Lecture 7 -- Chemical Equilibrium

Acids and Bases: Core Concepts ReviewChapter 151Textbook Chapters to Cover15.2, 15.3: Definitions of acids & bases15.4: Acid strength & Ka (next week!)15.5: Base solutions15.6, 15.7: Autoionization of water, [H3O]+, [OH]-, pH, pOH, pKw15.8: Acidic/basic salts (next week!)Omit the following sections:15.1 (intro), 15.9 (polyprotic acids), 15.10 (Lewis), 15.11 (molecular structure of acids/bases), 15.12 (acid rain)

2AgendaAcid-base definitionsStructure and functionStrengths of acids and basesAutoionisationpH, pOH, pKw

3Acid & Base Definitions4Acid-Base Definitions5

ArrheniusAcids increase the concentration of H3O+ in watere.g. HClO4 (aq) + H2O () H3O+ (aq) + ClO4- (aq)(Note: H3O+ is equivalent to H+)

Bases increase the concentration of OH-e.g. NaOH (s) Na+ (aq) + OH- (aq)

Arrhenius definition confines us to water.6Brnsted-LowryAcids transfer a proton (H+) to another substance

7ACIDBASECONJUGATEBASECONJUGATEACID

Brnsted-LowryBases accept a proton from another substance

Water is amphoteric--acts as an acid or a base

8ACIDBASECONJUGATEACIDCONJUGATEBASE

LC: Acid-Base PairsWhich of the following is the correct conjugate base for the acid HPO42-?

H3PO4H2PO4-PO43-OH-9LC: Acid-Base PairsWhich of the following is the correct conjugate base for the acid HPO42-?

PO43-

HPO42- + H2O PO43- + H3O+

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Structure/Function; Strength of Acids & Bases11Structure and FunctionAcids: H bound to an electronegative atom or groupe.g. H-Cl, H-ONO2, H-F, H-NH3+, H-OC(O)CH3electronegative atom polarizes bondeasy to break by solvationnote: H bound to C are not acidic

Bases have lone pair(s) that can accept protone.g. H3N: :OH- CO32- PO43- 12Acid StrengthSubstances ability to donate a protonStrong acid is completely ionized in waterno molecular acid exists in waterconjugate base has little tendency to accept protonse.g. HCl (g) + H2O (l) H3O+ (aq) + Cl- (aq)Weak acid is only partially ionized in waterconjugate base accepts protons readilye.g. H2S(g) + H2O(l) H3O+(aq) + HS-(aq)13

Base StrengthSubstances ability to accept H+ or form OH-Strong base reacts 100% with water to form OH-e.g. NaOH, KOH, Ca(OH)2 (dilute)Weak bases partially react with water to form OH-Portion of the base remains unreacted (molecular)e.g. NH3(aq) + H2O(aq) NH4+(aq) + OH-(aq)14

15Equilibrium positionEquilibrium favours proton transfer from stronger acid to the stronger base. Where will the equilibrium lie for the following?e.g. HSO4-(aq) + NH3(aq) NH4+(aq) + SO42-(aq)HSO4- is stronger than NH4+, so equilibrium favours productse.g.2 HS-(aq) + H2O(l) H2S(aq) + OH-(aq)H2S is stronger than H2O, so equilibrium favours reactants

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Autoionization of water, [H3O]+, [OH]-, pH, pOH, pKw

17Autoionization of WaterWater autoionizes to produce H3O+ and OH-2 H2O(l) H3O+(aq) + OH-(aq)orH2O(l) H+(aq) + OH-(aq)Note: hydrogen ion (H+) = hydronium ion (H3O+)Equilibrium constant (ion product): Kw = [H+][OH-] = [H3O+][OH-] = 1.010-14 at 25CKw valid for solutions as well as for pure water18

[OH-] and [H+] from Kw e.g. Find [OH-] & [H+] in pure water at 25C.H2O produces both H+ and OH-In pure water, [H+] = [OH-] = x Kw = [H+][OH-] = x2 = 1.010-14

and so, x = (1.010-14)1/2 = 1.010-7

[H+] = [OH-] = 1.010-7 molL-1

19LC: [OH-] and [H+] from Kw Find [OH-] & [H+] in pure water at 10C (Kw = 2.9310-15).[H+] = [OH-] = 1.0010-7 molL-1 [H+] = 5.4110-8 molL-1; [OH-] = 1.8510-7 molL-1[H+] = [OH-] = 5.4110-8 M[H+] = 1.8510-7 molL-1; [OH-] = 5.4110-8 molL-120LC: [OH-] and [H+] from Kw Find [OH-] & [H+] in pure water at 10C (Kw = 2.9310-15).H2O produces both H+ and OH-In pure water, [H+] = [OH-] = x Kw = [H+][OH-] = x2 = 2.9310-15and so, x = (2.9310-15)1/2 = 5.4110-8

[H+] = [OH-] = 5.4110-8 molL-1

Note: pH = 7.27 indicates neutral water at 10C21[H+] and [OH-] in 0.025 M HClHCl completely dissociates in waterTherefore, [H+] = 0.025 molL-1 H+ from water autoionization is negligibleFind [OH-] from Kw = [OH-][H+] = 1.010-14[OH-] = Kw/[H+] = (1.010-14)/(0.025) = 4.010-13 molL-1[OH-] is about 1/100 000 000 000th [H+]22LC: [OH-] in acidWhat is [OH-] in 10 mL of 1.5 molL-1 HCl at 25C?

1.5 molL-11.010-7 molL-16.710-13 molL-16.710-15 molL-1 23LC: [OH-] in acidWhat is the [OH-] in a 10 mL sample of 1.5 molL-1 HCl at a temperature of 25C?D. 6.710-15 M

[OH-] = Kw/[H+] = (1.010-14)/(1.5) = 6.710-15 molL-124The pH Scale[H+] expressed on a log scale:pH = -log([H+]) {or [H+] = 10-pH}pH of pure water at 298 K is pH = -log(1.010-7) = 7.00The pH of 0.025 M HCl ispH = -log(0.025) = 1.60

Reminder: Report as many decimal places in pH as significant figs in the argument of the log25pH at 298 K26

ExampleThe pH of Coca-Cola is about 3.0, while the pH of milk is about 6.4. By what factor does the [H+] of Coke exceed that of milk?

Answer has 1 sig. fig. because pH given to 1 decimal place

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pOH and pKwpOH = -log([OH-])

Because Kw = [OH-][H+],-log(Kw) = -log[OH-] - log[H+]pKw = pOH + pH

at 298 K, Kw=1.010-14; therefore at 298KpOH + pH = 14.00

28ExampleWhat is pOH and [OH-] of a pH 3.57 solution at 25 C?

pOH = 14.00 - pH = 14.00 - 3.57 = 10.43[OH-] = 10-pOH = 10-10.43 = 3.710-11 molL-1Note: 2 significant figures in [OH-]29LC: pOHAt 50C, Kw=5.4810-14. What is the pOH of a 0.50 molL-1 HBr solution at 50C?

0.3012.9613.5013.70

30LC: pOHWhat is the pOH of a 0.50 M HBr solution at 50 C?B. 12.96pKw = -log(Kw) = -log(5.4810-14) = 13.262pH = -log(0.50) = 0.30pOH = pKw pH = 13.262 - 0.30 = 12.9631SummaryGo home and review:pH, pOHKw Strong/weak acids & basesLots more calculations to come next week! Hooray!32Questions to solveReview questions 4 24. Problems by topic 35, 37, 39, 41, 43, 51, 53, 55, 57, 61, 65, 67, 69, 73, 77, 79, 83, 87, 91, 93, 95, 99, 101, 159.

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