Introduction to Acids and Bases Chapter 14 Sections 1-5, 7, 11 AP Chemistry Dr. Knorr

Arrhenius Acids and Bases Acids produce H+ in aqueous solutions water HCl H+(aq) + Cl- (aq) Bases produce OH- in aqueous solutions NaOH Na+(aq) + OH- (aq)

Brønsted-Lowry Acids and Bases Acids donate H+ in aqueous solutions water HCl H+(aq) + Cl- (aq) Bases accept H+ in aqueous solutions NH3 + H2O NH4+(aq) + OH- (aq) All Arrhenius A/B are B-L A/B

Lewis Acids and Bases Acids accept electron pairs in aqueous solutions Bases donate electron pairs in aqueous solutions H F H F H N + B F H N B F

Conjugate Acids and Bases A conjugate acid-base pair is related by loss/gain of one hydrogen ion. H2O + HA H3O+(aq) + A- (aq) gain H+ Ka = [H+][A-] [HA] base acid conjugate acid conjugate base lose H+

Acid Strength H2O + HA H3O+(aq) + A- (aq) B A CA CB Strength is related to position at equilibrium Strong acids have weak CBs and large Ka Weak acids have strong CBs and small Ka

Strong Acids Strong Acids (the big 7): HCl, HBr, HI, HNO3, H2SO4, HClO4, HClO3 Example: HCl(aq) H+ (aq) + Cl- (aq) **Strong acids completely ionize, so equilibrium constants >>1 and cannot be measured accurately Ka = [H+][Cl-] >>1 [HCl]

Water! Autoionization: H2O H+ + OH- Keq = Kw = [H+] x [OH-] = 1 x 10-14 Therefore [H+] = [OH-] = 1 x 10-7 Only @25°C

pH [H+]>[OH-] [H+] = [OH-] [OH-]>[H+] pH = - log [H+] [H+] pH State 1 x 10-5 M 5 Acidic 1 x 10-11 M 11 Basic 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 [H+]>[OH-] [H+] = [OH-] [OH-]>[H+] Neutral Acidic Basic

pOH Indicates the basicity (aka alkalinity) [OH-] of the solution pOH = - log [OH-] Another important relationship: pH + pOH = 14 [OH-] pOH pH State 1 x 10-5 M 5 9 Basic 1 x 10-11 M 11 3 Acidic

pH of Strong Acids Example: what is pH of 0.01M HNO3? HNO3H+ + NO3- What are the major species in solution? Major: H+, NO3-, and H2O Minor: HNO3, OH- pH = -log[H+ ] = -log(0.01) = 2.0

pH of Weak Acids Example: What is pH of a 1.0M sol’n of HOCl if Ka = 3.5 x 10-8? What are the species in solution? HOCl(aq) H+ (aq) + OCl-(aq) H2O(l) H+ (aq) + OH-(aq) Major species: HOCl & H2O

pH of Weak Acids (cont’d) HOCl(aq) H+ (aq) + OCl-(aq) Set up an ICE table! HOCl H+ OCl- I 1.00 0 0 E 1.00-x x x x = 1.9x10-4M = [H+ ], so pH = -log(1.9x10-4) = 3.72 Remember Kw = 10-14, so most H+ comes from HOCl Ka = [H+][OCl-] [HOCl] = __x2__ = 3.5 x 10-8 (1-x)

pH of Weak Acid Mixtures Example: What is pH of a sol’n of 0.10M HCN (Ka = 6.2 x 10-10) and 0.50M HNO2 (Ka = 4.0 x 10-4) What are the major species in solution? HCN(aq) H+ (aq) + CN-(aq) Ka = 6.2x10-10 HNO2(aq) H+ (aq) + NO2-(aq) Ka = 4.0x10-4 H2O(l) H+ (aq) + OH-(aq) Kw = 1.0x10-14 Major species: HCN, HNO2 & H2O HNO2 contributes the most H+ because of largest Ka

pH of Weak Acid Mixtures (cont’d) HNO2(aq) H+ (aq) + NO2-(aq) Set up an ICE table! HNO2 H+ NO2- I 0.5 0 0 E 0.5-x x x x = 0.014M = [H+ ], so pH = -log(0.014) = 1.85 Ka = [H+][NO2-] [HNO2] = x2___ = 4.0 x 10-4 (0.5-x)

Weak Acid Mixtures (cont’d) HCN(aq) H+ (aq) + CN-(aq) What is [CN-] in the same mixture? This is a small amount of HCN that dissociates when you consider [HCN]o Ka = [H+][CN-] [HCN] 6.2x10-10 = [0.014][CN-] … [CN-] = 4.4x10-9 [0.10]

Percent Dissociation (PD) PD = 100 x amount dissociated (M) initial concentration (M) For our previous mixture, HCNPD = 100 x 4.4x10-9 / 0.100 = 4.4x10-6 % HNO2PD = 100 x 0.014 / 0.500 = 2.8 %

PD Example In a 0.100M aqueous solution of lactic acid, 3.7% is dissociated. Calculate Ka. HC3H5O3 H+ C3H5O3- 0.1 0 0 0.1-x x x Ka = x2/(0.100-x) Ka = (3.7x10-3)2/(0.100-3.7x10-3) Ka = 1.4x10-4 PD = 3.7 = 100 * x/0.1 x = 3.7 x 10-3M

Polyprotic Acids Supply more than one proton, one at a time Conjugate base of 1st reaction becomes the acid in the 2nd reaction Usually Ka1 > Ka2 > Ka3, meaning loss of 2nd or 3rd proton is not as easy as 1st Examples: H3PO4; H2SO3

Polyprotic Acids (cont’d) Example: calculate the pH of a 5.0M H2SO3 solution and the concentration of all S species at equilibrium Ka1 = 0.015 and Ka2 = 1.0 x 10-7 The dominant equilibrium is H2SO3 H+ + HSO3- I 5.0 0 0 E 5.0-x x x Ka1 = [H+][HSO3-] [H2SO3] 0.015 = (x)(x) (5.0-x) x = 0.266 = [H+] =[HSO3-] pH = -log[H+] = -log (0.266) = 0.58

Polyprotic Acids (cont’d) If x = 0.266, [H2SO3] = 5-x, or ~4.7M [SO3-2] can be obtained for the expression for Ka2: Ka2 = [H+][SO3-2] [HSO3-] HSO3- H+ + SO3-2 Dissociation of HSO3- to form SO3-2 only negligibly affects [H+] and [HSO3-] because Ka2 is so small. 1.0 x 10-7 = [0.266][SO3-2] [0.266] [SO3-2] = 1.0 x 10-7M

Sulfuric Acid Ka1 is large (“strong”), Ka2 is small (“weak”) ~0.012 Major species are H+, HSO4-, and H2O Calculate the pH of a 2.0M H2SO4 solution and all ion concentrations at equilibrium After 1st dissociation, H+ ~ 2.0M, HSO4- ~ 2.0M HSO4-  H+ + SO4-2 2.0 2.0 0 2.0-x 2.0+x x Ka2 = [H+][SO4-2] [HSO4-] 0.012= (2+x)(x) … x = 0.012 (2-x) [H+] ~2.0M, so pH = -0.3 . H2SO4 ~0, [HSO4-] ~2.0M, [SO4-2] = 0.012M