# WEAK ACIDS AND BASES

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1 WEAK ACIDS AND BASES [MH5; Chapter 13] Recall that a strong acid or base is one which completely ionizes in water... In contrast a weak acid or base is only partially ionized in aqueous solution... The ionization of a weak acid or base in water is an equilibrium situation. It is, therefore, governed by an equilibrium constant; K a for acids and K b for bases. We write these equilibrium constants as: 163

2 Notice that... [H 2 O] (constant at M) does not appear in K a or K b. The larger the K a the stronger the acid; the larger the K b the stronger the base. EXAMPLE: CCR 3 COOH; K a = 2.0 x 10 1 is a stronger acid than CH 3 COOH; K a = 1.8 x 10-5 CH 3 NH 2 ; K b = 6.4 x 10-4 is a stronger base than NH 3 ; K b = 1.8 x 10-5 H s on C are not (generally) acidic in water - e.g. CH 3 COOH monoprotic (one H + ) and one K a We may define pk a analogous to ph and pk w : pk a = - log K a pk b = - log K b EXAMPLES: pk a for CCR 3 COOH = 0.70 [ - log (2.0 x 10-1 )] pk a for CH 3 COOH = 4.74 [ - log (1.8 x 10-5 )] pk b for CH 3 NH 2 = 3.19 [ - log (6.4 x 10-4 )] pk b for NH 3 = 4.74 [ - log (1.8 x 10-5 )] Because of the minus sign, the smaller the value of K a or K b the larger the value of pk a, or pk b OR The weaker the acid or base, the larger the pk a or pk b

3 Calculations Involving Weak Acids Suppose that we have a solution of acetic acid, CH 3 COOH (K a = 1.8 x 10 5 ); let its initial concentration be represented by c moll 1. At equilibrium, x mol L 1 have dissociated... CH 3 COOH º H + + CH 3 COO This is a quadratic equation in x. The quadratic equation may be simplified by assuming that x is negligible compared to c. CH 3 COOH is a weak acid; so we may assume that the small amount dissociation makes little difference to the amount of undissociated acid left. Now let c = 0.10 moll

4 If x is negligible compared to 0.10 moll 1 : So... In this case, x is about 1.3% of 0.10, a negligible error within the accuracy of the original data we were given. If we were to solve the equation exactly, x = M, so the approximate answer M only differs from the exact answer by , an error of less than 1%. General Guideline: This assumption; a negligible amount of the acid being ionized - is acceptable if x # 5% of the initial concentration of acid... This will generally be the case if the concentration of the acid, c, divided by the K a value is > 100. EXAMPLE: [CH 3 COOH] = 0.10 M; K a = Test the assumption that c - x. c: 166

5 The quantity [x /c] x 100 is called the degree of dissociation, expressed on a % basis; may also be called percent dissociation or ionization. Percent = Amount dissociated x 100% dissociation Original amount = We can also use x to calculate ph; because in a weak acid equilibrium; x = [H + ] [H + ] = M, therefore ph = 2.87 The degree of dissociation increases as the solution is diluted, although [H + ] decreases. EXAMPLE: For solutions of HF in water: K a for HF = 6.7 x 10 4 Initial [HF] Equil. [H + ] % Dissoc. ph x x x decreasing increasing increasing 167

6 What if we need to do the exact calculation? We have to solve the quadratic: x 2 + K a x K a C = 0 (from expansion of K a = x 2 /(C x) What else can we calculate using K a?? The K a expression says: Depending on the information given, we could also calculate the initial concentration of the weak acid or the actual value of K a... Strategy Always write the reaction for the weak acid given in the question. Write the equilibrium constant expression for the reaction. Identify what you know and what you are asked to find. Usually, it s a fairly straightforward matter of substituting for the appropriate variables in the Equilibrium Constant Expression. Sometimes there are a few intermediate calculations to perform

7 EXAMPLE 1: Nitrous acid, HNO 2 has a K a value of 6.0 x Calculate the initial concentration of HNO 2 if a solution of this acid has a ph of

8 EXAMPLE 2: A M solution of formic acid, HCOOH, has a ph of a) Calculate the % ionization of this solution. b) Calculate the K a for formic acid. 170

9 Calculations Involving Weak Bases The most commonly used weak bases are Ammonia (NH 3 ) and its derivatives. In these compounds one (or more) N atom has a pair of electrons not used (yet!) for bonding (a lone pair ). EXAMPLES: : NH 3 + H 2 O º NH OH CH 3 NH 2 + H 2 O º CH 3 NH OH We may write K b expressions for both of these... Problems using K b are treated in the same manner as problems using K a (weak acid). 171

10 EXAMPLE 1: What is the % ionization and ph of a solution of a M solution of NH 3? [K b = 1.8 x 10 5 ] 172

11 EXAMPLE 2: A solution of methylamine, CH 3 NH 2, has a ph of Calculate the initial concentration of methylamine in this solution. [K b of CH 3 NH 2 = 6.4 x 10 4 ] 173

12 EXAMPLE 3: Calculate the K b for the weak base B, if a M solution of that base is 8.50 % ionized. 174

13 Reactions of Conjugate Species Every weak acid (HA), will produce its conjugate base (A ) when it ionizes in water. The conjugate base of any acid is the species that is obtained from the acid by removal of one H + (or proton). Every weak base (B), will produce its conjugate acid (BH + ) when it ionizes in water. Similarly, the conjugate acid of any base is the species that is obtained from the base by addition of a proton (or H + ). It is essential to realize that in any conjugate acid/base pair, the acid always has one more H than the base! You must be able to recognize conjugate bases and acids, based on the original weak acid or base! 175

14 EXAMPLES: Consider the weak acid HNO 2 ; its conjugate base is NO 2. Now look at the weak base NH 3 ; its conjugate acid is NH HNO 2 / NO 2 and NH 3 / NH 4 pairs [differing by one H + ] are conjugate acid-conjugate base How do these conjugate species behave? What happens if you react them with water? Consider the generic weak acid, HA; its conjugate base is A. This species will act as a base: This reaction is often called hydrolysis. What is K b for A, and how is it related to K a for HA? Or...to get the same result, we could add the equations, and then (as we have seen earlier in these notes; p. 137) their K s are multiplied: 176

15 The relationship K a x K b = K w is always true for a conjugate acid-base pair. The weaker the acid, the more basic is its conjugate base. We can also use the log scale... pk a + pk b = pk w = 14.0 EXAMPLE 1: HF, a weak acid: K a = 7.24 x 10 4 ; pk a = 3.14 F, (conjugate base of HF) pkb = K b = EXAMPLE 2: CH 3 COOH, a weak acid: K a = 1.8 x 10 5 ; pk a = 4.74 CH 3 COO, (conjugate base of CH 3 COOH) pk b = K b = EXAMPLE 3: HCN, a very weak acid: K a = 4.00 x ; pk a = 9.40 CN, (conjugate base of HCN) pk b = K b = 177

16 Now look at the generic weak base B, whose conjugate acid is BH +. BH + will act like an acid... We also call this a hydrolysis reaction. We can determine K a for this conjugate acid in the same manner as we determined K b for a conjugate base previously... As was the case with the weak acids, the weaker the base, the more acidic is its conjugate acid. EXAMPLE 1: NH 3 is a weak base; K b = 1.8 x 10 5 ; pk b = 4.74 NH 4 + (conjugate acid of NH 3 ); pk a = K a (NH 4 + ) = EXAMPLE 2: Aniline, C 6 H 5 NH 2, is a very weak base; K b = 4.0 x ; pk b = 9.40 C 6 H 5 NH 3 + (conjugate acid of C 6 H 5 NH 2 ) ; pk a = K a = 178

17 Salts You may have noticed that the conjugate base of a weak acid, or the conjugate acid of a weak base is always an ion. So where does this ion come from? It is always produced when the parent weak acid or base ionizes, but these conjugate species can be also be found in ionic compounds. And ions are formed when an ionic solid is dissolved in water... Remember those solubility rules???? This is where they come in handy; so you will know whether or not a solid will dissolve in water to produce ions!! First we consider salts that yield the conjugate base of a weak acid. Recall that the conjugate base will always behave like a base when in aqueous solution: B + H 2 O º BH + + OH EXAMPLE 1: Consider the salt CH 3 COONa. In water... The CH 3 COO will now undergoes hydrolysis with water... The ion that is the conjugate species of a weak acid or base is the species that will undergo the hydrolysis with water. Remember that a conjugate species differs from its parent species by only one H +!! 179

18 EXAMPLE 2: Calculate the ph and % hydrolysis ( % base ionization) in M KF. [K a for HF = 7.24 x 10 4 ] 180

19 EXAMPLE 2: A M solution of the sodium salt, NaA of the weak monoprotic acid, HA, has a ph of Calculate K a for the acid, HA. 181

20 Salts that produce the conjugate acids of weak bases will exhibit acidic behaviour in solution... HA + H 2 O º H 3 O + + A EXAMPLE 1: The salt in question is NH 4 CR. In water: The NH 4 + is the conjugate acid of NH 3, so in water... EXAMPLE 2: Calculate the ph of a solution of M NH 4 NO 3. In water: Then

21 Equivalence Point of a Titration This is the point where the stoichiometric quantities of acid and base, defined by the equation, have been mixed. It is really important to note that the solution is NOT always neutral (i.e. ph = 7) at the equivalence point!! This is why, earlier, we used the term equivalence point rather than neutralization point. The ph at the equivalence point is only truly neutral (ph = 7) for a titration of a strong acid with a strong base. HCR + NaOH! NaCR + H 2 O strong strong neutral So why is the solution of NaCR neutral? Because neither Na + (aq) nor CR is the conjugate of a weak species (base or acid)... If both ions which form the salt (NaCR in this case) originally came from strong species, they may termed spectator ions and the solution will be neutral. In contrast: CH 3 COOH + NaOH! CH 3 COO Na + + H 2 O weak strong basic solution The salt formed in the reaction is CH 3 COO Na +. The CH 3 COO is a base ; the conjugate base of the weak acid CH 3 COOH. That means that this solution will be basic. 183

22 HCR + NH 3! NH 4 + CR + H 2 O strong weak acidic solution The salt formed in this reaction is NH 4 + CR. The NH 4 + is an acid; the conjugate acid of the weak base NH 3. So this solution will be acidic. There is one other possible combination; that of a weak acid + a weak base: acidic or basic at equivalence depending upon the relative strengths of the acid and base involved... Ignore! What is really happening at the equivalence point in titrations of a weak species with a strong species? 1) CH 3 COOH + NaOH! CH 3 COO Na + + H 2 O 2) HCR + NH 3! NH 4 + CR + H 2 O Because one of the species involved in the titration - CH 3 COOH in (1) and NH 3 in (2) - is a weak species, the reactions are not quite complete (because the weak species never completely ionize.) They are, however, close enough to complete that we can assume that the salt is by far the major species at equilibrium. So... as a first approximation, we start our calculation by assuming complete reaction to the salt. [As shown by equations 1) & 2).] To correct this not-quite-true first approximation, we correct it by allowing a back reaction; this is hydrolysis of water by the salt. 3) CH 3 COO + H 2 O º CH 3 COOH + OH 4) NH H 2 O º NH 3 + H 3 O + These hydrolysis reactions are actually the ionic equations corresponding to the reverse of (1) and (2)!! 184

23 How do we put all this information together to solve an equivalence point problem? The steps are: 1) Write an equation for the acid base reaction. (Similar to equations 1) or 2) on the previous page.) 2) Determine the number of moles of the acid and base...recall that at the equivalence point, these will be equal. 3) Determine the number of moles of salt formed (equal to the number of moles of acid or base). 4) Determine the total volume of the solution, and the Molarity of the salt solution. 5) Now write an equation for the hydrolysis reaction by the salt. (Similar to equations 3) and 4) on the previous page.) 6) Using K w, calculate K a or K b for the salt. 7) Now calculate the concentration of either the H 3 O + or the OH - formed as a result of hydrolysis. 8) Finally, calculate the ph. 185

24 EXAMPLE 1: Calculate the ph at the equivalence point of the titration of ml of M benzoic acid, C 6 H 5 COOH, with M KOH. [K a for C 6 H 5 COOH = 6.6 x 10 5 ] 186

25 EXAMPLE 2: Calculate the ph at the equivalence point of the titration of M methylamine, CH 3 NH 2, with M HNO 3. [K b for CH 3 NH 2 = 6.4 x 10 4 ] 187

26 Acid Base Indicators [MH5; 14.2] Indicators are used to detect the equivalence point of a titration. An indicator is a weak organic acid that has the particular property of being a noticeably different colour from its conjugate base. The indicator used must change colour at a ph that closely matches the ph expected at the equivalence point of the titration Over the small ph range where the acid changes over to the conjugate base, we see a change of colour... HIn º H + + In At the end-point, the indicator is changing colour; this is when [In ] = [Hin] At this point, K a (aka K HIn ) = [H + ] and pk a (aka pk HIn ) = ph Note: This is the pk a value of the indicator, not that of the acid being titrated! At 1 ph unit below this pk a value (acidic) the indicator is 90% in the HIn form. At 1 ph unit above this pk a value (basic) the indicator is 90% in the In form. The approximate range for colour change: 2 ph units. Selection of Indicator The indicator must change colour near the equivalence point. For a strong acid/strong base titration: the pk a of the indicator should be 5-9, although in practice the ph changes so rapidly at the end-point that any indicator is suitable. For a weak acid/strong base titration: the pk a of the indicator must be in the basic region, for example: phenolphthalein, pk a. 9 For a strong acid/weak base titration: the pk a of the indicator must be in the acidic region, for example: methyl orange, pk a

27 The Common Ion Effect Recall that the common ion effect refers to a system at equilibrium which has present an ion that is present as a result of that equilibrium, but is also present from some other source. Consider a solution of Acetic Acid, CH 3 COOH (K a = 1.8 x 10 5 ) CH 3 COOH º H + + CH 3 COO When we place the acid in water, equilibrium is established very rapidly (.10 (8-to-10) seconds)... What happens (non-quantitatively) to the percent dissociation of the acetic acid (CH 3 COOH) if... A strong acid is added to the same solution (so H + is the added common ion )? Or, equivalently, the CH 3 COOH ionizes into a solution of a strong acid, instead of into water? (For acid-base reactions in water, the order of addition is unimportant.) CH 3 COOH º H + + CH 3 COO 189

28 How do we treat such problems quantitatively? EXAMPLE 1: A solution is 0.20 M in CH 3 COOH (K a = 1.8 x 10 5 ) and M in HCR. What the % ionization of the CH 3 COOH? What is the ph of this solution? 190

29 A similar situation arises if a weak base dissociates into a solution of a strong base (or, equivalently, a strong base is added to a solution of a weak base). EXAMPLE 2: In a solution [NaOH] = M and [NH 3 ] = 0.15 M. (K b = 1.8 x 10 5 ) What is the % ionization of the NH 3? What is the ph of this solution? 191

30 Polyprotic Acids [MH5;13.4, page 364] Polyprotic acids are those which have more than one acidic hydrogen. This means that they dissociate (or ionize) in stages, with K a values for each step. EXAMPLE: Carbonic acid is H 2 CO 3 Step 1: H 2 CO 3 º H + + HCO 3 Step 2: HCO 3 º H + + CO 3 2 K 1 = 4.2 x 10 7 K 2 = 4.8 x Overall: H 2 CO 3 º 2 H + + CO 3 2 K overall = K 1 x K 2 These equilibria are typical of polyprotic acids; K 1 >> K 2. As a result:. all the [H + ] is due to the first acid ionization... These dissociations also illustrate simultaneous equilibria in the acid dissociation of a polyprotic acid; both are happening at the same time. So both bicarbonate anion, HCO 3, and carbonate anion, CO 3 2, are present in the equilibrium mixture in solution. 192

31 EXAMPLE: Consider the acid dissociation of 0.25 M H 2 CO 3. What are the concentrations of all species in the equilibrium mixture, the % dissociation in each stage, and the ph? [ K 1 = 4.2 x 10!7, K 2 = 4.8 x ] First acid dissociation: 193

32 Second acid dissociation: Because K 2 << K 1, the second acid dissociation does not affect either [H + ] or [HCO 3 ]: General Result: for a diprotic acid H 2 A in water; [A 2 ] = K 2 Since K 2 is very small (4.8 x ), CO 3 2 is quite a strong base K b = 2.08 x 10 4 (from K a x K b = K w ). 194

33 EXAMPLE: What is the ph of M Na 2 CO 3 solution? In water: Then the hydrolysis: Further reaction of HCO 3 with H 2 O is negligible; K b (HCO 3 ) = 2.38 x 10 8 << K b (CO 3 2 ) = 2.08 x 10 4 Note that HCO 3 may act as either an acid or a base: In water: HCO 3 + H 2 O º H 3 O CO 3 (acid) HCO 3 + H 2 O º H 2 CO 3 + OH (base) OR: If acid or base is added to HCO 3... HCO 3 + H +! H 2 CO 3 HCO 3 + OH! H 2 O + CO

34 BUFFER SOLUTIONS So far, we have looked at the behaviour of weak acids and bases in solution, as well as the behaviour of their conjugate species. But initially, we have had only the weak acid or base or only the conjugate species present (in water, of course!) What happens if we put a weak acid and its conjugate base in the same solution (or, a weak base and its conjugate acid)? This type of solution is the one which call a Buffer Solution; probably named as such because these solutions tend to resist changes in ph even when small amounts of strong acid or base are added to them. A Buffer Solution contains a weak acid and its conjugate base (or a weak base and its conjugate acid) in roughly equal amounts. A buffer solution can always be treated as a weak acid or base equilibrium, with the conjugate species behaving as a common ion (see p ). As there are several ways to combine solutions to produce a buffer solution, the trickiest part of dealing with buffer solutions is recognizing that a solution is, in fact, a buffer solution. Once recognized, the equilibrium can be set up and the calculation completed... Let s start with looking at buffers which contain a weak acid and its conjugate base. The first examples are straightforward; a weak acid and its conjugate base (in the form of a salt) are mixed together in one solution. 196

35 Calculation Involving Buffer Solutions [MH5; 14.1] EXAMPLE 1: A solution contains 0.20 mol acetic acid, CH 3 COOH, and 0.10 mol of sodium acetate, CH 3 COONa, made up to 1.0 L volume. Calculate the ph of the solution. [K a for CH 3 COOH = 1.8 x 10 5 ] 197

36 EXAMPLE 2: What is the ph of the previous solution if it was diluted to a volume of 10.0 L? What does this result tell us about buffer solutions? 198

37 EXAMPLE 3: What mass, in grams, of NaNO 2 must be added to 700 ml of M HNO 2 to produce a solution with a ph of 3.50? [K a (HNO 2 ) = 6.0 x 10 4 ] 199

38 But buffer solutions are not only made from weak acids... Let s look at a buffer solution made from a weak base and its conjugate acid: EXAMPLE 1: A solution was prepared by adding 21.5 g of NH 4 CR to 1.50 L of 0.25 M NH 3 (aq). Calculate the ph of this solution. [K b for NH 3 = 1.8 x 10 5 ] 200

39 EXAMPLE 2: What mass of CH 3 NH 3 CR must be added to 650 ml of a M solution of CH 3 NH 2 to produce a solution with ph = 10.50? [K b of CH 3 NH 2 = 6.4 x 10 4 ] 201

40 Another way of preparing a buffer is to mix an excess of weak acid with a strong base...the reaction produces the conjugate species. EXAMPLE 1: 100 ml of 0.50 M Benzoic Acid, C 6 H 5 COOH, is mixed with 100 ml of 0.20 M NaOH. What is the ph of the resulting solution? [K a of C 6 H 5 COOH = 6.6 x 10 5 ] 202

41 Remember that conjugate acids behave like acids too... EXAMPLE 2: Calculate the ph of a solution prepared by mixing 150 ml of M ammonium chloride, NH 4 CR, with 100 ml of M KOH. [K b for NH 3 = 1.8 x 10 5 ] 203

42 What if we could react an excess amount of weak base (which could also be a conjugate base) with a strong acid...the reaction produces the conjugate acid. EXAMPLE 1: 200 ml of a M solution of methylamine, CH 3 NH 2, are mixed with 150 ml of a M solution of HCRO 4. What is the ph of this solution? EXAMPLE 2: What is the reaction that would occur between excess potassium nitrite, KNO 2, and HNO 3? 204

43 On p. 196 we mentioned that buffer solutions tend to resist large changes in ph, even when small amounts of strong acid or base are added to them. How do they do this, and does one calculate any change in ph of a buffer solution? EXAMPLE 1: 0.20 mol of formic acid, HCOOH, and 0.15 mol of sodium formate, HCOONa, are made up to 1.0 L volume, to give a solution having ph = What happens to the ph if we add 0.03 mol of HCR to this solution? [K a of HCOOH = 1.9 x 10 4 ] 205

44 Obviously, we could also add strong base to this buffer solution. EXAMPLE 2: 0.20 mol of formic acid, HCOOH, and 0.15 mol of sodium formate, HCOONa are made up to 1.0 L volume, to give a solution having ph = Then 1.60 g of NaOH are added to the solution. What is the change in the ph? [K a of HCOOH= 1.9 x 10 4 ] 206

45 EXAMPLE 3: What change in ph would occur if we added the same amounts of HCR (as in Ex. 1) and NaOH (as in Ex. 2) to 1.0 L of just water (instead of buffer)? 207

46 EXAMPLE 2: A solution contains 0.60 mol of NH 4 CR and 0.30 mol of NH 3 (aq) in 1.5 L. The ph of the solution is What will be the change in ph if we add 0.07 mol of HCR to this solution? 208

47 Acid Base Titration Curves [MH5; 14.3] 1) Strong Acid - Strong Base: There is no buffering action and the ph changes abruptly at the equivalence point. 209

48 2) Weak acid - Strong Base: Adding strong base to a weak acid, the solution is buffered before the equivalence point, where we have the acid + conjugate base present. The ph changes slowly and the solution is basic at the equivalence point. The vertical step around equivalence is < in (1), so choice of indicator more important. 210

49 3) Strong Acid - Weak Base: Adding weak base to a strong acid, the solution is not buffered until after the equivalence point. Before the equivalence point, the ph of the solution is controlled by the strong acid. After the equivalence point, there exists the mixture of weak base and its conjugate acid. The ph changes very slowly before the equivalence point; at which the solution is acidic. The vertical step around equivalence is < in (1), so choice of indicator more important. 211

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