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 hydrogen 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)

(stoichiometrically =) end point: indicator color change persists equivalence point,(Veq): equal stoichiometric amounts react completely HBr + NaOH  mol titrant (stoichiometrically =) mol analyte 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

p. 733 SA with SB very low initial pH level pH rise

SA with SB rapid, steep jump in pH at Veq

SA with SB 7 at Veq, pH = ___ At Veq… moles added moles reacted (stoichiometrically =) moles reacted …the solution contains only water and salt. (irrelevant conjugate) SA with SB at Veq, pH = ___ 7

SA with SB after Veq, very high pH levels.

Titration Calculations Calculate unknown moles of the ANALYTE. (then calculate: M OR molar mass OR mass %) Stoich: L X  mol X  mol Y = mol Y L X mol X Calculate unknown VOLUME of TITRANT (mL) added to reach equivalence (all reacted). Stoich: L X  mol X  mol Y  L Y = L Y L X mol X mol Y Calculate unknown pH at any point in the titration (especially at equivalence). Stoich: What’s in the sol’n? Find excess H+, OH– HW p. 763 #38

Calculating pH for 4 parts of Titration Curve: pH BEFORE titration has begun: pH = –log [H+] for SA b/c [HA] = [H+] 1 HW p. 763 #40 STRONG with Strong pH DURING titration: MRP (More RICE Please!) Moles: L x M = mol H+ & mol OH– RICE: H+ + OH–  H2O + conj. mol÷L’s total = [H+] or [OH–] pH: –log [H+] or –log [OH–] 2 4 3 1 2 pH @ EQUIVALENCE (Veq): pH = 7.00 (only H2O & salt, Kw) 3 What’s in the flask? (write a rxn) pH AFTER equivalence (Veq): MRP (More RICE Please!) 4

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

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)

WB with SA HW p. 763 #33 At Veq, pH < 7 conjugate acid of base lowers pH of H2O by… HA + H2O ↔ H3O+ + A–

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)

In a solution of 0.10 M H3PO4, which species is in the least amount? HW p. 763 #36 Polyprotic Acids There’s a Ka and pKa for each dissociation, with a distinct Veq. In a solution of 0.10 M H3PO4, which species is in the least amount? 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 HW p. 763 #40 STRONG with Strong pH DURING titration: MRP (More RICE Please!) Moles: L x M = mol H+ & mol OH– RICE: H+ + OH–  H2O + conj. mol÷L’s total = [H+] or [OH–] pH: –log [H+] or –log [OH–] 2 4 3 1 2 pH @ EQUIVALENCE (Veq): pH = 7.00 (only H2O & salt, Kw) 3 What’s in the flask? (write a rxn) pH AFTER equivalence (Veq): MRP (More RICE Please!) 4

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 pH DURING titration: (buffer region) MREP (More RICE Equil. Please!) Moles: L x M = mol H+ & mol OH– RICE: H+ + OH–  H2O + conj. mol÷L’s total = [H+] or [OH–] Equil Rxn: HA + H2O ⇄ H3O+ + A– RICE in M’s & Ka w/ x (not x2) for [H+] pH = –log [H+] 2 What’s in the flask? (write a rxn) 2 1 (buffer) @ ½Veq: pH = pKa Ka = [H+][A–] [HA] b/c [HA]=[A–] @ ½Veq

Calculating pH for 4 parts of Titration Curve: and continued 3 4 HW p. 763 #42 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