Chapter 6 Problems 6.19, 6.21, 6.24 6-29, 6-31, 6-39, 6.41 6-42, 6-48, 6-19. A solution contains 0.0500 M Ca2+ and 0.0300 M Ag+. Can 99% of Ca2+ be precipitated by sulfate without precipitating Ag+? What will be the concentration of Ca2+ when Ag2SO4 begins to precipitate? 6-19. A solution contains 0.0500 M Ca2+ and 0.0300 M Ag+. Can 99% of Ca2+ be precipitated by sulfate without precipitating Ag+? What will be the concentration of Ca2+ when Ag2SO4 begins to precipitate? Ca2+ at equilibrium 0 . 0500 M * 0 . 01 0 . 000500 M 2+ + SO 22CaSO44 Ca2+ 44 = -5 2= -5 2+][SO 2-] Ksp [0.000500][SO sp 2.4 x 10 = [Ca 4 ] 4 [SO42-] = 0.048 M Ag2SO4 2Ag+ + SO42Q = [Ag+]2[SO42-] Ksp = 1.5 x 10-5 Q = [0.03]2[0.000500] Q>K Q = 4.3 x 10-5 6-19. A solution contains 0.0500 M Ca2+ and 0.0300 M Ag+. Can 99% of Ca2+ be precipitated by sulfate without precipitating Ag+? What will be the concentration of Ca2+ when Ag2SO4 begins to precipitate? Ag2SO4 2Ag+ + SO42K/[Ag2+]2 = [SO42-] Ksp = 1.5 x 10-5 1.5 x 10-5/[0.0300]2 = [SO42-] 1.67 x 10-2= [SO42-] Find Ca2+ CaSO4 Ca2+ + SO42[Ca2+] = 0.0014 M Ksp= 2.4 x 10-5 = [Ca2+][1.67 x 10-2] About 2.8 % remains in solution 6.21 If a solution containing 0.10 M Cl-, Br-, I- and Cr2O42- is treated with Ag+, in what order will the anions precipitate? AgCl Ag+ + ClAgBr Ag+ + BrAgI Ag+ + IAg2 CrO4 2Ag+ + CrO4- K x 10-10=[Ag][0.1] =[Ag][Cl] 1.8 x 1.8 10-9=[Ag] sp = -13=[Ag][Br] -12 K = 5.0 x 10 5.0 sp x 10 =[Ag]=[Ag][0.1] =[Ag][I] K 8.3-16x=[Ag] 10-17=[Ag][0.1] sp = 8.3 x 10 -6=[Ag] K 1.2 x 10-12 =[Ag]2[0.1] [Cl] 3.5 x 10 sp = SOLVE for Ag+ required at equilibrium 6-24. The cumulative formation constant for SnCl2(aq) in 1.0 M NaNO3 is b2=12. Find the concentration of SnCl2 for a solution in which the concentration of Sn2+ and Cl- are both somehow fixed at 0.20 M. b2=12 Sn2+ (aq) + 2Cl- (aq) SnCl2 (aq) b2=12 [SnCl 2 ( aq )] [Sn 2 ][ Cl ] 2 [SnCl 2 ( aq )] [ 0 . 20 ][ 0 . 20 ] [SnCl 2 ( aq )] 0 . 096 M 2 Complex Formation complex ions (also called coordination ions) Lewis Acids and Bases acid => electron pair acceptor (metal) base => electron pair donor (ligand) Effects of Complex Ion Formation on Solubility Consider the addition of I- to a solution of Pb+2 ions Pb2+ + I- PbI+ K1 [ PbI [ Pb 2 ] 1 . 0 x10 ][ I ] PbI+ + I- PbI2 K2 = 1.4 x 101 PbI2 + I- PbI3- K3 =5.9 PbI3+ I- PbI42- K4 = 3.6 2 Effects of Complex Ion Formation on Solubility Consider the addition of I- to a solution of Pb+2 ions Pb2+ + I- <=> PbI+ K1 [ PbI [ Pb 2 ] 1 . 0 x10 ][ I ] PbI+ + I- <=> PbI2 K2 = 1.4 x 101 Pb2+ + 2I- <=> PbI2 K’ =? Overall constants are designated with b This one is b2 2 Effects of Complex Ion Formation on Solubility Consider the addition of I- to a solution of Pb+2 ions Pb2+ + I- PbI+ K1 [ PbI [ Pb 2 ] 1 . 0 x10 ][ I ] PbI+ + I- PbI2 K2 = 1.4 x 101 PbI2 + I- PbI3- K3 =5.9 PbI3+ I- PbI42- K4 = 3.6 2 Acids and Bases & Equilibrium Section 6-7 Strong Bronsted-Lowry Acid A strong Bronsted-Lowry Acid is one that donates all of its acidic protons to water molecules in aqueous solution. (Water is base – electron donor or the proton acceptor). HCl as example Strong Bronsted-Lowry Base Accepts protons from water molecules to form an amount of hydroxide ion, OH-, equivalent to the amount of base added. Example: NH2- (the amide ion) Question Can you think of a salt that when dissolved in water is not an acid nor a base? Can you think of a salt that when dissolved in water IS an acid or base? Weak Bronsted-Lowry acid One that DOES not donate all of its acidic protons to water molecules in aqueous solution. Example? Use of double arrows! Said to reach equilibrium. Weak Bronsted-Lowry base Does NOT accept an amount of protons equivalent to the amount of base added, so the hydroxide ion in a weak base solution is not equivalent to the concentration of base added. example: NH3 Common Classes of Weak Acids and Bases Weak Acids carboxylic acids ammonium ions Weak Bases amines carboxylate anion Equilibrium and Water Question: Calculate the Concentration of H+ and OH- in Pure water at 250C. EXAMPLE: Calculate the Concentration of H+ and OH- in Pure water at 250C. Initial Change Equilibrium H2O H+ + OH- liquid - - -x +x +x +x +x Liquid-x Kw = [H+][OH-] = ? KW=(X)(X) = ? EXAMPLE: Calculate the Concentration of H+ and OH- in Pure water at 250C. Initial Change Equilibrium H2O H+ + OH- liquid - - -x +x +x +x +x Liquid-x Kw = [H+][OH-] = 1.01 X 10-14 KW=(X)(X) = 1.01 X 10-14 (X) = 1.00 X 10-7 Example What is the concentration of OH- in a solution of water that is 1.0 x 10-3 M in [H+] (@ 25 oC)? “From now on, + Kw = [H ][OH ] assume the to 1 x 10-14 = [1 x 10-3][OH-] temperature be 25oC unless 1 x 10-11 = [OH-] otherwise stated.” pH ~ -3 -----> ~ +16 pH + pOH = - log Kw = pKw = 14.00 Is there such a thing as Pure Water? In most labs the answer is NO Why? CO2 + H2O HCO3- + H+ A century ago, Kohlrausch and his students found it required to 42 consecutive distillations to reduce the conductivity to a limiting value. Weak Acids and Bases HA Ka H+ + A- Ka Ka’s ARE THE SAME [ H ][ A ] [ HA ] HA + H2O(l) H3O+ + A Ka [ H 3 O ][ A ] [ HA ] Weak Acids and Bases B + H2O Kb Kb [ BH BH+ + OH ][ OH [B] ] Relation Between Ka and Kb Relation between Ka and Kb Consider Ammonia and its conjugate acid. NH3 + H2O NH4+ + H2O H2O + H2O Kb Ka NH4+ + OH- Kb Ka [ NH 3 ][ H 3 O ] 4 ][ OH [ NH ] [ NH 3 ] NH3 + H3O+ OH- + H3O+ [ NH K 4 ] [ NH 3 ][ H 3 O ] [ NH 4 ][ OH ] K [ NH 3 ] [ NH 34 ] w [ H O ] [ OH ] Example The Ka for acetic acid is 1.75 x 10-5. Find Kb for its conjugate base. Kb Kw = Ka x Kb Kb 1 . 0 10 Kw Ka 14 1 . 75 10 5 5 . 7 10 10 Example Calculate the hydroxide ion concentration in a 0.0100 M sodium hypochlorite solution. OCl- + H2O HOCl + OH[ HOCl ][ OH ] Kb The acid dissociation constant = 3.0 x 10-8 [ OCl ] 1st Insurance Problem Challenge on page 120 Chapter 8 Activity Write out the equilibrium constant for the following expression Fe3+ + SCNK D Fe(SCN)2+ [ Fe ( SCN ) [ Fe 3 2 ][ SCN ] ] Q: What happens to K when we add, say KNO3 ? A: Nothing should happen based on our K, our K is independent of K+ & NO3- K decreases when an inert salt is added!!! Why? 8-1 Effect of Ionic Strength on Solubility of Salts Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions. Hg2(IO3)2(s) D Hg22+ + 2IO3I C E some -x some-x +x +x +2x +2x Ksp=1.3x10-18 2 2 K sp [ Hg ][ IO 3 ] 1 . 3 10 2 18 K sp [ x ][ 2 x ] 1 . 3 10 2 [ x ] 6 . 9 10 A seemingly strange effect is observed when a salt such as KNO3 is added. As more KNO3 is added to the solution, more solid dissolves until [Hg22+] increases to 1.0 x 10-6 M. Why? 7 18 Increased solubility Why? Complex Ion? No Hg22+ and IO3- do not form complexes with K+ or NO3-. How else? The Explanation Consider Hg22+ and the IO3Electrostatic attraction 2+ - The Explanation Consider Hg22+ and the IO3Electrostatic attraction 2+ - Hg2(IO3)2(s) The Precipitate!! The Explanation Consider Hg22+ and the IO3NO3NO3 NO3NO 3 NO3 - 2+ - NO3NO3- Add KNO3 K+ - NO Electrostatic attraction 3 K+ NO3- NO3 K+ K+ - K+ K+ The Explanation Consider Hg22+ and the IO3NO3NO3 NO3- NO3NO 3 K+ NO3- K+ 2+ - NO3 NO3- K+ - NO3- NO3 K+ - K+ K+ Hg22+ and IO3- can’t get CLOSE ENOUGH to form Crystal lattice Or at least it is a lot “Harder” to form crystal lattice The potassium hydrogen tartrate example OH O +- K O OH O OH po tassiu m hy drogen tartrate Alright, what do we mean by Ionic strength? Ionic strength is dependent on the number of ions in solution and their charge. Ionic strength (m) = ½ (c1z12+ c2z22 + …) Or Ionic strength (m) = ½ S cizi2 Examples Calculate the ionic strength of (a) 0.1 M solution of KNO3 and (b) a 0.1 M solution of Na2SO4 (c) a mixture containing 0.1 M KNO3 and 0.1 M Na2SO4. (m) = ½ (c1z12+ c2z22 + …) Alright, that’s great but how does it affect the equilibrium constant? Activity = Ac = [C]gc AND K c C a A A A A A d D b B [C ] g [ D ] g c c C a A d [ A] g [ B ] g a b d D b B Relationship between activity and ionic strength Debye-Huckel Equation 2 log g x 0 . 51 z x 1 3 . 3 x m m m = ionic strength of solution g = activity coefficient Z = Charge on the species x = effective diameter of ion (nm) 2 comments (1) What happens to g when m approaches zero? (2) Most singly charged ions have an effective radius of about 0.3 nm Anyway … we generally don’t need to calculate g – can get it from a table Activity coefficients are related to the hydrated radius of atoms in molecules Relationship between m and g Back to our original problem Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions. Hg2(IO3)2(s) D Hg22+ + 2IO3- K sp A Hg 2 A 2 2 IO 3 [ Hg Ksp=1.3x10-18 2 2 3 ]g Hg 2 [ IO ] g 2 At low ionic strengths g -> 1 2 2 IO 3 Back to our original problem Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions. Hg2(IO3)2(s) D Hg22+ + 2IO3- K sp A Hg 2 A 2 2 IO 3 [ Hg Ksp=1.3x10-18 2 2 3 ]g Hg 2 [ IO ] g 2 2 In 0.1 M KNO3 - how much Hg22+ will be dissolved? 2 IO 3 Back to our original problem Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions. Hg2(IO3)2(s) D Hg22+ + 2IO3- K sp A Hg 2 A 2 2 IO 3 [ Hg Ksp=1.3x10-18 2 2 3 ]g Hg 2 [ IO ] g 2 2 2 IO 3