Acids and Bases: Overview. Common Uses of Acids and Bases. Common Types of Acids. Common Types of Bases. Acids and Bases.

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1 Acids and Bases: Overview Common Uses of Acids and Bases Definitions of acids and bases Equilibria involving acids and bases Conjugate acid base pairs Autoionisation of water The p scale (ph, poh, pka, pkb) Weak acids and bases Calculations involving pka and pkb Strong acids and bases Polyprotic acids Salts of acids and bases Buffers Indicators Titrations Chem B 1 Chem B 2 Common Types of Acids Common Types of Bases Organic acids Organic acids have a carboxyl group Organic bases have a nitrogen atom with a lone pair Oxy Acids Organic bases Many acids are oxy acids where the proton is attached to an oxygen atom. Hydroxides and oxides LiOH, NaOH, K 2 O etc. Hydrohalic Acids H-F, H-Cl, H-Br and H-I Chem B 3 Chem B 4 Acids and Bases Acids were originally recognised by their sour taste, e.g., lemon (citric acid). Bases usually by their bitter taste and slippery feel, e.g., drain uncloggers (caustic soda or sodium hydroxide). Arrhenius was the first to recognise the nature of acids and bases. Strong acids and bases: HCl (aq) + NaOH (aq) H 0 = -56 kj mol -1 HI (aq) + NaOH (aq) H 0 = -56 kj mol -1 HNO3 (aq) + NaOH (aq) H 0 = -56 kj mol etc Why do these all have the same heat of reaction? Because the only species that actually react are: H + (aq) + OH - (aq) = H 2 O(l) H 0 = -56 kj mol -1 Chem B 5 Arrhenius: Definitions H + (aq) + OH (aq) H2O (l) ACID: H + producer in aqueous solution i.e. AH BASE: OH producer i.e. MOH Brønsted Lowry: ACID: proton donor (H + ) e.g. HCl BASE: proton acceptor e.g. NH3 Lewis ACID: electron pair acceptor e.g. BF3 BASE: electron pair donor e.g. NH3 H + + A HA A + :B A:B Chem B 6 1

2 NaOH and Dry Ice Brønsted Lowry Model acid = proton donor base = proton acceptor 1 M NaOH ph 12 CO2 (s) CO2 (g) CO2 (aq) H + (aq) + HCO3 (aq) Electron-pair acceptor (Lewis acid) H + (aq) + OH (aq) H2O (l) DEMO: 5.10 dry ice in water plus 1M HCL and NaOH + litmus paper Chem B 7 Note that water can act as Chem an acid B or a base - it is amphoteric 8 Acids, Bases & Equilibrium In water: an acid (e.g., HCl) ionises to produce H + (aq) NB: Actually H3O + (aq), but we usually just write H + (aq) + Acids, Bases & Equilibrium HA (aq) + H2O (l) H3O + (aq) + A - (aq) A STRONG acid has equilibrium to the right (HA mostly dissociated) A WEAK acid has equilibrium to the left (HA mostly intact) Equilibrium Equation: H 3 O + Ka is the ACID DISSOCIATION CONSTANT Note: it has the same form as if reaction was a simple dissociation: HA (aq) H + (aq) + A - (aq) Generally exists as [H(H2O)n] + (aq) Chem B 9 Chem B 10 Conjugate Acid Base Pairs NH4 + is the conjugate acid of NH3 NH3 is the conjugate base of NH4 + A conjugate base has one less proton than its conjugate acid. HSO4 : conjugate base is SO4 2 conjugate acid is H2SO4 H2SO4 is a dibasic or diprotic acid: (lies ~100% to right) H 2 SO 4 (aq) + H 2 O (l) H 3 O + (aq) + HSO 4 (aq) HSO 4 (aq) + H 2 O (l) H 3 O + (aq) + SO 2 4 (aq) Chem B 11 Questions: Write the formula of the Write the formula of the conjugate bases conjugate acids H 3 O + OH - H 2 SO 4 H 2 O HClO 4 CN - CH 3 COOH NH 3 HPO 2 4 HPO 2 4 Chem B 12 2

3 Acid Base Reactions Autoionisation of Water H2O (l) H + (aq) + OH (aq) Equilibrium constant given special symbol: Kw = [H + ][OH ] NB: [H2O (l)] = constant This is why pure water is a weak electrolyte At 25 C: Kw = REMEMBER THIS! The positions of these equilibria determine the strengths of the acids. Neutral solution: [H + ] = [OH ] = 10 7 M Acidic solution: [H + ] > 10 7 M Basic : [H + ] < 10 7 M Chem B 13 Chem B 14 The ph Scale Because the concentrations of acids and bases can vary over many orders of magnitude, it is convenient to define a logarithmic scale to compare them: ph = log10[h + ] e.g. If [H + ] = M then ph = log(10 6 ) = ( 6) = 6.0 Chem B 15 The p Convention ph = log 10 [H + ] poh = log 10 [OH ] pk w = log 10 [K w ] = 14 at 25 C Acid : ph < 7 Neutral: ph = 7 Basic: ph > 7 Since K w = [H + ][OH ]: log 10 K w = log 10 [H + ] + log 10 [OH ] log 10 [H + ] log 10 [OH ] = log 10 K w ph + poh = 14 poh = 14 ph Note that an acidic solution contains some OH -, and a basic solution contains some H3O + Chem B 16 Temperature Dependence of ph The ph Scale H 2 O E a H + + OH - ΔH 0 = 56 kj mol -1 Remember, this is a log scale - so 0.1 M HCl is ~100 more acidic than 0.1 M acetic acid. K w = only at 25 C For T > 25 C, K w > ph + poh < 14 if T > 25 C For T < 25 C, K w < neutral ph > 7 if T > 25 C Chem B 17 Chem B 18 3

4 Autoionisation of Water H2O (l) H + (aq) + OH (aq) Equilibrium constant given special symbol: Kw = [H + ][OH ] NB: [H2O (l)] = constant At 25 C: Kw = Neutral solution: [H + ] = [OH ] =10-7 M Acidic solution: [H + ] > 10-7 M Basic : [H + ] < 10-7 M This is why pure water is a weak electrolyte Acids, Bases & Equilibrium HA (aq) + H2O (l) H3O + (aq) + A (aq) A STRONG acid has equilibrium to the right (HA mostly dissociated) A WEAK acid has equilibrium to the left (HA mostly intact) Equilibrium Equation: Ka is the ACID DISSOCIATION CONSTANT Note: it has the same form as if reaction was a simple dissociation: HA (aq) H + (aq) + A (aq) Chem B 19 Chem B 20 Weak Acids HA(aq) Most acids and bases are weak they do not completely ionise in water. Acid dissociation constant H + (aq) + A (aq) pk a = log 10 K a The more positive the pka, the weaker the acid (and the stronger the conjugate base). Chem B 21 Acetic Acid: Vinegar acetic (ethanoic) acid, CH 3 COOH (HAc for short): K a = = 1.99 x 10 5 pk a = 4.7 ph of 0.1 M solution of acetic acid > 1 (ph would be log(0.1) = 1 only if it were completely ionised) Chem B 22 Relationship Between Ka and pka Examples of Ka Values The larger the value of Ka the stronger the acid and the lower the value of pka. Note effect of changing the organic part of carboxylic acids Ka = then pka = log10( ) = pka = then Ka = = Chem B 23 Chem B 24 4

5 Example Find the ph of 0.1 M acetic acid, CH3COOH (HAc) DATA: pka = 4.7, Ka = HAc (aq) H + (aq) + Ac (aq) K a = H 2 O (l) H + (aq) + OH (aq) K a = Concns. (M) HAc (aq) H + (aq) + Ac (aq) before eqm: after: 0.1 x x x make x either H + or Ac Chem B 25 Example (continued) Since the equilibrium constant is very small we assume x << 0.1, i.e. ( 0.1 x ) x 2 / 0.1 x = ph = log 10 [H + ] = log 10 x = log = 2.9 Check: x = = << 0.1, ( 0.1 x ) = M which is less than 5 % different to our assumption that x = 0.1, so our assumption was valid. [Note that the exact answer found by solving a quadratic would be x = ] Chem B 26 Demo: Strong Acids versus Weak Acids Example Find % ionisation of 0.50 M HF (pka = 3.1). Conc (M) HF (aq) H + (aq) + F (aq) before eqm.: after: 0.50 x x x DEMO: 5.9 strong and weak acid with metal Chem B 27 Hence Ka = x 2 / ( x) If x << 0.50, x 2 / 0.50 x 2 = x = M (check: indeed x << 0.50 M) % ionisation = x / = 4 % (indeed a weak acid!) Chem B 28 Weak Bases Weak Bases Ionisation of a weak base: NH3 (aq) + H2O (l) NH4 + (aq) + OH (aq) Equilibrium constant is called base ionisation constant, Kb : Note effect of changing what the N is attached to We can calculate poh and hence ph, given Kb. Chem B 29 Chem B 30 5

6 pka and pkb For conjugate systems (Brønsted Lowry acid base pairs) As acid (HA): HA (aq) H + (aq) + A (aq) As conjugate base (A ): A (aq) + H 2 O (l) HA (aq) + OH (aq) pk a + pk b = 14.0 Example Find the ph of 10 2 M NaHCO 2 (pk a of formic acid (HCO 2 H) is 4.1) (check: it will be basic so expect ph > 7) HCO 2- (aq) + H 2 O (l) OH - (aq) + HCO 2 H (aq) before: after: 10-2 x x x pk b = 14 pk a = = 9.9 Hence only need values of pk a, since pk b = 14 pk a Chem B 31 Chem B 32 Example (continued) (indeed x << 10-2 ) x = [OH - ] so poh = 5.95 ; ph = = 8.05 = 8.1 (to one significant figure) When we say 10-2 M, experimentally we mean M < conc < M Q. What would the ph be for M NaHCO 2? M NaHCO 2? Chem B 33 Weak Acids and Bases HA (aq) H + (aq) + A (aq) Acid dissociation constant pk a = log 10 K a More positive pk a weaker acid (and stronger conjugate base) B (aq) + H 2 O (aq) HB + (aq) + OH (aq) Base ionisation constant pk b = log 10 K b More positive pk b weaker base (and stronger conjugate acid) Chem B 34 Relative Strengths of Acids and Bases Relative Strengths of Acids and Bases Strongest acids lose their protons easily: For H-X bonds, larger X weaker H-X bonds stronger acids. For H-O-X bonds, more electronegative X weaker H-O bonds stronger acid Chem B 35 Chem B 36 6

7 Relative Strengths of Acids and Bases Strongest bases hold on to protons strongly: An acid-base equilibrium always lies in the direction of the weaker acid and weaker base. HCN (aq) + CH 3 COO (aq) CH 3 COOH (aq) + CN (aq) weaker acid & weaker base stronger acid & stronger base pk a = 9.2 pk b = 9.3 pk a = 4.7 pk b = 4.8 Strong Acids and Bases Completely ionise in water: e.g. HCl (aq) H + (aq) + Cl (aq) equilibrium lies completely to right, K a Strong acids: H 2 SO 4, HCl, HBr, HI, HNO 3, HClO 4 Learn these! Strong bases: All hydroxides of Groups 1 & 2 (except Be): NaOH, Ca(OH) 2, Chem B 37 Chem B 38 Example What is the ph of a 0.1 M HCl solution? Example: What is the ph of a M NaOH solution? What are the sources of [H + ]? HCl (aq) H + (aq) + Cl - (aq) H 2 O (l) H + (aq) + OH - (aq) [H + ] = 0.1 M [H + ] = 10-7 M Sodium hydroxide is fully soluble thus totally ionised The water equilibrium will be driven to the left by the high value of H + from the HCl hence [H + ] from auto-ionisation of water will be negligible and similarly [OH - ] is negligible. Thus [H + ] = 0.1 M and ph = log 10 [H + ] = 1.0 Thus [OH ] = M poh = log 10 [OH ] = log 10 (0.002) = 2.7 ph = = 11.3 Chem B 39 Chem B 40 Calculate the ph of: 1) M HNO 3 2) M NaOH 3) M Ca(OH) 2 More Examples Solutions 1) HNO 3 is a strong acid thus fully dissociated. HNO 3 + H 2 O H 3 O + + NO 3 [H 3 O + ] = 10 3 M Thus ph = log = 3.0 2) Sodium hydroxide is fully soluble [OH ] = 10 3 M and poh = log = 3.0 ph + poh = 14 therefore ph = = 11 3) Ca(OH) 2 gives 2 mole of OH for each mole thus [OH ] = 0.002M and poh = log = Chem B 41 Chem B 42 7

8 Alternate question? 1) What is the ph of a solution formed by mixing 400 ml of 0.05 M HCl with 600 ml of 0.05 M NaOH Solutions: 1) First what is the reaction? Neutralization thus is only H + + OH H 2 O We are given concentrations and volumes thus when the solutions are added final volume and concentrations must be calculated. n= C x V from HCl mol of H + = 0.05 M x 0.40L = 0.02mol From NaOH mol of OH = 0.05 M x 0.60L = 0.03mol Volume = 1.0 L Excess OH thus [OH ] = ( )/1.0L = 0.01M poh = log = 2.0 and ph = 12 Chem B 43 Chem B 44 Alternate question? 2) What is the [H + ] of a solution with a ph of 4.5? Solution: If ph= 4.5 then [H + ] = or 3.2 x 10 5 Harder Example What is the ph of a M HCl solution? What are the sources of [H + ]? HCl (aq) H + (aq) + Cl (aq) [H + ] = M H 2 O (l ) H + (aq) + OH (aq) [H + ] = 10 7 M So is [H + ] = M?? NO the water equilibrium will be driven to the left by the acid. Chem B 45 Chem B 46 Solution: Conc (M) H 2 O H + + OH Initial H 2 O autoionisation: ( x) x K w = [H + ][OH ] = ( x)(x) = Solve quadratic equation x = M, giving ph = 6.62 b x where b 4ac 2a ax 2 2 bx c 0 Polyprotic Acids H 3 PO 4 (aq) H + (aq) + H 2 PO 4 (aq)k a1 = H 2 PO 4 (aq) H + (aq) + HPO 4 2 (aq) K a2 = HPO 4 2 (aq) H + (aq) + PO 4 3 (aq) K a3 = K a1 > K a2 > K a3 Harder to remove +ve charge against increasing ve charge. consider one equilibrium at a time Chem B 47 Chem B 48 8

9 Polyprotic Acids Salts of Weak Acids and Bases Is a solution of NaNO 2 acidic or basic? Ionisable protons close together Ionisable protons a long way apart Take the ions apart Na + is from the strong base NaOH NO 2 is from the weak acid and HNO 2. The base wins ph > 7. Chem B 49 Chem B 50 Overall reaction H 2 O (aq) + NO 2 (aq) OH (aq) + HNO 2 (aq) We can discount the other reaction: H 2 O (aq) + Na + (aq) NaOH (aq) + H + (aq) Sodium hydroxide will not reform NaOH is a strong base and cannot coexist with H + Salts of Weak Acids and Bases Does a solution of NH 4 Cl have ph > 7 or < 7? Separate the salt ions NH + 4 comes from the weak base NH 4 OH Cl comes from the strong acid HCl. The acid wins ph < 7. Chem B 51 DEMO: ammonium chloride in water Chem B 52 Overall reaction H 2 O (aq) + NH 4 + (aq) NH 3 (aq) + H 3 O + (aq) We can discount the other reaction: H 2 O (aq) + Cl (aq) HCl (aq) + OH (aq) because HCl is a strong acid and cannot coexist with OH Salts of Weak Acids and Bases What is the ph of ammonium acetate at 25 C? 2 possible reactions are: CH 3 COO (aq) + H 2 O (l) CH 3 COOH (aq) + OH (aq) K b = Anions of weak acids hydrolyse OH NH + 4 (aq) + H 2 O (l) NH 3 (aq) + H 3 O + (aq) K a = Cations of weak bases hydrolyse H 3 O + In this case K a = K b so salt is neutral! LOOK at the K a and K b how different are they Chem B 53 DEMO ammonium acetate in water Chem B 54 9

10 Salts of Strong Acids and Bases Does a solution of NaCl have ph > 7 or < 7? We can discount the reaction: H 2 O (aq) + Na + (aq) NaOH (aq) + H + (aq) because NaOH is a STRONG BASE and cannot coexist with H +, and we can discount the reaction: H 2 O (aq) + Cl (aq) HCl (aq) + OH (aq) because HCl is a STRONG ACID and cannot coexist with OH Salts of strong acids and strong bases are always neutral pka of Weak Acid and its Salt What is the ph of a solution made up as 0.1 M in acetic acid (HAc) and 0.1 M in sodium acetate (pk a of HAc = 4.7)? Reactions are: CH 3 COOH (aq) + H 2 O (l) CH 3 COO (aq) + H 3 O + (aq) NaCH 3 COO (aq) + H 2 O (l) Na + (aq) + CH 3 COO (aq) We have COMMON ION Chem B 55 Chem B 56 Solution: take 0.1 mol of CH 3 COOH and 0.1 mol NaCH 3 COO and dilute to 1 L Can work either from acid or base dissociation constants. Take acid this time: HAc (aq) H+ (aq) + Ac (aq) Before: After: 0.1 x x x (x mol HAc dissociates and forms x mol of Ac ) could solve as quadratic, but make usual assumption that x << 0.1 The Common Ion Effect If you add the salt of an acid to a solution of the same acid then the equilibrium will shift towards neutral. CH 3 COOH (aq) + H 2 O (l) CH 3 COO (aq) + H 3 O + (aq) Addition of CH 3 COONa will boost [CH 3 COO ] By Le Châtelier s principle the first equilibrium will shift to the left to remove CH 3 COO and therefore decrease [H 3 O + ]. Same for a base IMPORTANT FOR BUFFER SOLUTIONS ph = log x = 4.7 = pk a Chem B 57 Chem B 58 Buffers A solution containing both: a weak acid + its salt OR a weak base + its salt withstands ph changes when (limited) amounts of acid or base are added. Reason: Le Châtelier s principle. if we add acid, then reaction HA H + + A goes to left to absorb change; vice versa if we add base Sodium bicarbonate is used as a buffer in swimming pools DEMO 5.11 Chem B 59 Buffers Consider the change in ph of pure water (ph = 7) if we add an equal amount of 10 3 M HCl: Take 500 ml of water ph = 7.0 ([H + ] = 10 7 M Add 500 ml of 10 3 M HCl [H + ] = M (can neglect amount already present in water), so ph goes from 7 to 3.3! Huge change! What about buffers? Chem B 60 10

11 Example Consider a buffer solution with 0.1 M each of sodium acetate (NaAc) & acetic acid (HAc): HAc H + + Ac initially: eqm: x x x i.e., suppose all but x of the added H + forms HAc, but x will be very small even in comparison to (found earlier that ph = 4.7) What is the ph when 10 3 M HCl is added? Chem B 61 x = 1.02 K a = << ph = log x = 4.69 the ph hardly changes from 4.7! Solution is buffered against ph change Chem B 62 Example (continued) Henderson Hasselbalch Equation For a buffer solution, which contains similar concentrations of a conjugate acid/base pair of a weak acid: Chem B 63 Chem B 64 Buffer Preparation Since the dissociation of HA or protonation of A doesn t lead to a significant change in the concentrations of these species. If the ph of a required buffer is the same as pk a of an available acid then use equimolar amounts of acid and conjugate base. If the required ph differs from the pk a then use the Henderson Hasselbalch equation. Chem B 65 Chem B 66 11

12 Buffer Preparation 1) Choose conjugate acid base pair chosen mostly by the desired ph. The buffer is most effective when the ratio of the acid/base is close to 1 ph close to pk a. 2) Calculate the ratio of the buffer component concentrations. Find the ratio of [A ]/[HA] that gives the desired ph. Use the Henderson Hasselbalch equation ph pka log10 [A ] [HA] Chem B 67 Example 1 In the H 3 PO 4 / NaH 2 PO 4 / Na 2 HPO 4 / Na 3 PO 4 system, how could you make up a buffer with a ph of 7.40? DATA: K a1 = , K a2 = , K a3 = pk pk pk H 3 PO 4 H 2 PO 4 HPO 4 2 PO 4 3 a1 a2 a3 log log log [7.2 x 10 [6.3 x 10 [4.2 x ] ] 7.20 ] Chem B 68 [A ] Example 2 ph pka log10 [HA] so the required ratio of Na 2 HPO 4 to NaH 2 PO :1 pk a1 = 2.14 pk a2 = 7.20 pk a3 = Must use mixture of H 2 PO 4 and HPO 4. Could go through whole procedure, or simply use Henderson Hasselbalch equation 2 [HPO4 ] ph pka log10 [H2PO4 ] 2 [HPO4 ] log10 [H2PO4 ] 2 [HPO4 ] [H2PO4 ] Chem B 69 [HCOO ] log [HCOOH] [HCOO ] log 0.16 [HCOOH] [HCOO ] [HCOOH] Prepare a buffer solution with a ph = The pk a of formic acid is 3.74 or K a = 1.8 x The buffer components can be formic acid HCOOH and the formate ion HCOO supplied by a soluble salt such as sodium formate, HCOONa. To calculate the component ratios use the equation For every 1.0 mol of HCOOH we need 1.4 mol of HCOONa Chem B 70 Buffer Capacity Buffer capacity is related to the amount of strong acid or base that can be added without causing significant ph change. Depends on amount of acid & conjugate base in solution: highest when [HA] and [A ] are large highest when [HA] [A ] (most effective buffers have acid/base ratio less than 10 and more than 0.1) ph range is ±1 Buffers in Natural Systems Biological systems, e.g. blood, contain buffers: ph control essential because biochemical reactions are very sensitive to ph Human blood is slightly basic, ph In a healthy person, blood ph is never more than 0.2 ph units from its average value ph < 7.2, acidosis ; ph > 7.6, alkalosis Death if ph < 6.8 or > 7.8 Chem B 71 Chem B 72 12

13 Buffers in Natural Systems To hold the ph of the blood close to 7.4 the body uses 3 buffer systems: Carbonate the most important Phosphate proteins The acid is carbonic acid, H 2 CO 3 the base is the carbonate HCO 3. The pk a of H 2 CO 3 is Since the ph of an equimolar mixture of acid and conjugate base is equal to its pk a, a buffer made of equal concentrations of H 2 CO 3 and HCO 3 has a ph of 6.37 Chem B 73 Blood Buffering Blood however has a ph of 7.4. The carbonate buffer can maintain this ph only if [H 2 CO 3 ] is not equal to [HCO 3 ] In Fact [HCO ] [H CO ] The normal concentrations in blood are about M H 2 CO 3 Chem B 74 [HCO ] [H CO ] BLOOD 3 10 This means that it is a better buffer system for acids which lowers the ratio than for bases which increases the ratio. This is in harmony with the functioning of the body as more acidic than basic substances enter the blood. This ratio can be easily maintained as the body can readily alter the amount of CO 2 entering the blood 2 3 Buffer System in Blood Extracellular buffer (outside cell) H + (aq) + HCO 3 (aq) H 2 CO 3 (aq) H 2 CO 3 (aq) H 2 O (l) + CO 2 (g) Removal of CO 2 shifts equilibria to right, reducing [H + ], i.e., raising the ph The ph can be reduced (made acidic) by: H 2 CO 3 (aq) + OH (aq) HCO 3 (aq) + H 2 O (l) Chem B 75 Chem B 76 Titrations How much acid or base is in a given amount of solution? If solution is ACIDIC add base until the acid is neutralised If solution is BASIC add acid until base is neutralised 1. The unknown solution goes in the conical flask 2. The solution of known CONCENTRATION goes in the burette 3. The INDICATOR is chosen to change colour at the appropriate ph. Equivalence Point: When number of moles of added base = original no. of moles of acid Strong acid/strong base ph = 7 Weak acid/strong base ph > 7 Strong acid/weak base ph < 7 End Point: When a colour change in the indicator is observed Choose an indicator that changes colour close to the equivalence point Chem B 77 Chem B 78 13

14 Indicators weak acid weak base Each form has a different colour. The ph at which acid base depends on the pka of the indicator. Chem B 79 Salts of Acids and Bases Salt acidic/basic/neutral NaOH BASIC STRONGLY NaHCO 3 SLIGHTLY BASIC Na 2 CO 3 BASIC NaCl NEUTRAL NH 4 Cl ACIDIC CH 3 COONH 4 NEUTRAL Al 2 (SO 4 ) 3 ACIDIC KHSO 4 ACIDIC CH 3 COOH ACIDIC HCl ACIDIC STRONGLY Chem B 80 Salts of Acids and Bases Reasoning NaHCO 3 Na + +HCO 3 HCO 3 + H 2 O? Titrations: Strong Acid / Strong Base Na 2 CO 3 Na + +CO 3 2 CO H 2 O HCO 3 + OH NH 4 Cl NH Cl NH H 2 O NH 3 + H 3 O + Al 2 (SO 4 ) 3 SO Al(H 2 O) H 2 O Al(H 2 O) 5 (OH ) 2+ + H 3 O + KHSO 4 K + + HSO 4 + H 2 O K + + H 3 O + + SO 4 2 Chem B ml of HCl solution is titrated with NaOH Chem B 82 Titrations: Weak Acid / Strong Base Titrations: Weak Base / Strong Acid Equivalence point ph > 7 (value depends on starting concentrations) Change is more gradual Chem B 83 Equivalence point ph < 7 (value depends on starting concentrations) Chem B 84 14

15 5. Acids and Bases: Summary You should understand: The Brønsted Lowry concept of an acid as a proton donor and a base as a proton acceptor. The difference between strong acid (or base) and a weak acid (or base); and the role of the conjugate base (or acid) in equilibrium. The temperature dependence of Kw exp( H 0 /RT) and therefore ph. Trends in the relative strengths of acids and bases due to bond strength and electronegativity. How a buffer solution works due to the presence of a weak acid (or base) and its salt. You should know: The definition of the acid dissociation constant Ka = [H + ][A ]/[HA]. The p convention, e.g., ph = log10[h + ], poh = log10[oh ], pka = log10[ka]. That ph + poh = pkw = 14, and that pka + pkb = pkw = 14. You should be able to determine The conjugate base of a given acid, and the conjugate acid of a given base. The ph of a strong or weak acid or base in solution. The extent of ionisation of a weak acid or base in solution. The ph of a buffer using Henderson Hasselbalch ph pka + log([added base]/[added acid]). The equivalence point of a titration. Chem B 85 Chem B 86 15