Unit 10 SUMMARY: Buffers & Acid-Base Titrations weak conjugate acid-base pair. resist pH changes (reacts added acid/base) Buffer Capacity: more moles (L & M) of buffer react more acid/base before significant pH changes. pH range: buffer works near it’s pKa Optimum Capacity and Range: - weak acid with a pKa near desired pH - [HA] = [A–] (so pKa = pH) Ka = [H+][A–] [HA]

Common Ion Effect: adding common ion shifts left causing the weak acid/base to be less ionized. pH of Buffers: 0.12 M lactic acid, HC3H5O3 (Ka = 1.4  10−4) 0.11 M sodium lactate, NaC3H5O3 HC3H5O3 + H2O  H3O+ + C3H5O3– [H3O+] [C3H5O3−] [HC3H5O3] Ka = I 0.12 M 0 M 0.11 M C –x + x +x E 0.12 – x x 0.11 + x ≈ 0.12 M ≈ 0.11 M (x)(0.11) (0.12) 1.4  10−4 = x = 1.5  10−4 M H3O+ pH = –log(1.5  10−4) = 3.82

Titration The analytical technique used to calculate the moles in an unknown soln. mass % (g /gtotal) molarity (mol/L) molar mass (g/mol) buret titrant end point: indicator color change known vol. (V) known conc. (M) equivalence point: equal stoich. amounts (mols) react completely Safari Montage (2 min) Titration analyte known vol. (V) unknown conc. (M) (or moles)

SA with SB At Veq, pH = 7 At Veq… moles added moles reacted (stoichiometrically =) moles reacted …the solution contains only water & salt. (irrelevant conjugate) SA with SB At Veq, pH = 7 H2O + H2O ↔ H3O+ + OH–

WA with SB At Veq, pH > 7 (only water & conj. base) A– + H2O ↔ HA + OH– Animation: (http://www.chembio.uoguelph.ca/educmat/chm19104/chemtoons/chemtoons9.htm)

WB with SA At Veq, pH < 7 (only water & conj. acid) HA + H2O ↔ H3O+ + A– Indicators: weak acids with diff color conjugates choose one that changes color (has pKa) near pH of equivalence point (Veq) of titration.

WA with SB Verbally describe the visual curve: 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”)

Titration Calculations Calculate unknown moles of the ANALYTE. (then calculate: M OR molar mass OR mass %) Stoich: L T  mol T  mol A = mol A L T mol T Calculate unknown VOLUME of TITRANT (mL) added to reach equivalence (all reacted). Stoich: L A  mol A  mol T  L T = L T L A mol A mol T Calculate unknown pH at any point in the titration (especially at equivalence). Stoich: What’s in the sol’n? Find M of H+ or OH–

Calculating pH for 4 parts of Titration Curve: pH BEFORE titration has begun: pH = –log [H+] for SA b/c [HA] = [H+] 1 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+] 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: M pH: –log [H+] 1 WEAK A with Strong B 0 M 0 M –x +x +x K = x2 [HA] x x 2 1 pH DURING titration: (buffer region) MREP (More RICE Eq. Please!) Moles: L x M = mol HA & mol OH– RICE: HA + OH–  H2O + A– mol÷L’s total = [HA] & [A–] Equil Rxn: HA + H2O ⇄ H3O+ + A– RICE in M’s & Ka w/ x for [H+] pH: –log [H+] (buffer) 2 What’s in the flask? (write a rxn) Ka = [H+][A–] [HA] @ ½Veq: pH = pKa b/c [HA]=[A–] @ ½Veq K = x[A–] [HA]

Calculating pH for 4 parts of Titration Curve: WEAK A with Strong B and continued 3 4 pH @ EQUIVALENCE (Veq): MREP (More RICE Eq. Please!) Moles: L x M = mol HA & mol OH– RICE: HA + OH–  H2O + A– mol÷L’s total = ONLY [A–] Equil Rxn: A– + H2O ⇄ HA + OH– RICE in M’s & Kb w/ x2 for [OH–] pH: pOH = –log [OH–] & 14 – pOH = 3 4 3 2 1 (buffer) Kb = x2 [A–] What’s in the flask? (write a rxn) pH AFTER equivalence (Veq): MRP (M & R give [OH–] excess) pOH = –log [OH–] & 14 – pOH = 4 MREP (NO Eq) b/c [A–] produces negligible [OH–]