Chapter 7 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)