Chapter 9 Aqueous Solutions and Chemical Equilibria 1 CaCO322 (s ) CO (gl ) + H O ( ) (limestone) (water falls) 2- Ca (aq ) 2HCO3 ( aq ) 2 Classifying Solutions of Electrolytes Electrolytes—solutes form ions when dissolved in water (or certain other solvents, e.g. acetonitrile) Strong (weak, non-) electrolytes -+ HNO32 + H O NO 3 + H 3 O (completely ionized) -+ CH3233 COOH+ H O CH COO + H O (partially ionized) sugar (s) + water = sugar aquesous soln (completely non-ionized) 3 1 HF 2+ - - Ca(OH)2 Ca +2OH , no Ca(OH) in solution HF is a weak electrolyte due to hydrogen bond. 4 Acids and Bases An acid donates protons, a base accepts protons An acid donates protons only in the presence of a proton acceptor (a base). A base accepts protons only in the presence of a proton donor (an acid) 5 Bronsted-Lowry Acids and Bases These two chemists pointed out that acids and bases can be seen as proton transfer reactions. According to the Bronsted-Lowry concept: • An acid donates a proton and the base accepts it. • Newly formed species are called conjugate base and conjugate acid, respectively. 6 2 acid + base conjugate base + conjugate acid - HCl + NH34 Cl + NH Acid12 + base base 1 + acid 2 A conjugate base is formed when an acid loses a proton (HClCl-); A conjugate acid is formed + when a base accepts a proton (NH3NH4 ) 7 Consider the reaction of NH3 with H2O: + - NH3 (aq) + H2O (aq) NH4 (aq) + OH (aq) (base1) (acid2) (acid1) (base2) Note: + NH3 & NH4 are a conjugate acid-base pair. - H2O & OH : also a conjugate acid-base pair 8 Example Identify the acid and base species in the following equations: 2- - - CO3 (aq) + H2O(l) HCO3 (aq) + OH -+ CH3233 COOH + H O CH COO + H O (Acetic acid) (base) (hydronium) 9 3 Amphiprotic species (“amphi-”: both): a species that can act as either an acid or a base, depending on the other reactant. Consider water: - - H2O + CH3O OH + CH3OH acid base + - H2O + HBr H3O + Br base acid 10 -- HPO24 + HO 2 HPO 34 + OH base1 acid2 acid1 base2 ()amphiprotic -2-+ H24 PO + H 2 O HPO 4 + H 3 O acid1 base2 base1 acid2 +- NH22 CH COOH NH 3 CH 2 COO glycine zwitterion “sweeterian” A zwitterion is an ion that bears both a positive and a negative charge 11 AUTOPROTOLYSIS In autoprotolysis a proton is transferred between two identical molecules, one of which acts as a Brønsted acid, releasing a proton which is accepted by the other molecule acting as a Brønsted base. 12 4 AUTOPROTOLYSIS Water undergoes self-ionization autoprotolysis, since H2O acts as an acid and a base. + - H2O + H2O H3O + OH The extent of autoprotolysis is very small. The equilibrium constant expression for this reaction is: + - [H3O ] [OH ] Kc = 2 [H2O] The concentration of water is essentially constant. 2 + - Therefore: [H2O] Kc = [H3O ] . [OH ] constant = Kw 13 Example: Calculate the hydronium and hydroxide ion concentration of pure water at 25 ◦C and 100 ◦C. 14 We call the equilibrium value of the ion product + - [H3O ][OH ] the ion-product constant of water. + - -14 Kw = [H3O ] [OH ] = 1.0 x 10 at 25°C + Using Kw you can calculate concentrations of H3O and OH- in pure water. + - -14 [H3O ] [OH ] = 1.0 x 10 + - But [H3O ] = [OH ] in pure water + - -7 [H3O ] = [OH ] = 1.0 x 10 M If you add an acid or a base to water the concentrations + - of H3O and OH will no longer be equal. But Kw will still hold. 15 5 Acid / Base Strengths The strength of an acid or base is determined by the extent it dissociates in water to form H+ or OH- respectively Strong acids: HA H+- + A (100% dissociation) + [H ] = [HA]initial Strong bases: MOH M+- + OH (100% dissociation) [OH ] = [MOH]initial 16 Weak Acid Weak acids only partially dissociate to give H+ in H2O Equilibrium Constant + - HA + H2O H3O + A + - Ka = [H ][A ] / [HA] Ka – acid dissociation constant The smaller the Ka value, the weaker the acid 17 Weak Base Weak bases react with water by abstracting a proton Equilibrium Constant + - B + H2O BH + OH + - Kb = [BH ][OH ] / [B] Kb – base hydrolysis constant Hydrolysis – reaction with water The smaller the Kb value, the weaker the base 18 6 • The strongest acids have the weakest conjugate bases and the strongest bases have the weakest conjugate acids. • A acid (base) can be cationic, anionic, neutral. 19 Differentiating/Leveling Solvents In water, perchloric and hydrochloric acids are strong acids and have the same strength. In acetic acid, perchloric acid is considerably stronger than HCl; its dissociation being ~5000 times greater. Acetic acid acts as differentiating solvent toward the two acids by revealing the inherent differences in their activities, and water is called leveling solvent for perchloric, hydrochloric (and nitric) acids because all three are 100% ionized in water and show no difference in strength. 20 Equilibrium Constant aA bB... cC dD ... [][]...CDcd K [AB ]ab [ ] ... 21 7 Standard State 1. Solution concentration ([X]) unit: M (mol/L). 2. Gas concentrations (partial pressure): atm. 3. The concentrations of pure solids, pure liquids, and solvents are omitted because they are (constant) unity. 4. K = f (T). 5. Only thermodynamic (tendency) rather than kinetic (reaction rate) conditions are considered. 22 23 Calculation of Ka or Kb from its Conjugate Kb or Ka [H+- ][A ] Acid: HA H+- + A K a [HA] Conjugate base of HA acid, A [HA][OH- ] A + H O HA + OH- K 2 b [A- ] [H+- ][A ] [HA][OH - ] KK = [H+- ][OH ] K ab[HA] [A- ] w 24 8 Manipulating Equilibrium Constants (1) AB CD K1 (2) EF GH K2 (3) (1) (2) KKK312 (3)ABEF CDGH K3 ? (4) (1) (2)KKK412 / (4)ABGH CDEF K4 ? [Class practice] 25 Solubility Product Constant (Ksp) In General: (For sparingly soluble compounds) +y -x AxBy (s) xA + yB [A+y]x [B-x]y K = ------------ [AxBy] +y x -x y [AxBy]K=Ksp =[A ] [B ] {[AxBy](solid)isconstant} 26 Examples of Ksp Expression AgCl() s Ag Cl Ksp [ Ag ][ Cl ] 22 2 CaF2 () s Ca2 F Ksp [ Ca ][ F ] 22 22 BaSO44() s Ba SO Ksp [ Ba ][ SO 4 ] 22 22 CaCO33() s Ca CO Ksp [ Ca ][ CO 3 ] 2 PbI2 () s Pb2? I Ksp 27 9 Calculating Ksp from the Solubility If the solubility of AgCl is 1.9x10-3 g/L, what would be the Ksp? +- AgCl( s ) Ag Cl , Ksp = [Ag ][Cl ] (sss ) +- 2 52 10 Kssp = [Ag ][Cl ] (1.3 10 ) 1.7 10 [1.910/sgLM3 ? 1.910/33gL 1.910/ gL = 1.3 105 M ] FW. AgCl 143.4 g/ mol 28 The solubility of lead(II) arsenate, Pb3(AsO4)2, is -5 3.0x10 g/L, calculate its Ksp. 23 2332 Pb342( AsO ) ( s ) 3 Pb 2 AsO 4sp K = [ Pb ] [ AsO 4 ] (sss ) 3 2 23 32 3 2 5 36 KPbAsOsssp = [ ] [ 4 ] (3 ) (2 ) 108 s 4.2 10 [3.010/sgLM5 ? 3.0 1055gL / 3.0 10 gL / = 3.3 108 M ] F.. W Pb342 ( AsO ) 899.44 gmol/ 29 Calculating Solubility from Ksp AgI() s Ag I (sss ) 2 Ksp [][]Ag I s s s sK sp 2 PbI2 () s Pb 2 I (sss ) 2 22 23 Ksp [][]Pb I =(2) s s = 4 s KK s 3 sp() sp 1/3 44 30 10 A general expression for a precipitate dissociation Axy B (s) xyA + B (sxsys ) xy x y xyxy() KABxsysxyssp [][] [ ][ ] KK1 ss()xy sp; ( sp ) xy xyxy xy xy 31 Solubility of a Precipitate in Pure Water How many grams of AgCl (fw = 143.32) can be dissolved in 100.00 mL of water at o -10 25 C? (Ksp [AgCl] =1.82X10 ) AgCl() s Ag Cl (sss ) 2 KAgClssssp =[ ][ ]= 1/2 10 1/2 sK()sp (1.8210) 1.82 1055 1.35 10 M 32 . How many grams of AgCl (fw = 143.32) can be dissolved in 100.00 mL of water at 25 oC? [from previous calculation: s =1.35X10-5 M] weight(# of moles) formula weight (CV ) formulaweight ( C s ) = [1.35 1053 (M ) 100.00 10 (Lgmol )] 143.32( / ) = 2.0 10-4 g = 0.20 mg 33 11 The Common Ion Effect In AgCl( s ) Ag Cl system, add a strong electrolyte containing Ag+- or Cl ions (the same as the dissociated ions--common ions) eg.., NaCl Na Cl or AgNO33Ag NO . In the presence of common ions, the solubility of the precipitate will dramatically decrease— Common Ion Effect. [Le Chatelier’s Law] 34 Calculate the solubility of slightly soluble barium fluoride, BaF2, in (a) pure water, and (b) 0.15 M -6 NaF. (Ksp = 1.0x10 ) (a) In pure water 2 BaF2 () s Ba 2 F (sss ) 2 22 23 Ksp [][](2)4Ba F s s s K 1.0 106 s ()sp 1/3 ( ) 1/3 44 6.3 103 M 35 (b) In 0.15 M NaF 2 BaF2 ( s ) Ba 2 F ; NaF Na F(0.15 M ) (s ) ssF 2 [ ]remaining [ F ] remaining 0.15 M 22 2 KBaFssp [ ][ ] (0.15) 0.0225 s K 1.0 106 sMsp 4.4 105 0.0225 0.0225 Check the assumption: 2s + [F-] = 8.8x10-5 + 0.15 = 0.15 M Assumption was valid. BaF2 solubility in 0.15 M NaF is ~1/140 times that in pure water. 36 12 Class Practice: Calculate the molar solubility of Ag2CO3 in a solution that is -12 0.0200 M in Na2CO3. (Ksp =8.1X10 ) (key: 1.0x10-5 M) 37 . EXAMPLE: Calculate the molar solubility of Ag2CO3 that results when 200. mL of 0.0100 M AgNO3 is mixed with 100.