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More About Chemical Equilibria Acid-Base & Precipitation Reactions Chapter 18

More About Chemical Equilibria Acid-Base & Precipitation Reactions Chapter 18. PLAY MOVIE. PLAY MOVIE. Acid-Base Reactions. What is relative pH before , during , & after reaction? Need to study: 1) Common ion effect 2) buffers. Strong acid + strong base

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More About Chemical Equilibria Acid-Base & Precipitation Reactions Chapter 18

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  1. More About Chemical EquilibriaAcid-Base & Precipitation ReactionsChapter 18 PLAY MOVIE PLAY MOVIE

  2. Acid-Base Reactions What is relative pH before, during, & after reaction? Need to study: 1) Common ion effect 2) buffers • Strong acid + strong base HCl + NaOH • Strong acid + weak base HCl + NH3 • Weak acid + strong base HOAc + NaOH • Weak acid + weak base HOAc + NH3

  3. The Common Ion EffectSection 18.1 QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) Here we are adding NH4+, an ion COMMON to the equilibrium. Le Chatelier predicts that the equilibrium will shift to the left (1), right (2), no change (3). The pH will go up (1), down (2), no change (3). NH4+ is an acid!

  4. pH of Aqueous NH3 QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) Let us first calculate the pH of a 0.25 M NH3 solution. [NH3] [NH4+] [OH-] initial 0.25 0 0 change -x +x +x equilib 0.25 - x x x

  5. pH of Aqueous NH3 QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) Assuming x is << 0.25, we have [OH-] = x = [Kb(0.25)]1/2 = 0.0021 M This gives pOH = 2.67 and so pH = 14.00 - 2.67 = 11.33 for 0.25 M NH3

  6. pH of NH3/NH4+ Mixture Problem: What is the pH of a solution with 0.10 M NH4Cl and 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) We expect that the pH will decline on adding NH4Cl. Let’s test that! [NH3] [NH4+] [OH-] initial change equilib

  7. pH of NH3/NH4+ Mixture Problem: What is the pH of a solution with 0.10 M NH4Cl and 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) We expect that the pH will decline on adding NH4Cl. Let’s test that! [NH3] [NH4+] [OH-] initial 0.25 0.10 0 change -x +x +x equilib 0.25 - x 0.10 + x x

  8. pH of NH3/NH4+ Mixture Problem: What is the pH of a solution with 0.10 M NH4Cl and 0.25 M NH3(aq)? NH3(aq) + H2O NH4+(aq) + OH-(aq) Assuming x is very small, [OH-] = x = (0.25 / 0.10)(Kb) = 4.5 x 10-5 M This gives pOH = 4.35 and pH = 9.65 pH drops from 11.33 to 9.65 on adding a common ion

  9. Buffer SolutionsSection 18.2 HCl is added to pure water. PLAY MOVIE HCl is added to a solution of a weak acid H2PO4- and its conjugate base HPO42-. PLAY MOVIE

  10. Buffer Solutions A buffer solution is a special case of the common ion effect. The function of a buffer is to resist changes in the pH of a solution. Buffer Composition Weak Acid + Conj. Base HOAc + OAc- H2PO4- + HPO42- NH4+ + NH3

  11. Buffer Solutions Consider HOAc/OAc- to see how buffers work ACID USES UP ADDED OH- We know that OAc- + H2O HOAc + OH- has Kb = 5.6 x 10-10 Therefore, the reverse reaction of the WEAK ACID with added OH- has Kreverse = 1/ Kb = 1.8 x 109 Kreverse is VERY LARGE, so HOAc completely uses up OH- !!!!

  12. Buffer Solutions Consider HOAc/OAc- to see how buffers work CONJ. BASE USES UP ADDED H+ HOAc + H2O OAc- + H3O+ has Ka = 1.8 x 10-5 Therefore, the reverse reaction of the WEAK BASE with added H+ has Kreverse = 1/ Ka = 5.6 x 104 Kreverse is VERY LARGE, so OAc- completely uses up H+ !

  13. Buffer Solutions Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? HOAc + H2O OAc- + H3O+ Ka = 1.8 x 10-5 0.700 M HOAc has pH = 2.45 The pH of the buffer will have 1. pH < 2.45 2. pH > 2.45 3. pH = 2.45

  14. Buffer Solutions Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? HOAc + H2O OAc- + H3O+ Ka = 1.8 x 10-5 [HOAc] [OAc-] [H3O+] initial change equilib 0.700 0.600 0 -x +x +x x 0.700 - x 0.600 + x

  15. Buffer Solutions Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? HOAc + H2O OAc- + H3O+ Ka = 1.8 x 10-5 [HOAc] [OAc-] [H3O+] equilib 0.700 - x 0.600 + x x Assuming that x << 0.700 and 0.600, we have [H3O+] = 2.1 x 10-5 and pH = 4.68

  16. Buffer Solutions Notice that the expression for calculating the H+ conc. of the buffer is Notice that the H3O+ or OH- concs. depend on (1) K and (2) the ratio of acid and base concs.

  17. Henderson-Hasselbalch Equation Take the negative log of both sides of this equation The pH is determined largely by the pKa of the acid and then adjusted by the ratio of acid and conjugate base.

  18. Adding an Acid to a Buffer Problem: What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (before HCl, pH = 7.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (a) Calc. [HCl] after adding 1.00 mL of HCl to 1.00 L of water C1•V1 = C2 • V2 C2 = 1.00 x 10-3 M = [H3O+] pH = 3.00

  19. Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH before = 4.68) Solution to Part (b) Step 1 — do the stoichiometry H3O+ (from HCl) + OAc- (from buffer) HOAc (from buffer) The reaction occurs completely because K is very large.

  20. Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (b): Step 1—Stoichiometry [H3O+] + [OAc-] [HOAc] Before rxn Change After rxn 0.00100 mol 0.600 mol 0.700 mol -0.00100 -0.00100 +0.00100 0.599 mol 0 0.701 mol

  21. Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (b): Step 2—Equilibrium HOAc + H2O  H3O+ + OAc- [HOAc] [H3O+] [OAc-] Before rxn (M) Change (M) After rxn (M) 0.701 mol/L 0 0.599 mol/L -x +x +x 0.701-x x 0.599 + x

  22. Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (b): Step 2—Equilibrium HOAc + H2O  H3O+ + OAc- [HOAc] [H3O+] [OAc-] After rxn 0.701-x 0.599+x x Because [H3O+] = 2.1 x 10-5 M BEFORE adding HCl, we again neglect x relative to 0.701 and 0.599.

  23. Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (b): Step 2—Equilibrium HOAc + H2O H3O+ + OAc- [H3O+] = 2.1 x 10-5 M pH = 4.68 The pH has not changed on adding HCl to the buffer!

  24. Preparing a Buffer You want to buffer a solution at pH = 4.30. This means [H3O+] = 10-pH = 5.0 x 10-5 M It is best to choose an acid such that [H3O+] is about equal to Ka (or pH ≈ pKa). —then you get the exact [H3O+] by adjusting the ratio of acid to conjugate base.

  25. Preparing a Buffer You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M POSSIBLE ACIDS Ka HSO4- / SO42- 1.2 x 10-2 HOAc / OAc- 1.8 x 10-5 HCN / CN- 4.0 x 10-10 Best choice is acetic acid / acetate.

  26. Preparing a Buffer You want to buffer a solution at pH = 4.30 or [H3O+] = 5.0 x 10-5 M Solve for [HOAc]/[OAc-] ratio Therefore, if you use 0.100 mol of NaOAc and 0.278 mol of HOAc, you will have pH = 4.30.

  27. Preparing a Buffer A final point — CONCENTRATION of the acid and conjugate base are not important. It is the RATIO OF THE NUMBER OF MOLES of each. Result: diluting a buffer solution does not change its pH

  28. Commercial Buffers • The solid acid and conjugate base in the packet are mixed with water to give the specified pH. • Note that the quantity of water does not affect the pH of the buffer.

  29. Preparing a Buffer Buffer prepared from 8.4 g NaHCO3 weak acid 16.0 g Na2CO3conjugate base HCO3- + H2O  H3O+ + CO32- What is the pH? PLAY MOVIE

  30. PRECIPITATION REACTIONSSolubility of SaltsSection 18.4 PLAY MOVIE Lead(II) iodide

  31. Types of Chemical Reactions • EXCHANGE REACTIONS: AB + CD AD + CB • Acid-base: CH3CO2H + NaOH NaCH3CO2 + H2O • Gas forming: CaCO3 + 2 HCl CaCl2 + CO2(g) + H2O • Precipitation: Pb(NO3) 2 + 2 KI  PbI2(s) + 2 KNO3 • OXIDATION REDUCTION • 4 Fe + 3 O2 2 Fe2O3 • Apply equilibrium principles to acid-base and precipitation reactions.

  32. Analysis of Silver Group All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions. PLAY MOVIE

  33. Analysis of Silver Group Although all salts formed in this experiment are said to be insoluble, they do dissolve to some SLIGHT extent. AgCl(s) Ag+(aq) + Cl-(aq) When equilibrium has been established, no more AgCl dissolves and the solution is SATURATED.

  34. Analysis of Silver Group AgCl(s)  Ag+(aq) + Cl-(aq) When solution is SATURATED, expt. shows that [Ag+] = 1.67 x 10-5 M. This is equivalent to the SOLUBILITYof AgCl. What is [Cl-]? [Cl-] = [Ag+] = 1.67 x 10-5 M

  35. Analysis of Silver Group AgCl(s)  Ag+(aq) + Cl-(aq) Saturated solution has [Ag+] = [Cl-] = 1.67 x 10-5 M Use this to calculate Kc Kc = [Ag+] [Cl-] = (1.67 x 10-5)(1.67 x 10-5) = 2.79 x 10-10

  36. Analysis of Silver Group AgCl(s)  Ag+(aq) + Cl-(aq) Kc = [Ag+] [Cl-] = 2.79 x 10-10 Because this is the product of “solubilities”, we call it Ksp = solubility product constant • See Table 18.2 and Appendix J

  37. Some Values of Ksp Table 18.2 and Appendix J

  38. Lead(II) Chloride PbCl2(s)  Pb2+(aq) + 2 Cl-(aq) Ksp = 1.9 x 10-5 = [Pb2+][Cl–]2 PLAY MOVIE

  39. Solubility of Lead(II) Iodide Consider PbI2 dissolving in water PbI2(s) Pb2+(aq) + 2 I-(aq) Calculate Kspif solubility = 0.00130 M Solution 1. Solubility = [Pb2+] = 1.30 x 10-3 M [I-] = ? [I-] = 2 x [Pb2+] = 2.60 x 10-3 M

  40. Solubility of Lead(II) Iodide Consider PbI2 dissolving in water PbI2(s) Pb2+(aq) + 2 I-(aq) Calculate Kspif solubility = 0.00130 M Solution 2. Ksp = [Pb2+] [I-]2 = [Pb2+] {2 • [Pb2+]}2 Ksp = 4 [Pb2+]3 = 4 (solubility)3 Ksp = 4 (1.30 x 10-3)3 = 8.79 x 10-9

  41. Precipitating an Insoluble Salt Hg2Cl2(s)  Hg22+(aq) + 2 Cl-(aq) Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2 If [Hg22+] = 0.010 M, what [Cl-] is req’d to just begin the precipitation of Hg2Cl2? That is, what is the maximum [Cl-] that can be in solution with 0.010 M Hg22+ without forming Hg2Cl2?

  42. Precipitating an Insoluble Salt Hg2Cl2(s)  Hg22+(aq) + 2 Cl-(aq) Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2 Recognize that Ksp = product of maximum ion concs. Precip. begins when product of ion concs. EXCEEDS the Ksp.

  43. Precipitating an Insoluble Salt Hg2Cl2(s)  Hg22+(aq) + 2 Cl-(aq) Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2 Solution [Cl-] that can exist when [Hg22+] = 0.010 M, If this conc. of Cl- is just exceeded, Hg2Cl2 begins to precipitate.

  44. Precipitating an Insoluble Salt Hg2Cl2(s)  Hg22+(aq) + 2 Cl-(aq) Ksp = 1.1 x 10-18 Now raise [Cl-] to 1.0 M. What is the value of [Hg22+] at this point? Solution [Hg22+] = Ksp/ [Cl-]2 = Ksp / (1.0)2 = 1.1 x 10-18 M The concentration of Hg22+ has been reduced by 1016 !

  45. The Common Ion Effect Adding an ion “common” to an equilibrium causes the equilibrium to shift back to reactant.

  46. Common Ion Effect & Solubility PbCl2(s)  Pb2+(aq) + 2 Cl-(aq) Ksp = 1.9 x 10-5 PLAY MOVIE

  47. Barium SulfateKsp = 1.1 x 10-10 (b) BaSO4 is opaque to x-rays. Drinking a BaSO4 cocktail enables a physician to exam the intestines. (a) BaSO4 is a common mineral, appearing a white powder or colorless crystals.

  48. The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2. Ksp for BaSO4 = 1.1 x 10-10 BaSO4(s)  Ba2+(aq) + SO42-(aq) Solution Solubility in pure water = [Ba2+] = [SO42-] = x Ksp = [Ba2+] [SO42-] = x2 x = (Ksp)1/2 = 1.1 x 10-5 M Solubility in pure water = 1.0 x 10-5 mol/L

  49. Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2. Ksp for BaSO4 = 1.1 x 10-10 BaSO4(s) Ba2+(aq) + SO42-(aq) The Common Ion Effect Solution Solubility in pure water = 1.1 x 10-5 mol/L. Now dissolve BaSO4 in water already containing 0.010 M Ba2+. Which way will the “common ion” shift the equilibrium? ___ Will solubility of BaSO4 be less than or greater than in pure water?___

  50. Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2. Ksp for BaSO4 = 1.1 x 10-10 BaSO4(s) Ba2+(aq) + SO42-(aq) The Common Ion Effect Solution [Ba2+] [SO42-] initial change equilib. 0 0.010 + y + y 0.010 + y y

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