.00 Systematic treatment of equilibrium:?? How can the strong base KOH produce an acidic solution?? The contribution of OH - from the dissociation of water (1.0 x 10-7 ) has to be taken into account! 1. Reactions: H 2 O H + + OH - 2. Charge balance equation: [H + ] + [K + ] [OH - ] 3. Mass balance equation: [K + ] 1.0 x 10-8 M 4. Equilibrium constant expression: K w [H + ] [OH - ] 1.0 x 10-14 5. Count equations and unknowns: 3 of each 6. Solve! Let s set [H + ] x, then [OH - ] x + 1.0 x 10-8 and: x (x + 1.0 x 10-8 ) 1.0 x 10-14 or: x 2 + (1.0 x 10-8 ) x (1.0 x 10-14 ) 0 x 9.6 x 10-8 M ph 7.02 2

Concentration of strong acid or base: High (> 10-6 M): use regular formulas Low (< 10-8 M): ph 7.00 Intermediate (10-6 > [ ] > 10-8 ): systematic treatment necessary Note: water almost never produces 10-7 M H + and 10-7 M OH - e.g. in 10-4 M HBr, the ph 4, and [OH - ] 10-10 M But the only source of OH - is the dissociation of water, thus if the dissociation of water produces 10-10 M OH -, it also produces 10-10 M H + K w [H + ] [OH - ] in any aqueous solution 3 Weak Acids and Bases Weak acids are weak proton donors, only partly dissociated in aqueous solution, yielding an equilibrium mixture of weak acid, conjugate base and H + ; i.e. the following equilibrium reaction does not go to completion: HA H + + A - K a HA + H 2 O H 3 O + + A - [H + ] [A - ] [HA] (Acid dissociation ct.) HA and A - are a conjugate acid-base pair: they are related by the gain or loss of a proton Typical weak acids are carboxylic acids, conjugate acids of weak bases (e.g. NH + 4 NH 3 + H + ) which also include protonated anions from polyprotic acids (e.g. H 2 PO - 4 HPO 2-4 + H + ) Often, pk a is used instead of K a ; a typical pk a value for carboxylic acids is ~ 4.5 the higher the pk a, the smaller K a and the weaker the acid 4

Likewise, for weak bases the following equilibrium does not go to completion and a mixture of weak base, conjugate acid and OH - is obtained: [BH + ] [OH - ] B + H 2 O BH + + OH - K b [B] (Base hydrolysis ct.) Again, B and BH + are a conjugate acid-base pair Typical weak bases are amines (R-NH 2 + H 2 O R-NH 3+ + OH - ) and conjugate bases of weak acids (e.g. CH 3 COO - + H 2 O CH 3 COOH + OH - ) Relationship btw K a and K b for a conjugate acid-base pair: K a K b K w Note: most handbooks list pk a values for acids and bases, i.e. for bases the pk a of the conjugated acid is listed (see Appendix G in Harris 8 th Ed.) e.g.: pk a of NH 4+ is listed as 9.25; the pk b of NH 3 is found as 14.00 9.25 4.75 pk a of CH 3 COOH is 4.76; the pk b of CH 3 COO - is 14.00 4.76 9.24 5 The conjugate base of a weak acid is a weak base; the conjugate acid of a weak base, is a weak acid The stronger the acid (base), the weaker its conjugate base (acid) Strong acids (e.g. HCl) have such weak conjugate bases that they are not considered bases at all (Cl - is not a base) Weak-Acid Equilibria How to find the ph of a solution of a weak acid HA, given its formal concentration F and its pk a? K a K w HA H + + A - H 2 O H + + OH - Charge balance: [H + ] [A - ] + [OH - ] Mass balance: F [HA] + [A - ] [H + ] [A - ] Equilibria: K a K w [H + ] [OH - ] [HA] 4 equations 4 unknowns solvable in principle, but cubic equation 6

In very good approximation: [H + ] from HA >> [H + ] from H 2 O Then: [H + ] [A - ] Let [H + ] x [A - ]; [HA] F - x K a x 2 F - x This last expression also follows from: HA H + + A - Initial conc. F - - Final conc. F x x x Solve quadratic equation for x (reject neg. root) Check your assumptions!! EXAMPLE 1: What is the ph of a 0.0500 M solution of o-hydroxybenzoic acid (K a 1.07 x 10-3 )? 1.07 x 10-3 x 2 0.05000 - x The assumption here is that [H + ] [A - ] x 6.80 x 10-3 M ph 2.17 If ph is 2.17, then poh 14.00 2.17 11.83 and [OH - ] 10-11.83 1.5 x 10-12 M All the OH - comes from dissociation of H 2 O, and this produces an equal amount (1.5 x 10-12 M) of H + 7 The amount of H + produced by dissociation of H 2 O (1.5 x 10-12 M) is negligible compared to that produced by HA dissociation (6.80 x 10-3 M) Alternative (easier) procedure: instead of solving the quadratic equation, you can at first neglect the x in the denominator, which in some cases will be small compared to F (i.e. < 5% of F): [H + ] K a F Check that in the above example, however, this procedure is not allowed since x > 0.05 F EXAMPLE 2: What is the ph of a 0.10 M ammonia (pk a 9.25) solution? NH 3 + H 2 O NH 4+ + OH - K b [NH 4+ ] [OH - ] [NH 3 ] Initial conc. F - - Final conc. F x x x Find K b for NH 3 first: pk b 14.00 9.25 4.75 K b 1.78 x 10-5 1.78 x 10-5 x 2 0.10 - x x [OH - ] [NH 4+ ] 1.3 x 10-3 M poh 2.88 ph 11.12 Check assumption: [NH 4+ ] [OH - ] 8

Fraction of dissociation α (for an acid): [A - ] α [HA] + [A - ] For Example 1: α x (F x) + x 6.80 x 10-3 M 0.0500 M x F α 0.14 i.e. the acid is 14% dissociated at a formal concentration of 0.0500 M With increasing dilution, the fraction of dissociation increases (Why?) At equal formal concentrations, the stronger acid is always more dissociated than the weaker acid 9 Fraction of association α (for a base): For a base: α [BH + ] [B] + [BH + ] x F For Example 2: α 1.3 x 10-3 M 0.10 M 0.013 (or 1.3% associated) Buffers A buffer is a mixture of a weak acid (base) and a comparable amount of its conjugate base (acid) (within a factor of 10). A buffered solution resists changes in ph when acid or base is added, or when dilution occurs. The central equation in buffer calculations is the Henderson-Hasselbalch equation, which is merely a rearranged form of the K a expression. The HH equation always applies, but the correct equilibrium concentrations have to be used! 10

Mixing a weak acid and its conjugate base: derivation of the HH equation Weak acid HA (formal conc. F HA ) and its conjugated base NaA (F A -) HA H + + A - x [H + ] F HA x x x A - + H 2 O HA + OH - y [OH - ] F A - y y y [HA] F HA x+ y F HA [H + ]+ [OH - ] [A - ] F A - y + x F A - + [H + ] - [OH - ] A similar result is obtained from the mass balance and charge balance equations For a weak base and its conjugate acid, similar expressions are obtained: [BH + ] F HB + -[H + ] + [OH - ] [B] F B + [H + ] [OH - ] 11 The conjugate base of a weak acid will hydrolyze to a very small extent (e.g. if K a is 10-4, K b 10-10 ), so the solution will be acidic, i.e. [H + ] >> [OH - ]. We can safely ignore the contribution of OH - in previous equations: [HA] F HA [H + ] [A - ] F A - + [H + ] We replaced [H + ] by x So K a now is: [H + ] [A - ] K a [HA] x (F A -+ x) (F HA x) Solve quadratic equation to find x [H + ], and ph Further simplification: in most cases, [H + ] << F HA and F A -, except if the weak acid is a not-so-weak acid (pk a < ~3), or if the formal concentrations are small So most of the time we can write the equilibrium concentrations as: [HA] F If a moles of a weak acid are mixed with b moles of its conjugate HA [A - base, the moles of acid remain close to a, and the moles of base ] F A - close to b. Very little reaction occurs! We can use the usual expression for K a : K a [H + ] [A - ] [HA] 12

log K a log [H + ] + log [A- ] [HA] Henderson-Hasselbalch equation where [A - ] F A - and [HA] F HA If the buffer is prepared from a weak base B and its conjugate acid BH + : Note that when [A - ] [HA], ph pk a For every power of 10 change in the [A - ]/ [HA] ratio, the ph changes by 1 unit My advice: Use the above HH equations to solve a buffer related problem Calculate [H + ] from the ph value, and compare to F HA and F A - if [H + ] < ~1% of F HA and F A - then accept the answer if [H + ] > ~1% of F HA and F A -, then solve the quadratic equation on previous slide 13 An illustration of what happens when a weak acid is mixed with its conjugate base: Weak acid HA with pk a of 4.00 (conjugate base has pk b of 10.00) at a formal concentration F of 0.10 M HA H + + A - 0.10 x x x 10-4 α x 2 0.10 - x 0.0031 M 0.10 M When the conjugate base is dissolved in H 2 O (0.10 M): x 0.0031 M 0.031 (3.1% dissociation) A - + H 2 O HA + OH - 10-10 x 3.2 x 10 0.10 - x -6 M 0.10 x x x α 3.2 x 10-5 (0.0032% association) HA dissociates very little, and the addition of A - will shift the equilibrium to the left (even less HA will be dissociated). If 0.050 mol of HA and 0.036 mol of A - are mixed in water, there will be close to 0.050 mol HA and 0.036 mol of A - in solution! x 2 14

EXAMPLE: The conjugate acid of the base tris(hydroxymethyl)aminomethane ( tris ) has a pk a of 8.072. What is the ph of a solution prepared by dissolving (HO-CH 2 ) 3 -C-NH 2 (HO-CH 2 ) 3 -C-NH + 12.43 g tris (MM 121.135 g/mol) and 4.67 g tris 3 hydrochloride (MM 157.596 g/mol) in 1.00 L of water? [B] [BH + ] 12.43 g/l 121.135 g/mol 0.1026 M 4.67 g/l 157.596 g/mol 0.0296 M ph 8.072 + log 0.1026 M 0.0296 M ph 8.61 And [H + ] 10-8.61 << [B] and [BH + ] Notice that the volume of the solution is irrelevant, because volume cancels in numerator and denominator. Does this mean the ph of a buffer is independent of the dilution? Nearly so, but dilution affects the activity of solutions see further. If 12.0 ml of 1.00 M HCl is added to the above buffer, what will be the new ph? Strong acid reacts completely with the weak base to give BH + 15 12.0 ml x 1.00 M HCl 0.0120 mol HCl B + H + BH + Initial # moles: 0.1026 0.012 0.0296 Final # moles: 0.0906-0.0416 ph 8.072 + log 0.0906 M 0.0416 M ph 8.41 In general, a buffer resists ph changes because the strong acid or base added consumes B or BH + ; this does not change the log term in the HH equation too much, as long as you don t use up the B or BH +. The ph range pk a ± 1 ph unit is considered useful for preparation of a buffer outside this range, there is often not enough B or BH + to react with acid or base. Note: why does a weak base react essentially completely with a strong acid? 1 B + H + BH + K K a (for BH + ) For the above example of tris: pk a 8.072, or K a 8.47 x 10-9 or K 1.2 x 10 8 The same reasoning applies to the reaction of a weak acid and a strong base. Of course, for the reaction of a strong acid and a strong base, K is even larger: H + + OH - H 2 O K 1/K w 10 14 16

Example of mixing a not-so-weak weak acid with its conjugate base: Calculate the ph of a mixture in water of 0.0100 mol HA (pk a 2.0) and 0.0100 mol A - (1.00 L of solution) [A - ] 0.01 ph pk a + log 2.0 + log 2.0 [HA] 0.01 If ph 2.0, then [H + ] 0.01 M not negligible compared to formal conc. of 0.01!! We have to use the quadratic formula: HA H + + A - 0.0100 x x 0.0100 + x K a [H + ] [A - ] [HA] (x) (0.0100 + x) (0.0100 x) x 0.00414 M, or ph 2.38 [HA] F HA [H + ] 0.0100 0.00414 0.00586 M [A - ] F A- + [H + ] 0.0100 + 0.00414 0.0141 M 10-2 Plugging these correct values for [HA] and [A - ] into the HH equation gives us the same ph as calculated with the quadratic equation: 0.0141 ph 2.0 + log 2.38 0.00586 17 How to prepare a buffer: How to prepare 1.00 L of a 0.100 M tris buffer ph 7.60? Weigh 0.100 mol tris and dissolve in 800-900 ml H 2 O Put a calibrated ph electrode in the solution, and add NaOH solution (~ 1 M) until the ph is 7.60 (note: a ph meter measures activities, not conc.) Transfer to a 1.000 L volumetric flask, wash the beaker a couple of times and add the washings to the flask, dilute to volume and mix The reason why we not calculate how much tris and NaOH solution to mix is that we have not included activity coefficients. As an example: a 1/1 mixture of H 2 PO 4- /HPO 4 2- at 0.1 M ionic strength: [A - ] γ (A-) [A - ] Real HH ph pk a + log ph pk [HA] a + log equation γ (HA) [HA] 1 ph 7.20 + log ph 7.20 + log 0.355 + log 1 1 0.775 1 ph 7.20 ph 6.86 Act. Coeff. from (ph pk a ) Table 7.1 Be also aware that ph depends on the temperature this is buffer dependent (e.g. tris: ph 8.07 at 25 C ph 7.7 at 37 C) 18

Buffer Capacity: measure of how well a solution resists changes in ph when strong acid or base is added dc β b -dc a C a and C b are the # moles of strong acid or dph dph base needed to change the ph by 1 unit Solution containing 0.100 F HA (pk a 5.00) Ordinate (C b ) is formal concentration of strong base that has to be mixed with 0.100 F HA to give indicated ph. Derivative of upper curve β β is maximal when ph pk a When choosing a buffer, choose one with a pk a as close as possible to the desired ph! β can be increased by increasing the buffer concentration β increases at high ph because it gets increasingly difficult to increase the ph of an already very basic solution 19 POLYPROTIC ACID-BASE EQUILIBRIA (Chapter 9) Diprotic Acids and Bases Can accept or donate two protons: H 2 A, HA -, A 2- The intermediate form HA - is said to be amphiprotic: it can accept and donate a proton Amino acids: R O NH 2 OH R O NH 3 + O Zwitterion: molecule with both pos. and neg. charges We will discuss how to calculate the ph of solutions of these diprotic acids/bases using the amino acid leucine as an example: H 3 N + CH(R)COOH H 3 N + CH(R)COO - H 2 NCH(R)COO - pk a1 2.328 pk a2 9.744 H 2 L + HL L - R: H 2 C H CH C 3 CH 3 Diprotic acid: H 2 L + HL + H + K a1 4.70 x 10-3 HL L - + H + K a2 1.80 x 10-10 20

Diprotic base: L - + H 2 O HL + OH - K b1 HL + H 2 O H 2 L + + OH - K b2 K a1 K b2 K w K a2 K b1 K w (verify this) Fully protonated form H 2 L + H 2 L + is a weak acid, so will dissociate only partly; HL is an even much weaker acid that will hardly dissociate at all (and: Le Châtelier!) we can think of H 2 L + as a monoprotic acid! Calculation of the ph of a 0.0500 M H 2 L + solution (see also slide 7): H 2 L + HL + H + 4.7 x 10-3 x 2 0.0500 - x x 0.0132 M [H + ] [HL] ph 1.88 [H 2 L + ] 0.0500 0.0132 0.0368 M F [H 2 L + ] + [HL] + [L - ] We assume that the dissociation of HL is negligible: F [H 2 L + ] + [HL]; or [H 2 L + ] F - [HL] 21 Using K a2, and with [H + ] [HL] 0.0132 M, we can also calculate [L - ] (which we assumed to be negligible compared to [HL]): 0.0132 [L - ] 1.80 x 10-10 0.0132 [L - ] 1.80 x 10-10 M K a2 ~ 8 orders of magnitude < [HL] The approximation that the 2 nd dissociation is negligible is valid for most diprotic acids even if is just 10 times smaller than, [H + ] calculated by ignoring the 2 nd ionization would be in error by just a few percent. A solution of a fully protonated diprotic acid behaves like a solution of a monoprotic weak acid with K a K a1 The fully deprotonated (basic) form L - L - + H 2 O HL + OH - K b1 K w / K a2 5.55 x 10-5 HL + H 2 O H 2 L + + OH - K b2 K w / K a1 2.13 x 10-12 From K b1, we notice that L - will not hydrolyze to a large extent, and from K b2 we see that the little HL that is formed hardly reacts at all. We treat L - as a monobasic species, with K b K b1 : 22

L - + H 2 O HL + OH - 0.0500 x x x The concentration of H 2 L + can be found from the K b2 equilibrium: 5.55 x 10-5 x 2 0.0500 - x x 0.00164 M [HL] [OH - ] (ph 11.21) [L - ] 0.0484 M 0.00164 [H 2 L + ] 2.13 x 10-12 [H 2 L + ] 2.13 x 10-12 M K 0.00164 b2 Indeed insignificant compared to [HL] A solution of the fully deprotonated form of a diprotic acid behaves like a solution of a monoprotic weak base with K b K b1 The intermediate form HL: HL L - + H + K a K a2 1.80 x 10-10 HL + H 2 O H 2 L + + OH - K b K b2 2.13 x 10-12 From the magnitude of the K a and K b, we expect a solution of HL to be slightly acidic. However, both reactions will proceed to nearly equal extent, because the H + formed in the first one, reacts with OH - formed in the second one, thereby driving that reaction to the right. 23 Systematic treatment of equilibrium is required: Charge balance: [H + ] + [H 2 L + ] [L - ] + [OH - ] or: [H + ] + [H 2 L + ] - [L - ] - [OH - ] 0 Substituting from the acid dissociation equilibria (slides 20 and 21), and the expression for K w : [HL] [H + ] K [H + ] + - 2 [HL] K - w [H + ] [H + ] 0 [H + ] 2 + [HL] [H + ] 2 - [HL] - K w 0 Further rearrangement leads to: [H + ] [HL] + K w + [HL] (Eq. 9-10) What is [HL]? Since HL is both a weak acid and a weak base, neither the dissociation, nor the hydrolysis proceed very far, so we can substitute [HL] for the formal concentration F (here: 0.0500 M) 24

[H + ] F + K w + F (Eq. 9-11) (Note that and are the acid dissociation constants (slide 20)) With the appropriate substitutions, we find for 0.0500 M HL: [H + ] 8.80 x 10-7 M, or ph 6.06 Finally, we can also calculate [H 2 L + ] and [L - ] from the and equilibria: [H + ] [HL] (8.80 x 10-7 ) (0.0500) [H 2 L + ] 4.70 x 10-3 9.36 x 10-6 M [K (1.80 x 10-10 ) (0.0500) [L - ] 2 ] [HL] 1.02 x 10-5 M [H + ] 8.80 x 10-7 Our assumption was that [HL] 0.0500 M ( formal conc.); i.e. that a negligible fraction of HL reacted to L - and H 2 L +. They are both present at ~0.02% of the value of [HL], so this assumption was warranted. Also note that [H 2 L + ] [L - ]; i.e. both reactions proceed to ~ the same extent (slide 20). 25 What if our assumption does not turn out to be valid? This happens when and are nearly equal, and F is small. In that case, the method of successive approximations can be used. Here, the calculated values of [H 2 L + ] and [L - ] are used to calculate a more accurate value of [HL] via: [HL] F - [H 2 L + ] - [L - ]. This value of [HL] is then used to calculate a new ph (Eq. 10-10), which is then used to calculate new values of [H 2 L + ] and [L - ], which are used to calculate a new [HL], which is used to calculate a new ph. This is repeated until the ph value converges. Box 9-2 in Harris (8 th Ed.) gives an example of this procedure. 26

More approximations to calculate the ph for the intermediate form: Usually, F >> K w, then: [H + ] F + K w + F Then, if << F: [H + ] F + F F F log [H + ] ½ (log + log ) ph ½(p + p ) (Eq. 9-12) The ph of the intermediate form of a diprotic acid is close to midway between p and p (independent of the concentration) Always use Eq. 9-11 to to calculate the ph of the intermediate form of a diprotic acid, use Eq. 9-12 as a quick check (should be close) 27 Diprotic Buffers: We can write two HH-equations (both of which are always true!): ph p + log [HA- ] [H 2 A] ph p + log [A2- ] [HA - ] EXAMPLE 1: What is the ph of a solution prepared by dissolving 1.00 g KHP (MM 204.221 g/mol) and 1.20 g Na 2 P (MM 210.094 g/mol) in 50.0 ml of water? p 2.950 and p 5.408 We know [HP - ] and [P 2- ], so we use p : (1.20 g) / (210.094 g/mol) ph p + log 5.47 (1.00 g) / (204.221 g/mol) (No need to use the volume ) EXAMPLE 2: How many mls of 0.800 M KOH should be added to 3.38 g oxalic acid to give a ph of 4.40 when diluted to 500 ml? MM (oxalic acid) 90.035 g/mol; p 1.27, p 4.266 28

The ph we are aiming for is above p. When ph p, there is a 1:1 mole ratio of HOx - and Ox 2-, so we need more Ox 2- than HOx -. First, we have to convert all H 2 Ox to HOx - H 2 Ox + OH - HOx - + H 2 O We have (3.38 g) / (90.035 g/mol) 0.0375 4 mol H 2 Ox, and require this many moles of OH -, or (0.0375 4 mol) / (0.800 mol/l) 46.9 ml. Next, we need to convert some of the HOx - to Ox 2- : HOx - + OH - Ox 2- + H 2 O Initial moles: 0.0375 4 x - Final moles: 0.0375 4 x - x ph p + log [A2- ] [HA - ] x 4.40 4.266 + log 0.0375 4 x Or: x 0.0216 4 mol (not M) The volume of KOH solution that contains this many moles OH - (0.0216 4 mol) / (0.800 mol/l) 27.1 ml. Total volume of KOH needed 46.9 + 27.1 74.0 ml 29 Polyprotic Acids and Bases e.g. triprotic system: H 3 A H 2 A - + H + K a1 H 2 A - HA 2- + H + K a2 HA 2- A 3- +H + K a3 K 3 A 3- + H 2 O HA 2- + OH - K b1 K w / K a3 HA 2- + H 2 O H 2 A - + OH - K b2 K w / K a2 H 2 A - + H 2 O H 3 A + OH - K b3 K w / K a1 Strategy: H 3 A is treated as a monoprotic weak acid with K a H 2 A - is treated as the intermediate form of a diprotic acid (Eq. 9-11): [H + ] F + K w + F 30

HA 2- is also treated as the intermediate form of a weak acid, but it is bracketed by H 2 A - and A 3-, so we have to use and K 3 instead of and : K 3 F + K w [H + ] + F A 3- is treated as monobasic, with K b K b1 K w / K a3 EXAMPLE: Find the ph of 0.10 M H 3 His 2+, 0.10 M H 2 His +, 0.10 M HHis, and 0.10 M His - Most acidic proton ( ) HO O HN N + H NH 3 + Least acidic proton (K 3 ) Intermediate acidic proton ( ) 31 H 3 His 2+ fully acidic form p 1.6 H 2 His + p 5.97 HHis pk 3 9.28 amphiprotic form amphiprotic form His fully basic form 0.10 M H 3 His 2+ : treat as monoprotic weak acid H 3 His 2+ H 2 His + + H + F x x x 10-1.6 x 2 0.10 - x x 0.039 M; ph 1.41 0.10 M H 2 His + : use Eq. 10-11, with 10-1.6 and 10-5.97 : [H + ] (10-1.6 )(10-5.97 )(0.10) + (10-1.6 )(10-14 ) 10-1.6 + 0.10 [H + ] 1.47 x 10-4 M ph 3.83 (This is close to ph ½(p + p ) 3.78) 32

0.10 M HHis: use Eq. 10-11, with 10-5.97 and K 3 10-9.28 : We find: ph 7.62, which is the same as ½ (p + pk 3 ) 0.10 M His - : treat as monobasic: His - + H 2 O HHis + OH - K b1 K w / K 3 10-14 /10-9.28 1.91 x 10-5 F x x x 1.91 x 10-5 x 2 0.10 - x x 0.00137 M; poh 2.86 or ph 11.14 Principal Species in Solution: Monoprotic: e.g. benzoic acid: pk a 4.20 What is the principal form at ph 8.20: HA or A -? Using: ph pk a + log [A- ] [HA] 33 We know that at ph 4.20, [HA] [A - ]. For every increase of 1.00 ph unit in ph, the [A - ] / [HA] increases by a factor of 10. At ph 8.20, there is 10 4 times more [A - ] (benzoate anion) then [HA]. Polyprotic Species: ph p: [H 3 A] [H 2 A - ] p : [H 2 A - ] [HA 2- ] pk 3 : [HA 2- ] [A 3- ] The principal species is determined by comparing the ph of the solution with the pk a values. The closest pk a value indicates the dominant acid/conjugate base pair if ph < pk a then acid dominates, if ph > pk a then base dominates. Fractional Composition Equations Give the fraction of each species of acid (or base) at a given ph 34

For a monoprotic acid HA we can write: HA H + + A - K a [H + ] [A - ] [HA] Mass balance: F [HA] + [A - ] or: [A - ] F - [HA] K a [H + ] (F - [HA]) [HA] or: [HA] [H + ] F [H + ] + K a The fractions of acid in the forms HA (α HA ) and A - (α A- ) are: [HA] α HA [HA] + [A - ] [HA] F or: α HA [H + ] [H + ] + K a α A- [A - ] [HA] + [A - ] [A - ] F or: α A- K a [H + ] + K a α HA + α A- 1 35 Fractional composition of an acid with pk a 5.00 as a function of ph We can extend this to diprotic (polyprotic) systems: H 2 A HA - + H + HA - A 2- + H + [H + ] [HA - ] [H 2 A] [HA - ] [H 2 A] [H + ] 36

[H + ] [A 2- ] [HA - ] [A 2- ] [HA - ] [H + ] [H 2 A] [H + ] 2 Mass balance: F [H 2 A] + [HA - ] + [A 2- ] [H 2 A] + [H 2 A] + [H + ] [H 2 A] [H + ] 2 [H 2 A] 1 + + [H + ] [H + ] 2 α H2 A [H 2 A] F α HA- [HA - ] F [H + ] 2 [H + ] 2 + [H + ] + [H + ] [H + ] 2 + [H + ] + 37 α A 2- [A 2- ] F [H + ] 2 + [H + ] + This can be extended to polyprotic acids, and also to bases (for bases:, are the acid dissociation constants of the conjugated acid): 38

α Hn A [H + ] n D K α Hn-1 A - 1 [H + ] n-1 D α Hn-i A i- K i [H + ] n-i D K n α A n- D where D [H + ] n + [H + ] n-1 + [H + ] n-2...+...k n 39 Choosing an acid/base pair for a buffer of a desired ph: Select an acid whose pk a is as close as possible to the desired ph. This allows the buffer to be made up with nearly equal concentrations of acid and conjugate base. Such a buffer is nearly equally effective in buffering against added acid and added base. [conj. base] Useful ph range for a buffer pk a ± 1; i.e. 10 > > 0.1 acid CO 2 H CO 2 H CO 2 H acid/base pair pk a of acid useful ph range H 3 A/H 2 A 2.88 1.88 to 3.88 H 2 A /HA 2 4.75 3.75 to 5.75 HA 2 /A 3 7.13 6.13 to 8.13 Make sure that the concentration of the buffer is adequate for its intended purpose! 40

ACID BASE TITRATIONS (Chapter 10) Some general remarks on titrations (Chapter 1-5 and 1-6) Titration: reagent solution (titrant) incrementally added to a solution of analyte until complete reaction tells us amount of analyte that was present Usually volumetric (volume of titrant measured) but can also be gravimetric (mass of titrant is measured) Types of titrations: acid-base; oxidation-reduction; complex formation; precipitation Assumption: large K, rapid (quantitative reaction: > 99.9%), so before equivalence point (EP) only negligible amount of titrant remains unreacted EP Exact amount of titrant needed for stoichiometric reaction. Not necessarily exactly the same as the end point, which is determined by a sudden physical change (e.g. colour change). If EP is the 'true value', then the end point is an experimental estimate of the EP; any difference between EP and end point titration error e.g. acid-base indicator consumes some of the titrant to complete its colour change - do blank titration (same procedure without analyte) and subtract blank volume from volume of titrant for unknown 41 Titrant can be a solution of a primary standard: a compound that can be weighed and dissolved in a known volume of solution so that a solution is obtained of accurately known concentration. Requirements for primary standards: purity 99.9%, stable to drying temperatures, no decomposition under storage, prefer high MM to minimize relative error of weighing process (e.g. KHP) Relatively few substances meet these requirements. When no primary standard is available, the titrant is prepared with the approximate concentration, and then standardized with a primary standard e.g. NaOH can be standardized with KHP Graphing titration curves: A graph of analyte or titrant concentration versus volume of titrant added is impractical as the concentrations of both analyte and titrant change by several orders of magnitude near the equivalence point. 42

It is much more instructive to plot the p function of analyte or titrant concentration versus volume of titrant added: p function of X px - log [X] Br - titrated with Ag + mirror images pag + pbr - V e V Ag + V e V Ag + The negative sign in front of log X essentially produces an inverse relationship between px and [X] EP point of max. slope (dx/dy max.; d 2 x/dy 2 0) for a 1:1 stoichiometry only (in practice, titration curves are steep enough that the inflection point is a good estimate of the EP) 43 Reminder: how to calculate the V e Calculate the equivalence point for the titration of 20.0 ml of 0.100 M Na 3 PO 4 with 0.200 M Pb(C 2 H 3 O 2 ) 2 3 Pb 2+ + 2 PO 3-4 Pb 3 (PO 4 ) 2(s) initial moles PO 3-4 CV 0.100 mol L 1 x 0.0200 L 2.00 x 10-3 mol moles Pb 2+ needed 2.00 x 10-3 mol PO 3-4 (3 mol Pb 2+ / 2 mol PO 3-4 ) 3.00 x 10-3 mol -3 2+ 3.00 x 10 mol Pb V 0.0150L e -1 0.200 mol L Useful equations/calculations: Before EP, the fraction of analyte consumed V V e -V thus, the fraction of analyte remaining 1 V e V e V Dilution factor: i V volume of titrant added V i + V V i initial volume of analyte solution V e equivalence volume of titrant V V e 2 reasons why [analyte] decreases: consumption and dilution 44

Acid-base titrations: Allows us to determine concentrations and pk a values of acids and bases in a solution Titration curves: Plot of ph vs. volume of titrant added Strong base with strong acid e.g.: titration of 50.00 ml of 0.02000 M KOH with 0.1000 M HBr Reaction: OH - + H + H 2 O K 1/K w 10 14 Volume to equivalence point V e : (V e ml) (0.1000 M) (50.00 ml) (0.02000 M) V e 10.00 ml Before EP: ph determined by excess OH - in solution When for instance 3.00 ml HBr is added: V [OH - ] e -V V [OH - ] i 10.00 3.00 i (0.02000 M) V i + V 10.00 V e (see slide 44) 50.00 50.00 + 3.00 45 [OH - ] 0.0132 M ph pk w poh 14 1.88 12.12 At EP: ph determined by dissociation of water H 2 O OH - + H + x x Only for the titration of a strong acid (base) with a strong base K w x 2 x 10-7 ph 7.00 (acid) will the ph be 7.00 at the EP After EP: ph determined by excess H + When for instance a total of 10.50 ml HBr is added: [H + ] [H + ] i V V e 10.50 10.00 (0.1000 M) 8.26 x 10-4 M V i + V 50.00 + 10.50 ph 3.08 46

At very low (high) ph, solutions of strong acids (bases) resist ph changes upon addition of H + or OH -, but not upon dilution. Titration of a weak acid with a strong base: e.g. titration of 50.00 ml 0.02000 M MES (pk a 6.27) with 0.1000 M NaOH HA + OH - A - + H 2 O H O N + O S K 1/K b 1/(K w /K a ) 1/(10-14 /10-6.27 ) 5.4 x 10 7 MES O O 2-(N-morpholino)ethanesulfonic acid 47 (see also slide 16: strong + weak react completely) Volume to EP: (V e ml) (0.1000 M) (50.00 ml) (0.02000 M) V e 10.00 ml Before addition of base: Solution of 0.02000 M HA in water: weak acid problem: HA H + + A - 10-6.27 0.02000 - x 0.02000 x x x x 2 x 1.03 x 10-4 ph 3.99 Before EP: Mixture of HA and A - : Buffer! Can use HH-equation only need to know the relative amounts of HA and A -, since it is their ratio that determines the ph When for instance 3.00 ml NaOH is added: HA + OH - A - + H 2 O Rel. initial quantities: 1 3/10 - Rel. final quantities: 7/10-3/10 48

ph pk a + log [A- ] [HA] 3/10 6.27 + log 5.90 7/10 Alternatively, one can also calculate the actual concentrations of HA and A - : [HA] [HA] i V e -V V e (0.02000 M) 0.01321 M V i V i + V 10.00 ml 3.00 ml 10.00 ml 50.00 ml 50.00 ml + 3.00 ml [A - ] [HA] i V V e V i V i + V 3.00 ml 10.00 ml 50.00 ml (0.02000 M) 50.00 ml + 3.00 ml 0.005660 M [A - ] as 0.005660 / 0.01321 3/7 [HA] calculated before with simplified procedure 49 When 5.00 ml of NaOH is added (half of V e ): HA + OH - A - + H 2 O Rel. initial quantities: 1 5/10 - Rel. final quantities: 5/10-5/10 ph pk a when V ½ V e (but true pk a requires activity coefficients) At EP: Exactly enough NaOH is added to consume all HA: HA + OH - A - + H 2 O Rel. initial quantities: 1 1 - Rel. final quantities: - - 1 This is essentially the same problem as calculating the ph of a solution of NaA in water: A - + H 2 O HA + OH - F x x x The only pitfall is that F is no longer 0.02000 M since the original solution has been diluted: 50

F (0.02000 M) 50.00 0.01667 M 50.00 + 10.00 K b K w /K a 10-14 /10-6.27 0.01667 - x x 2 x 1.76 x 10-5 poh 4.75 ph 9.25 Note that ph at the EP 7.00. The weak acid at the EP is converted to its conjugate base (which hydrolyzes to give OH - ) After EP: addition of NaOH to a solution of A - ; ph determined by excess OH - When for instance 10.10 ml NaOH is added: [OH - ] (0.1000 M) (10.10 10.00) 1.66 x 10-4 M 50.00 + 10.10 poh 3.78 ph 10.22 51 EP: steepest part of the curve; inflection point with maximum slope ½V e : ph pk a : also inflection point, but of minimum slope here the solution is most resistant to ph changes a buffer is most resistant to ph changes at ph pk a Titration curve depends on pk a and on [HA] It is not practical to titrate an acid or base when its strength is too weak, or its concentration is too low. pk a or pk b has to be > ~7 or 8 52

Titration of a weak base with a strong acid: Essentially the reverse of the previous case. B + H + BH + K 1/K a (reverse of acid dissociation) Before addition of acid: We simply have a weak base dissolved in water: B + H 2 O BH + + OH - initial : F - - final: F - x x x K b [BH + ] [OH - ] [B] Before EP: Mixture of B and BH + : Buffer! Can use HH-equation only need to know the relative amounts of B and BH + ph pk a + log [B] [BH + ] (where pk a is for BH + ) 53 At EP: All B converted into BH + : problem of calculating ph is same as for dissolving BH + : BH + B + H + K a K w /K b F x x x F is calculated from F by taking into account the dilution. ph at EP < 7.00, because of the above reaction After EP: ph determined by excess H + Titrations in diprotic systems: EXAMPLE: titration of 10.00 ml of 0.1000 M B with 0.1000 M HCl B is dibasic with pk b1 4.00 and pk b2 9.00 B + H + BH + EP 1 BH + + H + BH 2+ 2 EP 2 V e1 10.00 ml, and V e2 20.00 ml 2 V e1 (always) pk a1 5.00 pk a2 10.00 54

Point A: Solution only contains weak base B B + H 2 O BH + + OH - 0.1000 - x x x x 2 K b1 10 0.1000 - x -4 x 3.11 x 10-3 poh 2.51 ph 11.49 Point B: buffer B / BH + [B] ph pk a2 + log [BH + ] The pk a is for: BH + B + H + Point B is halfway to the EP, so ph pk a2 14 pk b1 10.00 At any other point in this buffer region, we simply have to find the fraction of the way from point A to C that the titration has progressed. 55 e.g. if 1.50 ml HCl is added: B + H + BH + Relative quantities 10/10 1.50/10-8.50/10-1.50/10 ph 10.00 + log 8.5/10 1.50/10 10.75 Point C: First EP, all B converted into BH +, which is amphiprotic We use Eq. 10-11 (slide 25): [H + ] F + K w + F V i 10.00 ml F F (0.1000 M ) 0.0500 M V i + V 10.00 ml + 10.00 ml ph 7.50 ( ½ (p + p )) Notice that point C is on the steepest part of the titration curve, which means that it is the least buffered point of all worst choice for a buffer! 56

Point D: buffer BH + / BH 2+ 2 [BH + ] ph pk a1 + log [BH 2+ 2 ] Point D is at 15.00 ml, and ph pk a1 5.00 At an intermediate point of e.g. 17.20 ml, the ratio in the log term would be (20.00 17.20) / (17.20 10.00) 2.80 / 7.20 Point E: 2 nd EP; all BH + converted to BH 2 2+ - equivalent to dissolving BH 2 Cl 2 in water BH 2+ 2 BH + + H + F x x x V i 10.00 ml F F (0.1000 M ) 0.0333 M V i + V 10.00 ml + 20.00 ml K a1 10-5 x 2 0. 0333 - x x 5.72 x 10-4 M or ph 3.24 57 Beyond point E, the ph is determined by the excess H + from the addition of the strong acid; this is simply a dilution problem: e.g. a total of 25.00 ml HCl added: 25.00 ml 20.00 ml [H + ] (0.1000 M) 0.0143 M, or ph 1.85 10.00 ml + 25.00 ml The lower curve in the figure on slide 55 is the titration curve of 10.00 ml of 0.1000 M nicotine (pk b1 6.15; pk b2 10.85). There is no clear EP 2, because BH + is too weak a base (or in other words: BH 2 2+ is too strong an acid). The approximation that BH + completely reacts with HCl is not justified. A systematic treatment of the equilibrium would be required. 58

Endpoint detection in acid-base titrations: Using indicators: Acid-base indicator acid or base whose various protonated species have different colors R Y - + H + ph pk a1 + log [Y - ] [R] ph [Y - ] / [R] Color 0.7 1 / 10 red 1.7 1 / 1 orange 2.7 10 / 1 yellow Solution appears red when [Y - ] / [R] 1/10 (ph p 1), and yellow when [Y - ] / [R] 10/1 (ph p + 1; and [B 2- ] / [Y-] p 1). The ph range over which the indicator changes from red to yellow is called the transition range. 59 As a general rule of thumb, the ph range for color change of indicator HIn is ph pk HIn ±1 An indicator is chosen such that its transition range overlaps the steepest part of the titration curve as close as possible. Steep ph change btw ph 7 and 4. Bromocresol purple is the better choice. Difference btw endpoint (seen as a colour change) and the equivalence point indicator error Since the indicators here are acids/bases, they react with the titrant. Never use more than a few drops of dilute indicator solution (#moles of indicator has to be negligible compared to #moles of analyte). 60

Using derivatives: End point of titration: steepest point of the slope of the titration curve. Here, dph/dv is maximal, and d 2 ph/dv 2 is zero. Δ 61 A few notes on standards in acid-base titrations: Some acids and bases are suitable as primary standards (can be obtained in high purity, and are stable). O e.g. HCl, KHP O OH O K + Na 2 CO 3, Tris OH HO NH 2 HO NaOH and KOH are not primary standards; they contain carbonate (from atmosphere) and adsorbed water. Their solutions have to be standardized against e.g. KHP. When prepared, they have to be protected from the air to prevent: OH - + CO 2 HCO 3-. They are also best stored in polyethylene bottles because they react with glass (dissolve silica). 62

Non-Aqueous Titrations: Reasons for carrying out a titration in a solvent other than water: Analyte is not very soluble in H 2 O Analyte is too weak an acid or too weak a base to be titrated in H 2 O pk a or pk b > ~7-8 (see slide 52) i.e. the equilibrium constant for the reaction is too small to yield a distinct end-point N OH 8-hydroxyquinoline (oxine) H 2 N O C urea NH 2 K b 8.1x10-10 K b 1.3x10-14 Classification of non-aqueous solvents in acid-base chemistry: Amphiprotic solvents: can act both as an acid and a base; undergo selfionization (autoprotolysis) General: 2 SH SH 2+ + S - K S [SH 2+ ] [S - ] S: Solvent 63 pk HS Examples: 2 H 2 O H 3 O + + OH - 14.00 78.5 2 CH 3 OH CH 3 OH 2+ + CH 3 O - 16.7 32.6 2 HOAc H 2 OAc + + OAc - 14.45 6.1 2 NH 3 NH 4+ + NH 2-33.0 (at 50 C) 22 Non-ionizable solvents, with basic properties: ethers, pyridine Aprotic solvents: 'inert' solvents such as toluene, CCl 4, hexane Dielectric ct. Characteristics of amphiprotic solvents: Brønstedt-Lowry theory: ionization of an acid HA in an amphiprotic solvent SH is the sum of 2 half-reactions: HA H + + A - : measure of intrinsic acidity of HA SH + H + SH + 2 : measure of intrinsic basicity of SH HA + SH SH 2+ + A - 2 K a 64

Qualitatively, we see that the more basic the solvent, the more acidic HA will be in that solvent (the larger K a will be). If an acid (base) is too weak to be titrated in water, dissolve it in a more basic (acidic) solvent so that its K a (K b ) increases. e.g. oxine can be dissolved in HOAc, so that it becomes a much stronger base than when it is dissolved in water, and can be titrated with a distinct end point. C 9 H 7 ON + H 2 O C 9 H 7 ONH + + OH - K b 8.1x10-10 C 9 H 7 ON + HOAc C 9 H 7 ONH + + OAc - K >> K b The acid-base character of an amphiprotic solvent is also responsible for the so-called leveling effect: For very strong acids (HX), the reaction: HX + SH SH 2+ + X - goes essentially to completion. For instance: in water, strong acids such as HCl and HClO 4 are completely ionized. Therefore, the strongest acid that can exist in water is H 3 O + (the strongest base in water is OH - ). If an acid stronger than H 3 O + is dissolved in water, it protonates H 2 O to make H 3 O + (a stronger base than OH - will deprotonate H 2 O to give OH - ). Strong acids such as HCl and HClO 4 dissolved in water behave as if they have the same acid strength; they are both leveled to H 3 O +. 65 Consider now HOAc as a solvent for HCl and HClO 4 ; it is much less basic than H 2 O: HCl + HOAc H 2 OAc + + Cl - K 2.8 x 10-9 HClO 4 + HOAc H 2 OAc + + ClO 4 - K 1.3 x 10-5 Now, HClO 4 is a much stronger acid than HCl (they are not leveled to the same strength). The reaction at the bottom of slide 60 is actually the sum of two reactions: HA + HS H 2 S + A - formation of an ion-pair H 2 S + A - H 2 S + + A - dissociation of the ion-pair This is where the dielectric constant of the solvent plays an important role. Dielectric constant 'measure for how well the solvent is able to separate ions of opposite charge' In water, with its large dielectric constant, the dissociation step is virtually complete, and it is the ion-pair formation step that determines the overall ionization. In acetic acid, the dissociation step proceeds to limited extent, e.g. for HClO 4 : 66

HClO 4 + HOAc H 2 OAc + ClO 4 - H 2 OAc + ClO 4 - H 2 OAc + + ClO 4 - HClO 4 + HOAc H 2 OAc + + ClO 4 - K for the overall process is ~ 10-5. However, proton transfer is essentially complete, but the separation of the ion pairs is the limiting step. 67