Acid-Base Equilibria PDF

Chapter-7 Acid-Base Equilibria ILO’s 1- calculate the pH scale of the hydrogen ion concentration 2- evaluate pH values of salts of simple weak acidic and basic drugs 3- use the Henderson-Hasselbalch equation to estimate the pH of a buffer solution and find the equilibrium pH in acid-base reactions...

Chapter-7 Acid-Base Equilibria ILO’s 1- calculate the pH scale of the hydrogen ion concentration 2- evaluate pH values of salts of simple weak acidic and basic drugs 3- use the Henderson-Hasselbalch equation to estimate the pH of a buffer solution and find the equilibrium pH in acid-base reactions of weak drugs 4- recognize the concept of buffering capacity as a quantitative measure of resistance to pH change upon the addition of H+ or OH- ions 5- explain polyprotic acidic and basic drugs and their salts 6- calculate the pH of the solution of a polyprotic acidic and basic drugs 7- construct fractions of polyprotic species as a function of pH  Carboxylic acid drugs, a ibuprofen, b naproxen, c aspirin and d nicotinic acid Amine-containing drugs  Aqueous Solution Equilibria Electrolytes: Substances which form ions in solution 1. Strong: Mostly in ionic form in solution – HCl, HNO3, NaOH, KOH – HCl + H2O  H+(aq) + Cl-(aq) 2. Weak: Mostly not in ionic form in solution – H2CO3, Acetic acid (CH3COOH), NH3 – NH3 + H2O  NH4+(aq) + OH-(aq) Kb – CH3COOH + H2O  CH3COO-(aq) + H3O+(aq) Ka  Aqueous Solution Equilibria Acid-Base Theories: 1. Arrhenius Theory (Swedish scientist, Nobel Prize 1894) Acid: any substance that ionizes in water to give hydrogen ions (H+) that associate with the solvent to increase H3O+ in solution. Base: ionizes in water to give hydroxyl ions (OH-) Shortcoming: lack explanation of acids and bases in non-aqueous media.  Aqueous Solution Equilibria 2. BrØnsted-Lowry Theory (1923) Acid = proton donor [Acid = H+ + base] Base = proton acceptor acid base conj acid conj base HCl + H2O  H3O+ + Cl- (strong) (weak) H2CO3 + H2O  H3O+ + HCO3- (weak) (strong) BrØnsted-Lowry Theory  Lewis Acids and Bases (same Lewis who is behind the electron-dot formulas) Lewis acid: is an electron pair acceptor. A Lewis base: is an electron pair donor.  Aqueous Solution Equilibria Amphiprotic solvents: exhibit both acidic and basic properties such as Water, methanol, ethanol, glacial acetic acid H2O, CH3OH, CH3CH2OH, CH3COOH Some amphiprotic solutes: As an acid HCO3- + H2O  H3O+ + CO32- As a base HCO3- + H2O  H2CO3 + OH- As an acid HPO42- + H2O  H3O+ + PO43- As a base HPO42- + H2O  H2PO4- + OH-  Aqueous Solution Equilibria Amphiprotic solvent self-ionization: Water: H2O + H2O  H3O+ + OH- Methanol: CH3OH + CH3OH CH3OH2+ + CH3O- methonium ion / methoxide ion Glacial acetic acid: CH3COOH + CH3COOH  CH3COOH2+ + CH3COO-  The pH Scale pH = -log a(H ) or +  pH  -log[H+] p Anything = - log Anything pKw = -logKw at 25oC pKw = 14.00 Kw = [H+][OH-] pKw = pH + pOH =14 pH of the blood at body temperature (37oC) is 7.35-7.45 (slightly basic) Alkalosis Example: Find the pH of a 0.1 M HCl solution. Solution: HCl is a strong acid that completely dissociates in water, therefore we have HCl  H+ + Cl- H2O D H+ + OH- [H+]Solution = [H+]from HCl + [H+]from water However, [H+]from water = 10-7 in absence of a common ion, therefore it will be much less in presence of HCl and can thus be neglected as compared to 0.1 ( 0.1>>[H+]from water) [H+]solution = [H+]HCl = 0.1 pH = -log 0.1 = 1 We can always look at the equilibria present in water to solve such questions, we have only water that we can write an equilibrium constant for and we can write: Kw = (0.1 + x)(x) However, x is very small as compared to 0.1 (0.1>>x) 10-14 = 0.1 x x = 10-13 M Therefore, the [OH-] = 10-13 M = [H+]from water Relative error = (10-13/0.1) x 100 = 10-10% [H+] = 0.1 + x = 0.1 + 10-13 ~ 0.1 M pH = 1 Example Find the pH of a 10-7 M HCl solution. Solution HCl is a strong acid, therefore the [H+]from HCl =10-7 M Kw = (10-7 + x)(x) Let us assume that x is very small as compared to 10-7 (10-7>>x) 10-14 = 10-7 x x = 10-7 M Therefore, the[OH-] = 10-7 M = [H+]from water Relative error = (10-7/10-7) x 100 = 100 % Therefore, the assumption is invalid and we have to solve the quadratic equation. Solving the quadratic equation gives: [H+] = 7.62x10-7 M pH = 6.79 Salts of Strong Acids and Bases: The pH of the solution of salts of strong acids or bases will remain constant (pH = 7) where the following arguments apply: NaCl dissolves in water to give Cl- and Na+. For the pH to change, either or both ions should react with water. However, is it possible for these ions to react with water? Let us see: Cl- + H2O D HCl + OH- (wrong equation) Now, the question is whether it is possible for HCl to form as a product in water!! Of course this will not happen as HCl is a strong acid which is 100% dissociated in water. Therefore, Cl- will not react with water, but will stay in solution as a spectator ion. The same applies for any metal ion like K+ where if we assume that it reacts with water we will get: K+ + H2O D KOH + H+ (wrong equation) Now, the question is whether it is possible for KOH to form as a product in water!! Of course this will not happen as KOH is a strong base which is 100% dissociated in water. Therefore, K+ will not react with water, but will stay in solution as a spectator ion. It is clear now that neither K+ nor Cl- react with water and the hydrogen ion concentration of solutions of salts of strong acids and bases comes from water dissociation only and will be 10-7 M (pH =7).  Weak Acids and Bases A weak acid or base is an acid or base that partially (> x 1.75*10-5 = x2/0.10 x = 1.3x10-3 Relative error = (1.3x10-3/0.10) x 100 = 1.3% The assumption is valid and the [H+] = 1.3x10-3 M pH = 2.88 pOH = 14 – 2.88 = 11.12 Now look at the value of [OH-] = 10-11.12 = 7.6x10-12 M = [H+]from water. Therefore, the amount of H+ from water is negligible 28  Salts of Weak Acids and Bases: The salts of weak acids and bases are not neutral i.e., (pH = 7) like in the case of NaCl for example. The conjugate base of the weak acetic acid HOAc is the acetate anion salt (OAc-). It hydrolyzed in water is as follows: OAc- + H2O D HOAc + OH- Kb = [HOAc][OH-]/[OAc-] (Kb is also called KH) (Kb = Kw/Ka) 29 Sodium acetate +Na-OAC is a weak base (conj. base of the weak acid HAC) KH = Ammonium chloride NH4 +Cl- is a weak acid (conj. acid of the weak base NH3)  Salts of acidic and basic drugs Mephedrone (ammonium salt) Synthetic stimulant Tolmetin (sodium salt hydrate) Non-steroidal anti-inflammatory drug Heterocyclic acetic acid derivative class  ‫تلخيص‬ pH  -log[H+] Strong acid Weak Base or basic salt Weak Acid or acidic salt Buffer p 7.4-7.12 = 0.28 Antilog (2.8) = 10 2.8 =1.9 Example Find the pH of a 0.10 M H3PO4 solution. Solution H3PO4 D H+ + H2PO4- ka1 = 1.1 x 10-2 H2PO4- D H+ + HPO42- ka2 = 7.5 x 10-8 HPO42- D H+ + PO43- ka3 = 4.8 x 10-13 Since ka1 >> ka2 (ka1/ka2 > 104) the amount of H+ from the second and consecutive equilibria is negligible if compared to that coming from the first equilibrium. 46 Therefore, we can say that we only have: H3PO4 D H+ + H2PO4- ka1 = 1.1 x 10-2 47 Ka1 = x * x/(0.10 – x) Assume 0.10>>x since ka1 is small (!!!) 1.1*10-2 = x2/0.10 x = 0.033 Relative error = (0.033/0.10) x 100 = 33% The assumption is invalid and thus we have to use the quadratic equation. If we solve the quadratic equation we get: X = 0.028 Therefore, [H+] = 0.028 M pH = 1.55 48   Values Fractions of dissociating species at given pH How much of each species? Analytical concentration: C(H PO ) 3 4 C(H PO ) = [PO43-] + [HPO42-] + [H2PO4-] + [H3PO4] 3 4 0 = [H3PO4] /C(H PO ) , 1 = [H2PO4-] /C(H PO ) , 3 4 3 4 2 = [H PO42-] /C(H PO ) , 3 = [PO43-] /C(H PO ) 3 4 3 4 0 + 1 + 2 + 3 = 1 0 = [H+]3 / ([H+]3 + Ka1[H+]2 + Ka1Ka2[H+] + Ka1Ka2Ka3) 1 = Ka1[H+]2 / ([H+]3 + Ka1[H+]2 + Ka1Ka2[H+] + Ka1Ka2Ka3) 2= Ka1Ka2[H+] / ([H+]3 + Ka1[H+]2 + Ka1Ka2[H+] + Ka1Ka2Ka3) 3= Ka1Ka2K3 / ([H+]3 + Ka1[H+]2 + Ka1Ka2[H+] + Ka1Ka2Ka3)  The overlapping curves represent buffer regions. The pH values where 1 and 2 are 1 represent the end points in titrating H3PO4. These curves were calculated using a spreadsheet (next slide). PH=1/2[PKa1+PKa2] PH=1/2[PKa2+PKa3] 1st eq. point 2nd eq. point  PKa1 PKa2 PKa3 Fig. 7.1. Fractions of H3PO4 species as a function of pH. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)  Chapter 8 Acid-Base Titrations: 1. Strong acid vs strong base 2. Strong acid vs weak base 3. Strong base vs weak acid 4. Strong acid vs weak polyprotic base 5. Strong base vs weak polyprotic acid 6. Amino acid titration 7. Strong base vs mixture of weak acids  Acid-base titration ILO’s ‫مخرجات التعلم‬ 1- calculate and construct titration curves for strong acids with strong bases 2- recognize the concept of acid-base color Indicators 3- calculate and construct titration curves of weakly acidic drugs vs strong base. 4- calculate and construct titration curves of weakly basic drugs vs strong acid 5- calculate the equations governing weak acidic and basic drugs titration 6- calculate and construct polyfunctional acidic/basics including drugs titration curves  Titration Titrant standard solution (in buret) Titration: A standard solution (titrant) with a known [conc] is used to determine the [conc] of an unknown solution (analyte or API). The reaction that occurs is a neutralization reaction. unknown analyte (API) solution  Acidimetry involves the quantitative determination of basic drugs. Alkalimetry involves the quantitative determination of acidic drugs.  Volumetric Analysis Acid-Base Titrations Titration Curves for Strong Acids and Bases Strong acids & bases completely dissociated in HOH (e.g.) HCl, HClO4, NaOH, KOH) Consider only one equilibrium, Kw = [H3O+][OH-] Know the stoichiometric Ration of the acid-base reaction (S.R.) (e.g. HCl + NaOH  HOH + Na+ Cl- S.R. =1:1  Volumetric Analysis Acid-Base Titrations Titration Curves for Strong Acids & Bases What is the pH before a titration begins? 100.0 mL of 0.1000M HCl (anayte) [H3O+] = 1.000 x 10-1M [OH-] = Kw/[H3O+] = 1.000 x 10-13 M pH = - Log [H3O+] = 1.0000 (4s.f.) pOH = 13.0000 (4 s.f.) How do we find the pH of a titration before the equivalence point is reached? When add 1.00 mL of 0.1000M NaOH to 100.0mL of 0.1000M HCl [H3O+] = ((CacidVacid – CbaseVbase)/(Vacid + Vbase) [H3O+] = (100.0 mLx 0.1000M–1.0mL x 0.1000M)/(100+1mL) [H3O+] = 0.09802M pH = 1.0087 (4s.f.) [OH-] = Kw/[H3O+] = 1.020 x 10-13M pOH = 12.9913 (4s.f.)  Volumetric Analysis Acid-Base Titrations Titration Curves for Strong Acids & Bases How do we find pH at equivalence point? HCl + NaOH  HOH + Na+ Cl- Autodissociation of water governs the pH at equivalence point. HOH + HOH  H3O+ + OH- Kw = [H3O+][OH-] = 10-14 CH+ =(Kw)1/2 = 1.000 x 10-7M pH = 7.0000 (4s.f.) How do we find pH after equivalence point? Excess base add [OH-] = ((CbaseVbase – CacidVacid)/(Vbase + Vacid) [H3O+] = Kw /[OH-] pH = -Log [H3O+] pOH = -Log [OH-] then pH = 14 - pOH The text page 283 shows Table summarizes the equations governing the different portions of the titration curve. A strong acid – strong base titration curve has a large end point break. Phenolphthalein is used as an indicator because the colorless to pink transition is easy to see. This titration curve was constructed using a spreadsheet (next slide). ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) Fig. 8.1. Titration curve for 100 mL of 0.1 M HCl versus 0.1 M NaOH. Compare the species and equations for the strong acid with the calculations in the previous spreadsheet. A strong base-strong acid titration is treated similarly, but we start with excess base, and end with excess acid. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) As the concentrations of acid and titrant decrease, the end point break decreases. So the selection of indicator becomes more critical. 0.001 M 0.01 M 0.1 M ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) Fig. 8.2. Dependence of the magnitude of end-point break on concentration. The concentrations of acid and titrant are the same.  Volumetric Analysis Acid-Base Titrations Titration Curves for Strong Acids & Bases add standard HCL solution to NaOH solution: HCl + HOH  H3O+ + Cl- [H3O+] = C(HCl) = Ca NaOH  Na+ + OH- [OH-] = C(NaOH) = Cb HCl + NaOH  Na+ + Cl- + HOH H3O+ + OH-  HOH net ionic reaction When add 1.00 mL of 0.1000M HCl to 100 mL of 0.1000M NaOH (pH = 13.0000)  Volumetric Analysis Acid-Base Titrations Titration Curves for Strong Acids & Bases What is the pH before titration begins? 100.0 mL of 0.1000M NaOH [OH-] = 1.000 x 10-1M [H3O+] = Kw/[OH-] = 1.000 x 10-13 pH = - Log [H3O+] = 13.0000 (4s.f.) How do we find the pH of a titration before the equivalence point is reached? When add 1.00 mL of 0.1000M HCl to 100.0mL of 0.1000M NaOH [OH-] = ((CbVb – CaVa)/(Vb + Va) [OH-] = (100.0mL x 0.1000M – 1.0mL x 0.1000M)/(101mL) [OH-] = 0.09802M [H3O+] = Kw/[OH-] = 1.020 x 10-13M pH = 12.9913 (4s.f.)  Volumetric Analysis Acid-Base Titrations Titration Curves for Strong Acids & Bases How do we find pH at equivalence point? HCl + NaOH  HOH + Na+ Cl- Autodissociation of water governs the pH at equivalence point. HOH + HOH  H3O+ + OH- Kw = [H3O+][OH-] = 10-14 c =(Kw)1/2 = 1.000 x 10-7 pH = 7.0000 (4s.f.) How do we find pH after equivalence point? Excess acid added [H3O+] = ((CaVa – CbVb)/(Va + Vb) pH =-Log [H3O+] pOH = 14 – pH  This is the mirror image of the HCl titration curve. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) Fig. 8.3. Titration curve for 100 mL of 0.1 M NaOH versus 0.1 M HCl.  Volumetric Analysis Acid-Base Titrations Acid-Base Color Indicators Organic weak acids or bases Strongly colored acid and/or conjugate base form(*) Distinct color change from acid to base form Acid conj. Base HIn + OH-  In- + HOH Color A Color B (*) Only very small amount HIn required to give color ( 10-3 M  Volumetric Analysis Acid-Base Titrations Acid-Base Color Indicators Color A Color B HIn + HOH  In- + H3O+ Ka = [H3O+][In-]/[HIn] [H3O+] = Ka x [HIn]/[In-] NOTE: [HIn]/[In-] determines solution color [H3O+] determines [HIn]/[In-] High [HIn]/[In-] (>10:1) gives Color A Low [HIn]/[In-] ( 10:1 pH > pKa + Log(10/1) Color B, pH = pKa + 1 Indicator color change occurs at pH = pKa ± (1)  pH transition range = pKa ± 1. We select an indicator with a pKa near the equivalence point pH. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) Fig. 8.4. pH transition ranges and colors of some common indicators. Titration of weak acid and strong base @ midpoint pH=pKa  Example: Titrating 50.0mL 0.1M HOAc with 0.1M NaOH: HOAc + NaOH  HOH + OAc- + Na+ Initial Concentration of HOAc = CHA = 0.1M Equilibrium Concentration HOAc = Ca Equilibrium Concentration OAc- = Cs (a) Initial pH: [H3O+]  (Ka x CHA)1/2 = (1.75 x 10-5 x 0.1)1/2 [H3O+] = 1.323 x 10-3 M pH = 2.8785 (b) Buffer Region: (e.g. 10.0 mL NaOH added): [OAc-] = Cs = CbVb/(Va + Vb); Ca = (CHAVa – CbVb)/(Va + Vb) Cs = (0.1M x 10.0mL)/(60.0mL) = 0.0167M Ca = 0.0667M [H3O+] = Ka[HOAc]/[OAc-] = Ka Ca/Cs [H3O+] = (1.75 x 10-5)(0.0667/0.0167) = 6.99 x 10-5M pH = 4.1556  Acid-Base Buffers Buffer Region of Titration Curves Titrating 50.0mL 0.1M HOAc with 0.1M NaOH: HOAc + NaOH  HOH + OAc- + Na+ Buffer Region (e.g. 25.0mL NaOH added): [OAc-] = Cs = CbVb/(Va+Vb); Ca = (CHAVa- CbVb)/(Va+Vb) Cs= (0.1M)(25.0mL)/(75.0mL) = 0.0333M; Ca = 0.0333M [H3O+] = Ka[HOAc]/[OAc-] = KaCa/Cs [H3O+] = (1.75 x 10-5)(0.0333/0.0333) = 1.75 x 10-5 M pH = - Log[H3O+] = 4.757 (3s.f.) Note: at 50% titration, ½ neutralization point, pH = pKa  Volumetric Analysis Acid-Base Titrations Titrating 50.0mL 0.1M HOAc with 0.1M NaOH: HOAc + NaOH  HOH + OAc- + Na+ (c) Equivalence Point pH: [HOAc]  0 = Ca [OAc-]  CbVb/(Va+Vb) = Cs = 0.1M/2  0.05M OAc- + HOH  HOAc + OH- Kb = Kw/Ka = 5.71x 10-10 [HOAc][OH-] = Kb[OAc-] = KbCs; [HOAc]  [OH-] [OH-]  (KbCs)1/2  (5.71 x 10-10(0.05))1/2 = 5.34 x 10-6M [H3O+] = Kw/[OH-] = 1.87 x 10-9M ; pH = 8.728 (3 s.f.)  Volumetric Analysis Acid-Base Titrations Titrating 50.0mL 0.1M HOAc with 0.1M NaOH: HOAc + NaOH  HOH + OAc- + Na+ (d) pH After Equivalence Point: Bases present are OAc- & OH- (from excess NaOH) Excess NaOH is stronger base and determines pH Excess [OH-] = (CbVb – CHAVa)/(Va + Vb) At 0.1 mL past equivalence point. Vb = 50.1 mL [OH-] = ((0.1M)(50.1mL) – (0.1M)(50.0mL))/(100.1mL) [OH-] = 9.99 x 10-5M; [H3O+] = Kw/[OH-] = 1.001 x 10-10 pH = 9.9996 Titration of weak base and strong acid  Titration of weak acid and strong base Salt is basic so, equivalence point comes at a pH > 7. The start of the graph shows a relatively rapid rise in pH but this slows down  Titration of weak base & strong acid Salt formed is acidic, hence, equivalence point comes at a pH < 7. At the very beginning of the curve, the pH starts by falling quite quickly as the acid is added, but the curve very soon gets less steep. A weak acid gives a smaller end point break.. A stong base titrant is always used. We start with HOAc. Then we have a buffer mixture of OAc- and HOAc. At the equivalence point, we have OAc-, a weak base. Beyond the equivalence point, we have excess NaOH (suppresses OAc- hydrolysis), and the curve follows that for a strong acid titration. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) Fig. 8.5. Titration curve for 100 mL 0.1 M HOAc versus 0.1 M NaOH.  For a weak acid, we start with ionization of HA. In the buffer region, use the Henderson-Hasselbalch equation. At the equivalence point, A- hydrolyzes as a weak base. Beyond the equivalence point, excess OH- dominates. A weak base is treated similarly, but beginning with base and ending with acid. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)  The buffer regions are about the same for all concentrations. The equivalence point pH increases with increasing concentration. 0.1 M 0.01 M 0.001 M ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) Fig. 8.6. Dependence of titration curve of 100 mL acetic acid on concentration. NaOH concentration the same as HOAc concentration. The weaker the acid, the smaller the break and the more alkaline the equivalence point. Visual indicators can be used for Ka of 10-6. A pH meter provides better precision for weaker acids. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) Fig. 8.7. Titration curves for 100 mL 0.1 M weak acids of different K values versus 0.1 M NaOH. This is the reverse of the HOAc titration curve. We start with NH3 weak base. Then we have a buffer mixture of NH4+ and NH3. At the equivalence point, we have NH4+, a weak acid. Beyond the e.p., we have excess HCl, which suppresses the hydrolysis of NH4+, and the curve follows that for a strong base titration. Fig. 8.8. Titration curve for 100 mL 0.1 M NH3 versus 0.1 M HCl. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) The weaker the base, the smaller the break and more acid the equivalence point. Visual indicators can be used for Kb of 10-6. A pH meter provides better precision for weaker bases. ©Gary Christian, Fig. 8.9. Titration curves for 100 mL 0.1 M weak Analytical Chemistry, 6th Ed. (Wiley) bases of different Kb values versus 0.1 M HCl.  Polyfunctional Acids/Bases Equilibria Polyfunctional Acids: H3PO4 , H2CO3 , H2SO4 , H2SO3 , H2C2O4 , H2S H2S + HOH  H3O+ + HS- Ka1 = 5.7 x 10-8 HS- + HOH  H3O+ + S2- Ka2 = 1.2 x 10-15 Polyfunctional Bases: ethylenediamine (NH2C2H4NH2), CO32-, PO43-, HPO42-, S2- PO43- + HOH  HPO42- + OH- Kb1 = Kw/Ka3 = 2.4 x 10-2 HPO42- +HOH H2PO4- + OH- Kb2 = Kw/Ka2 = 1.6 x 10-7 H2PO4- +HOH H3PO4 + OH- Kb3 = Kw/Ka1 = 1.4 x10-12  Polyfunctional Acids/Bases Titration Curves Polyfunctional acids titrated with strong base: Separate equivalence points observed if ratios of successive dissociation constants > ~104 Equivalence point can be observed for dissociation step where Ka > ~10-8 e.g. H3PO4 Ka1 = 7 x 10-3, Ka2 = 6 x 10-8, Ka3 = 4 x 10-13 – First two end points can be observed – Third end point is not observed  Polyfunctional Acids/Bases Titration Curves Polyfunctional acids titrated with strong base: [H3O+] in solution of amphiprotic anion (HA-) OH- + H2A  HA- + HOH At first equivalence point, [HA-] = Cs  added OH- [H3O+] = ((Ka2Cs + Kw)/(1+ Cs/Ka1))1/2 If Cs/Ka1 >> 1; Ka2Cs >> Kw; then [H3O+]  (Ka1Ka2)1/2 Or pH = (pKa1 + pKa2)/2 We start with a weak acid, H2A, followed by a buffer region of HA- and H2A. The first equivalence point is HA- ([H+] ≈ constant). Then we have a A2-/HA- buffer region, and A2- (a fairly strong base) at the second equivalence point, followed by excess titrant. Fig. 8.12. Titration of diprotic acid, H2A, with sodium hydroxide. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) We start with ionization of H2A, a weak to moderately strong acid. In the two buffer regions, use the Henderson-Hasselbalch equation. At the first equivalence point, HA- has [H+] ≈ √Ka1Ka2. At the second equivalence point, A2- hydrolyzes as a fairly strong base. Then excess OH- dominates. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) The strong acid titrates first. At its equivalence point, we have a mixture of NaCl and HOAc, and the pH is acidic. This is followed by a buffer region of OAc- and HOAc, and then the HOAc equivalence point, where we have OAc-, a weak base. The weak acid Ka must be no larger than 10-5 to give a sharp second end point. For two weak acids, the Ka’s should differ by 104 or more. Fig. 8.13. Titration curve for 50 mL of mixture of ©Gary Christian, Analytical Chemistry, 0.1 M HCl and 0.2 M HOAc with 0.2 M NaOH. 6th Ed. (Wiley) Titration of weak triprotic acid vs strong base  We start with CO32-, a quite strong base. Then we have a HCO3-/CO32- buffer. At the first equivalence point, we have HCO3- ([H+] = √Ka1Ka2). Then we have a HCO3-/H2CO3 buffer, and H2CO3 at the second equivalence point. The first e.p. is used to approximate the second, which is more accurately used. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) Fig. 8.10. Titration curve for 50 mL 0.1 M Na2CO3 versus 0.1 M HCl. Dashed line represents a boiled solution with CO2 removed. Titrate till the methyl red indicator (gradually) changes from yellow through orange to red (occurs just before the equivalence point). Then boil to remove CO2 and continue titration for a sharp end point to a pink color. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) Fig. 8.11. Titration of 50 mL 0.1 M Na2CO3 with 0.1 M HCl using methyl red indicator.  Amino acids are polyprotic The resulting structure, with positive and negative sites, is called a zwitterion Titrations of polyprotic amino acids Titration of glutamic acid The isoelectric point is the pH at which there is zero net charge pI = (pKa1+pKa2)/2 Titration of lysine  Appendix: Aqueous Titrations from USP ‫المعايرة كما في دساتير الصيدلة‬  Direct titration methods Sodium bicarbonate chemical compound with the formula NaHCO₃ Sodium salicylate C7H5NaO3. Its chemical structure:  Indirect titration methods (residual or back titrations) Calamine: ZnO Sodium lactate: Ephedrine: Alkalimetry (direct titration method)  Some of the acidic drugs structure Benzoic acid Chlorpropamide Indomethacin Ibuprofen Citric acid Frusemide Non-aqueous titration of basic drugs  Some of the basic drugs structure Nicotinamide Metronidazole benzoate Noscapine Hydrochloride Hydrate  Chapter 9 Complex-Formation Titrations  ILO’s 1- explain metal chelation in nature, e.g., potassium ion channels in cell membranes 2- recognize the complex-formation titrations and the chelating agents used in the analysis. 3- calculate and construct the fraction of EDTA species as a function of pH 4- construct metal-EDTA titration curves and applications in pharmaceutical analytical analysis such as determination of K in slow-K tablet supplement  Complex-Formation Titrations General Principles Most metal ions form coordination compounds with electron-pair donors (ligands) Mn+ + qLm-  MLqn-mq Kf = [MLqn-mq]/[Mn+][Lm-]q The number of covalent bonds formed is called the “coordination number” (e.g. 2,4,6) e.g., Cu2+ has coordination number of 4 Cu2+ + 4 NH3  Cu(NH3)42+ Cu2+ + 4 Cl-  Cu(Cl)42-  Metal Chelation in Nature Potassium Ion Channels in Cell Membranes  Electrical signals are essential for life  Electrical signals are highly controlled by the selective passage of ions across cellular membranes - Ion channels control this function - Potassium ion channels are the largest and most diverse group - Used in brain, heart and nervous system K+ is chelated by O in channel channel contains pore that only allows K+ to pass K+ channel spans membrane Opening of potassium channel allows K+ to exit cell and change the electrical potential across membrane Current Opinion in Structural Biology 2001, 11:408–414 http://www.bimcore.emory.edu/home/molmod/Wthiel/Kchannel.html  Complex-Formation Titrations General Principles The most useful complex-formation reactions for titrimetry involve chelate formation A chelate is formed when a metal ion coordinates with two of more donor groups of a single ligand (forming a 5- or 6- membered heterocyclic ring)  Complex-Formation Titrations General Principles Chelate Formation Titrations Ligands are classified regarding the number of donor groups available: e.g., NH3 = “unidentate” (one donor group) Glycine = “bidentate” (two donor groups) (also, there are tridentate, tetradentate, pentadentate, and hexadentate chelating agents) Multidentate ligands (especially with 4 and 6 donors) are preferred for titrimetry. – react more completely with metal ion – usually react in a single step – provide sharper end-points Chelating Agents in Analytical Chemistry  Ethylenediamenetetraacetic acid (EDTA) EDTA forms 1:1 complexes with metal ions by with 6 ligands: 4 O & 2N. EDTA is the most used chelating agent in analytical chemistry, e.g. water hardness.  Complex-Formation Titrations General Principles Aminopolycarboxylic acid ligands The most useful reagents for complexometric titrations are aminopolycarboxylic acids – (tertiary amines with carboxylic acid groups) e.g., ethylenediaminetetraacetic acid (EDTA) EDTA is a hexadentate ligand EDTA forms stable chelates with most metal ions  Complex-Formation Titrations Solution Chemistry of EDTA(H4Y) EDTA has four acid dissociation steps pKa1= 1.99, pKa2= 2.67, pKa3= 6,16, pKa4= 10.26 5 forms of EDTA, (H4Y, H3Y-, H2Y2-, HY3-, Y4-) EDTA combines with all metal ions in 1:1 ratio Ag+ + Y4-  AgY3- Fe2+ + Y4-  FeY2- Al3+ + Y4-  AlY- KMY = [MYn-4]/[Mn+][Y4-]  Complex-Formation Titrations Formation Constants for EDTA Complexes Cation KMY Log KMY Cation KMY Log KMY Ag+ 2.1 x 107 7.32 Cu2+ 6.3 x 1018 18.80 Mg2+ 4.9 x 108 8.69 Zn2+ 3.2 x 1016 16.50 Ca2+ 5.0 x 1010 10.70 Cd2+ 2.9 x 1016 16.46 Sr2+ 4.3 x 108 8.63 Hg2+ 6.3 x 1021 21.80 Ba2+ 5.8 x 107 7.76 Pb2+ 1.1 x 1018 18.04 Mn2+ 6.2 x 1013 13.79 Al3+ 1.3 x 1016 16.13 Fe2+ 2.1 x 1014 14.33 Fe3+ 1.3 x 1025 25.1 Co2+ 2.0 x 1016 16.31 V3+ 7.9 x 1025 25.9 Ni2+ 4.2 x 1018 18.62 Th4+ 1.6 x 1023 23.2  Complex-Formation Titrations Equilibrium Calculations with EDTA For Mn+ + Y4-  MYn-4 KMY = [MYn-4]/[Mn+][[Y4-] Need to know [Y4-], which is pH-dependent pH dependence of Y4-: Define: a4 = [Y4-]/CT CT = [Y4-] + [HY3-] + [H2Y2-] + [H3Y-] + [H4Y] Conditional Formation Constant, KMY’ [MYn-4]/[Mn+][a4CT] = KMY KMY’ = a4 KMY = [MYn-4]/[Mn+][[CT]  Complex-Formation Titrations Equilibrium Calculations with EDTA Computing free metal ion concentrations: Use conditional formation constants, KMY’ a4 values depend on pH Thus, KMY’ are valid for specified pH only a4 values have been tabulated vs. pH  a4 = (K1K2K3K4) / ([H+]4 + K1[H+]3 + K1K2[H+]2 + K1K2K3[H+] + K1K2K3K4)  Y4- complexes with metal ions, and so the complexation equilibria are very pH dependent. Only the strongest complexes form in acid solution, e.g., HgY2-; CaY2- forms in alkaline solution. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) Fig. 9.1. Fraction of EDTA species as a function of pH. pH Dependence of αY4-  Complex-Formation Titrations Metal-EDTA Titration Curves Titration curve is: pM vs. EDTA volume e.g., 50.0mL 0.020M Ca2+ with 0.050M EDTA, pH 10.0 at pH 10.0, K(CaY )’ = (a4)(KCaY )=(0.35)(5.0 x 1010)=1.75 x 1010 2- 2- (pCa2+ values @ zero addition = -log 0.02= 1.69) (a) pCa2+ values before the equivalence point (10.0mL) Ca2+ + Y4-  CaY2- [CaY2-]formed = added EDTA [Ca2+]left = unreacted Ca2+ [Ca2+] =((50.0 x 0.020) –(10.0 x 0.050))/(60.0) = 0.0083M pCa2+ = 2.08 at 10.0mL EDTA  Complex-Formation Titrations Metal-EDTA Titration Curves (b) pCa2+ value at the equivalence point (20.0mL): For Ca2+ + Y4-  CaY2- KCaY2- = [CaY2-]/[Ca2+][[Y4-] In this region, virtually all the metal is in the form of CaY2- [CaY2-] = added EDTA [Ca2+] = dissociated chelate [CaY2-] = ((20.0mL x 0.050M)/(70.0mL))  0.0142M K(CaY )’ = [CaY2-] / [Ca2+] [CT] = (0.0142)/[Ca2+]2 2- [Ca2+] = ((0.0142)/(1.75 x 1010))1/2 = 9.0 x 10-7M; pCa2+ = 6.05 at 20.0mL EDTA  Complex-Formation Titrations Metal-EDTA Titration Curves (c) pCa2+ value after the equivalence point (25.0mL): CT = [excess EDTA] [excess EDTA] = ((25.0 x 0.050)-(50.0 x 0.020))/(75.0) = 0.0033M [CaY2-] = (50.0mL x 0.020M)/(75.0mL)  0.0133M K(CaY )’ = [CaY2-] / [Ca2+] [excess EDTA]; 2- [Ca2+] = (0.0133)/(0.0033)(K(CaY )’) 2- [Ca2+] = 2.30 x 10-10 pCa2+ = 9.64 at 25.0mL EDTA  As the pH increases, the equilibrium shifts to the right. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) Fig. 9.3. Titration curves for 100 mL 0.1 M Ca2+ versus 0.1 M Na2EDTA at pH 7 and 10. Figure 17-7 Influence of pH on the titration of 0.0100 M Ca2+ with 0.0100 M EDTA. Note that the end point becomes less sharp as the pH decreases because the complex formation reaction is less complete under these circumstances.  Metal Ion Indicators To detect the end point of EDTA titrations, we usually use a metal ion indicator or an ion-selective electrode Metal ion indicators change color when the metal ion is bound to EDTA: MgEbT + EDTA  MgEDTA + EbT (Red) (Clear) (Clear) (Blue) – Eriochrome black T (EbT) is an organic ion H2In- The indicator must bind less strongly with the metal ion than EDTA  EDTA Titrations Metal Ion Indicators Titration of Mg2+ by EDTA Eriochrome Black T (EbT) Indicator Addition of EDTA Before Near After Equivalence point Metal Ion Indicator Compounds  EDTA Titration Techniques Direct titration: analyte (metal) is titrated with standard EDTA with solution buffered at a pH where Kf’ is large Back titration: known excess of EDTA is added to analyte metal. Excess EDTA is titrated with 2nd metal ion.  Chapter 10 Precipitation Titration ILO’s 1- explain the foundation of argentometric methods that are based on AgNO3 reagent 2- determine drugs containing halides or pseudohalides by the argentometric method 3- construct precipitation titration curves 4- describe Mohr, Volhard, and Fajan’s methods of titrations  General Principles Involves formation of a precipitate Must determine the volume of standardized titrant in ml’s (AgNO3) to just precipitate all of the ions. Need an indicator or electrode to determine when the precipitation is complete  Argentometric Methods Titrations based upon silver nitrate (AgNO3) Most useful precipitation titration because of the low solubility product of halide salts or Pseudohalides (S2, HS, CN, SCN) Ksp AgCl 1.82 × 10-10 Ksp AgSCN 1.1 ×10-12 Ksp AgBr 5.0 × 10-13 Ksp AgCN 2.2 ×10-16 Ksp AgI 8.3 × 10-17  Precipitation Titration Curves involving silver ion The most common method of determining the halide ion concentration of aqueous solution is titration with a standard solution of silver nitrate. To construct titration curve 4 type of calculation are required, each of which corresponds to a distinct stage in the reaction： (1) Prior Ag+ addition （2）pre-equivalence, （3）equivalence, and （4）post-equivalence. Precipitation titration curve for 50.0 mL of 0.0500 M Cl– with 0.100 M Ag+. (a) pCl versus volume of titrant; (b) pAg versus volume of titrant.  The effect of Concentration on Titration curves Titration curve for A 50.00mL of 0.0500M NaCl with 0.100M AgNO3, and B 50.00mL of 0.00500M NaCl with 0.0100M AgNO3.  The effect of Reaction Completeness on Titration Curves Figure: Effect of reaction completeness on precipitation curves. For each curve , 50.00 mL of a 0.0500M solution of the anion was titrated with 0.1000 M AgNO3. Note that smaller values of Ksp give much sharper breaks at the end point Argentometric Methods Mohr Volhard Fajans  Mohr Method Direct titration Basis of endpoint: formation of a colored secondary precipitate Ag2CrO4(s) Indicator: soluble chromate salt (Na2CrO4)  Mohr Method Has to be performed at a neutral or weak basic solution of pH 7-9 In a lower pH (acid solution) CrO42-(aq) + H+(aq)  H2CrO4 H2CrO4 ↔ 2H+(aq) + CrO42-(aq) Part of the indicator is present as H2CrO4 and more Ag+ is needed to form Ag2CrO4 In a higher pH (basic solution) > 10 Ag+(aq) + OH-(aq)  AgOH(s) silver hydroxide may precipitated. Mohr Method for Cl- determination Relies on Ksp differences for two insoluble silver salts. Ag+(aq) + Cl-(aq)  AgCl(s) (titration rxn) 2Ag+(aq) + CrO42-(aq)  Ag2CrO4(s) (indicator rxn) AgCl is less soluble than Ag2CrO4 so it will precipitate first Ag2CrO4 is brick red in color so a color change is observed at the endpoint  Volhard Method Indirect method used as a procedure for titrating Ag+; determination of Cl- requires a back-titration of the excess Ag+ – First, Cl- is precipitated by excess AgNO3 Ag+ (aq) + Cl-(aq)  AgCl(s) – Excess Ag+ is titrated with KSCN in the presence of Fe3+ indicator Ag+(aq) + SCN-(aq)  AgSCN(s) – When Ag+ has been consumed, a red complex forms Fe3+(aq) + SCN-(aq)  FeSCN2+(aq)  Fajans Titration Uses adsorption indicator (In-) -O O O Cl Cl CO 2- Dichlorofluorescein is green in solution but pink when absorbed on AgCl  Fajans Titration Prior to the equivalence point, there is excess Cl- in solution. Some is adsorbed on the surface of the crystal, giving a partial negative charge. Repulsion with the –ve charge of the In- After the equivalence point, there is excess Ag+ in solution. Some adsorbs to the surface imparting a partial positive In- charge to the precipitate. In- Attract In- Fajans Titration  Chapter 11 Gravimetric Analysis ILO’s 1-analyze precipitates using gravimetric analysis is considered one of the most accurate and precise macro-quantitative analysis methods 2-explain calculations using Gravimetric Factor (GF) and calculating the % of substance sought  Gravimetric Analysis 1. A weighed sample is dissolved 2. An excess of a precipitating agent is added to this solution 3. The resulting precipitate is filtered, dried (or ignited) and weighed 4. From the mass and known composition of the precipitate, the amount of the original ion can be determined 3  Gravimetric Analysis Gravimetric Analysis – one of the most accurate and precise methods of macro- quantitative analysis. Analyte selectively converted to an insoluble form. Measurement of mass of material Correlate with chemical composition  Gravimetric Factor (GF): represents the weight of analyte per unit weight of precipitate GF unknown i.e., GF x  Gravimetric Calculations summary Gravimetric Factor (GF): GF = (fwt analyte (g/mol)/fwt precipitate(g/mol))x(a(moles analyte/b(moles precipitate)) GF = g analyte/g precipitate % analyte = (weight analyte (g)/ weight sample (g)) x 100% % (w/w) analyte (g) = ((wt ppt (g) x GF)/wt sample) x 100%