Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Unit 10 (Chp 17): Buffers & Acid-Base Titrations (more Ka, Kb, Kw, pH, pOH) John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc.

Buffers: H+ + HCO3– → H2CO3 3 HCl 1 free H+ weak conjugate acid-base pair. resist pH changes by reacting with added acid/base. 3 3 HCl How many H+’s added? How many H+’s remain? 1 H+ + HCO3– → H2CO3 Cl– H2CO3 HCO3– HCO3– H+ Na+ H2CO3 Na+ H2CO3  HCO3– buffer solution 1 free H+

Buffers: H2CO3 H+ + HCO3– H2CO3 H+ HCO3–

Buffers: animation Animation: http://mhhe.com/physsci/chemistry/essentialchemistry/flash/buffer12.swf

HW p. 762 #17 Buffers: If acid is added, the F− reacts to form HF and water.

Common-Ion Effect Consider a solution of acetic acid: If NaC2H3O2 (acetate ion) is added, the equilibrium will shift to the left. (Le Châtelier) HC2H3O2(aq)  H+(aq) + C2H3O2−(aq) “The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has a common ion with the weak electrolyte.” OR adding common ion shifts left (less ionized)

Common-Ion Effect First a review of what you can do already: Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF. Ka for HF is 6.8  10−4 [H+] [F−] [HF] Ka = = 6.8  10–4

Common-Ion Effect Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF. HF(aq)  H+(aq) + F−(aq) [HF], M [H+], M [F−], M Initial 0.20 Change −x +x Equilibrium 0.20 − x  0.20 x x2 (0.20) 6.8  10−4 = [H+] = 0.012 M pH = 1.92 x = 0.012

Common-Ion Effect NOW, calculate the [F–] and pH of 0.20 M HF and 0.10 M NaF. NaF (strong electrolyte) so [F–]init ≠ 0. HF(aq)  H+(aq) + F−(aq) [HF], M [H+], M [F−], M Initial 0.20 0.10 Change −x +x Equilibrium 0.20 − x  0.20 x 0.10 + x  0.10 (x)(0.10) (0.20) 6.8  10−4 = [H+] = 0.0014 M pH = 2.85 x = 0.0014

Common-Ion Effect “The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has a common ion with the weak electrolyte.” NaF HF(aq)  H+(aq) + F−(aq) Previously 0.20 M HF: [H+] = 0.012 M pH = 1.92 With NaF (common ion F–) 0.20 M HF and 0.10 M NaF: [H+] = 0.0014 M pH = 2.85 HW p. 761 #11

pH of Buffers HW p. 762 #19a What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.11 M in sodium lactate (NaC3H5O3)? Ka for lactic acid is 1.4  10−4 HC3H5O3 + H2O  H3O+ + C3H5O3– [H3O+] [C3H5O3−] [HC3H5O3] Ka = ICE gives: (x)(0.11) (0.12) 1.4  10−4 = pH = 3.82 x = 1.5  10−4 M H3O+ pH = –log(1.5  10−4)

Buffer Capacity & pH Range buffer capacity: the amount of acid or base neutralized before significant pH changes. pH range: the range of pH values over which a buffer works effectively. Best Capacity and Range: - weak acid with a pKa near desired pH - [HA] = [A–] (so that pKa = pH) if Ka = [H+][A–] [HA] , then pKa = pH[A–] [HA]

When Strong Acids or Bases Are Added to a Buffer… …it is safe to assume that all of the strong acid or base is consumed in the reaction.

Adding Strong Acid or Base to Buffer Add OH– consumes [HA] and produces [A–]. Add H+ consumes [A–] and produces [HA]. Use M’s in K exp. to find new [H+] & pH. add 0.020 mol OH– add 0.020 mol H+ HW p. 762 #21, 24, 26

Titration (video clip) The analytical technique used to calculate the moles in an unknown soln. mass % (g /gtotal) molarity (mol/L) molar mass (g/mol) buret titrant known vol. (V) known conc. (M) Safari Montage (2 min) Titration analyte known vol. (V) unknown conc. (M) (or moles)

equivalence point,(Veq): end point: indicator color change persists equivalence point,(Veq): equal stoichiometric amounts react completely HBr + NaOH  mol Acid = mol Base 2 HA + Ca(OH)2  H2SO4 + 2 KOH 

Titration Standard: solution of known conc. (M) Vanalyte (known) animation Standard: solution of known conc. (M) Vanalyte (known) Vtitrant (known) Mtitrant (known) Animation (2 min): http://chem-ilp.net/labTechniques/TitrationAnimation.htm Manalyte (unknown) 17

Indicators: Pink in BASE Phenolphthalein Colorless in ACID

Add H3O+ (titrate w/ acid) Indicators: weak acids with conj. bases of diff. color Example: Phenolphthalein HIn + H2O   H3O+  + In– Add H3O+ (titrate w/ acid) HIn + H2O  H3O+  + In– Demo: Magic Pitcher colorless Remove H3O+ (titrate w/ base) HIn + H2O   H3O+  + In– pink

From start to near the equivalence point (Veq), the pH goes up very slowly. SA with SB p. 733

Just before and after the Veq, the pH increases rapidly (steep jump). SA with SB

SA with SB 7 At Veq, pH = ? At Veq, moles acid = moles base, the solution contains only water and salt. SA with SB At Veq, pH = ? 7

SA with SB After Veq, pH levels off.

WA with SB p. 736 At Veq, pH > 7 Conjugate base of acid raises pH of H2O by… p. 736 Animation: Titration A– + H2O ↔ HA + OH– Animation: (http://www.chembio.uoguelph.ca/educmat/chm19104/chemtoons/chemtoons9.htm)

Indicators & pH Ranges animation Choose indicator that changes color (has pKa) near pH of equivalence point (Veq) of titration.

(stoich. calc. with mol-to-mol) WA with SB Before Veq, the pH is determined from the amounts of the acid and its conjugate base present at that particular time. (stoich. calc. with mol-to-mol)

WA with SB Weak acids: higher initial pH gradual pH rise (not flat then jump) subtle pH change at equiv. point (less steep) Weak bases: (same but drop not rise)

WB with SA At Veq, pH < 7 conjugate acid of base lowers pH of H2O by… HA + H2O ↔ H3O+ + A–

In a solution of 0.10 M H3PO4, which species is in the least amount? Polyprotic Acids In a solution of 0.10 M H3PO4, which species is in the least amount? There’s a Ka and pKa for each dissociation, with a distinct Veq. H3PO4 H2PO4– HPO42– PO43–

Calculating pH for 4 parts of Titration Curve: pH BEFORE titration has begun: pH = –log [H+] for SA b/c [HA] = [H+] 1 STRONG A with Strong B pH DURING titration: Titrn Rxn: H+ + OH–  H2O + conj. RICE in moles, ÷L’s total = [H+] left pH = –log [H+] left 2 4 3 1 2 pH @ EQUIVALENCE (Veq): no work (pH = 7.00, only H2O , Kw) 3 pH AFTER equivalence (Veq): Titrn Rxn: H+ + OH–  H2O + conj. RICE in moles, ÷L’s total = [OH–] excess pOH = –log [OH–] excess 4 What’s in the flask? (write a rxn)

Calculating pH for 4 parts of Titration Curve: pH BEFORE titration has begun: Equil Rxn: HA + H2O ⇄ H3O+ + A– RICE in M’s & Ka w/ x2 for [H+] pH = –log [H+] 1 WEAK A with Strong B 2 pH DURING titration: (buffer region) Titrn Rxn: HA + OH–  H2O + A– RICE in moles, ÷L’s total = [HA],[A–] Equil Rxn: HA + H2O ⇄ H3O+ + A– RICE in M’s & Ka w/ x (not x2) for [H+] pH = –log [H+] 2 1 (buffer) What’s in the flask? (write a rxn) @ ½Veq: pH = pKa Ka = [H+][A–] [HA] b/c [HA]=[A–] @ ½Veq and on next slide 3 4

Calculating pH for 4 parts of Titration Curve: and continued 3 4 WEAK A with Strong B pH @ EQUIVALENCE (Veq): Titrn Rxn: HA + OH–  H2O + A– RICE in moles, ÷L’s total = [A–] only Equil Rxn: A– + H2O ⇄ HA + OH– RICE in M’s & Kb w/ x2 for [OH–] pOH = –log [OH–] 3 4 3 2 1 (buffer) What’s in the flask? (write a rxn) pH AFTER equivalence (Veq): Titrn Rxn: HA + OH–  H2O + A– RICE in moles, ÷L’s total = [OH–] excess pOH = –log [OH–] excess ( [A–] produces negligible [OH–] ) 4