CHAPTER 15 AP CHEMISTRY

COMMON ION EFFECT If you have HC2H3O2(aq) <=> H+ (aq) + C2H3O2-(aq) What would happen if you added the salt NaC2H3O2? Addition of C2H3O2- would cause the system to shift to the left This is called the COMMON-ION EFFECT Suppose that we add 8.20 g, or 0.100 mol, of sodium acetate, NaC2H3O2, to 1 L of a 0.1000 M of acetic acid, HC2H3O2. What is the pH of the resultant solution? HC2H3O2(aq) <=> H+ (aq) + C2H3O2-(aq) Initial 0.1000 0 0.100 Change -X +X +X Equilibrium 0.1000-X X 0.100+X X(0.100+X)/(0.1000-X) 0.100X/ 0.1000 = 1.8 X 10-5 X= 1.8 X 10-5 = [H+] pH = 4.74 Page 699 example

BUFFER Prepared by adding both the weak acid HB and its conjugate base B- to water [H+] = ka[HB] = ka nHB [B-] nB- Effect of adding strong acids or bases to a buffer HB(aq) + OH-(aq) ---> H2O(l) + B-(aq) B-(aq) + H+(aq) ---> HB(aq) H+ and OH- ions are consumed so the pH is not affected Buffered solutions are most effective when concentrations of H+ and base B- are about the same

CONTINUED Henderson-Hasselbach equation is used in biology to calculate the pH of the buffer pH = pka+ log [base] [acid] What is the pH of a buffer mixture composed of 0.12 M lactic acid (HC3H5O3, ka = 1.4 X 10-4 ) and 0.10 M sodium lactate. HC3H5O3 <=> H+ + C3H5O3- -log(1.4 X 10-4 )= 3.85 Log(C3H5O3-/HC3H5O3) = log(0.10/0.12) = -0.079 pH = 3.85 - 0.079 = 3.77 How many moles of NH4Cl must be added to 1.0L of 0.10 M NH3 to form a buffer whose pH is 9.00? NH3 + H2O <=> NH4+ + OH- 1.8 X 10-5 = [NH4+] [OH-] 14-9 = 5 for pOH [NH3] 1.8 X 10-5 [NH3]/[OH-] = [NH4+] 1.8 X 10-5(0.10)/(1 X 10-5) 0.18mol

ACID-BASE TITRATION Read and study pages 701-703,706-708, 709-714 Titration curve Graph of pH to concentration strong acid - strong base page 715 H+(aq) + OH-(aq) <=> H2O(l) k = 1/kw = 1/(1 X 10-14) pH = 7, change rapidly as you near the end point When 50.00 mL of 1.000 M HCl is titrated to 1.000 M NaOH, the pH increases, find the pH of the solution after the following volumes of 1.000 M NaOH has been added a) 49.99 mL b) 50.01 mL nH+ = 0.05000L X 1.000molH+/1L = 0.05000 mol H+ nOH- = 0.04999L X 1.000mol OH-/1L = 0.04999 mol OH- The OH- and H+ will form water but in this problem there would still be H+ left over n H+ excess = 0.00001 mol The volume has changed it now has 100 mL, so the H+ concentration would be 1 X 10-5mol H+/0.100L = 1 X 10-4 M H+ pH = 4

CONTINUED Weak acid-strong base HA(aq) + OH-(aq) <=> A-(aq) + H2O(l) k =1/kb pH > 7 at end point, change occurs slowly at end point Strong acid-weak base H+ (aq) + A-(aq) <=> HA(aq) k = 1/ka pH < 7, at end point, change occurs slowly at end point Titration of polyproptic acids When neutralization steps are separated the substance will show a titration curve with multiple equivalence points

INDICATORS HIn(aq) <=> H+(aq) + In-(aq) In = indicator HIn and In- have different colors, the wavelengths that bounce back are different when hydrogen is attached Color of a solution depends on the concentration ratio [HIn] = [H+] [In- ] ka Endpoint occurs when [H+] = ka About pH 5 for methyl red, pH 7 for bromothymol blue, and pH 9 for phenolphthalein Read pages 732, 735-736

SOLUBILITY EQUILIBRIA If Q(ion product) is compared with the ksp, the following can be said Q > ksp ppt occurs until Q = ksp Q = ksp equilibrium (saturation has occured) Q < ksp solid dissolves until Q = ksp AgCl(s) <=> Ag+ (aq) + Cl- (aq) Slightly soluble ksp found on page 736 ksp = [Ag+][Cl-] Will a precipitate form when 0.100L of 3.0 X 10-3M Pb(NO3)2 is added to 0.400L of 5.0 X 10-3M Na2SO4? Find moles first 3.0 X 10-4mol Pb(NO3)2 = Pb2+ 2.0 X 10-3mol Na2SO4 = SO42- Add the two volumes together to get new volume of 0.500L 4.0 X 10-3M SO42- and 6.0 X 10-4M Pb(NO3)2 Q = (4.0 X 10-3)(6.0 X 10-4) = 2.4 X 10-6 If the ksp = 1.6 X 10-8 then a ppt would form

CONTINUED What is [Pb2+] in a solution at equilibrium with PbCl2 (ksp = 1.7 x 10-5) if [Cl-] = 0.020M? PbCl2 <=> Pb2+ + 2Cl- ksp= [Pb2+][Cl-]2 0.042M Pb2+ the solubility of slightly soluble salts containing basic anions with increase as the [H+] increases Looking at the solubility of a salt In pure water it is the same as the reaction PbCl2(s) <=> Pb2+(aq) + 2Cl-(aq) [Pb2+] = s [Cl-] = 2s [s][2s]2 = 4s3 = 1.7 x 10-5 s = 1.6 x 10-2M Pb2+ and 2(1.6 x 10-2) M Cl-

CONTINUED In a solution containing a common-ion 0.100 M HCl [Pb2+] = s [Cl-] = 2s + 0.10 = 0.10 s(0.10)2 = 1.7 X 10-5 s = 1.7 X 10-3M Pb2+ Read pages 739, 742, 744-748 Complex ions Metal ions with Lewis bases bonded to it page 754 Amphoterism Behavior of water molecules around a metal ion Read pages 749-750, 757 Read chapters 18, 19, and 22 notes are due the first day of fourth term