1 / 52

Chapter 17

Chapter 17. Additional Aspects of Equilibria. Salt Solutions in Water Salt solutions completely ionize in H 2 O. If their ions can react with H 2 O to form H + of OH - , the pH of the solution will change.

noah-lowe
Download Presentation

Chapter 17

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 17 Additional Aspects of Equilibria

  2. Salt Solutions in Water Salt solutions completely ionize in H2O. If their ions can react with H2O to form H+ of OH-, the pH of the solution will change.

  3. Anions that are the conjugate bases of weak acids will react with H2O to form OH- ions. Anions that are conjugate bases of strong acids are very weak bases and aren’t strong enough to react with H2O. No pH change.

  4. Anions that have ionizable protons (HSO4-) are amphoteric. Can act as an acid or a base and must be determined based on the Ka and Kb for the ion.

  5. 3- +2 +1 Will K2HC6H5O7 form an acidic or basic solution in H2O? or Must look at Ka and Kb. From table 16.3 in book get Ka3 = 4.0x10-7 Must calculate Kb from appropriate Kax.

  6.  Kw = Ka x Kb Now compare for our situation Ka = 4.0x10-7 Kb = 5.9x10-10 Ka is  700 times larger than Kb so the acid reaction will predominate and solution will be acidic.

  7. All cations (except alkali and heavy alkaline earth metal ions) act as weak acids. • Rules • Salts derived from strong acid and strong base from neutral solutions in H2O. • ex. NaCl Na+ from NaOH • Cl- from HCl • Salt derived from strong base/weak acid. Anion will be relatively strong conjugate base.  pH > 7. • ex. Ba(C2H3O2)2

  8. Salt derived from weak base/strong acid. Strong conjugate acid.  pH < 7. • ex. Al(NO3)3 • Al3+ + 3H2O  Al(OH3) + 3H+ • Salt derived from weak base/weak acid. ex. NH3C2H3O2. Both conjugate acid and base will be fairly strong in solution. pH depends on which ion hydrolyzes H2O better. Must compare Ka and Kb.

  9. Additional aspects of Aqueous Equilibria 1. Common Ion Effect RememberLeChatlier What happens if I add more C2H3O2- to the solution?

  10. The dissociation of a weak electrolyte is decreased by adding a strong electrolyte that has an ion in common with the weak electrolyte. Ex. ? pH of solution of 0.30 mol acetic acid (HC2H3O2) and 0.30 mol NaC2H3O2 (sodium acetate) in 1.0L?

  11. Steps 1. Identify major species and are they acid or base. HC2H3O2  Weak acid Won’t dissociate completely so: C2H3O2- + H+ + HC2H3O2 NaC2H3O2  Strong * electrolyte so: Na+ + C2H3O2- HC2H3O2 C2H3O2- C2H3O2- H+ C2H3O2- Na+ H+ C2H3O2- C2H3O2- HC2H3O2 H+ Na+ Na+ C2H3O2- * How do I know that?

  12. 2) Identify important equilibrium reaction. H+ is the same thing as H3O+ in aqueous solution (at this point in your chemistry life) 3) Calculate concentrations of equilibrium species. Remember: C2H3O2- comes from HC2H3O2 and NaC2H3O2

  13. Can we make the assumption that ‘x’ can be ignored here? How do you know?

  14. 2) Acid-Base Titrations Quantitatively add an acid (or base) to a base (or acid). Strong acid/strong base titrations: The graph of pH vs. ml of titrant is called a titration curve.

  15. pH before equivalence point is determined by concentration of acid NOT YET NEUTRALIZED. pH at equivalence point is pH of the salt solution. pH past equivalence point is determined by concentration of excess base.

  16. ? pH when 0.100M NaOH solution is added to 20.0 ml 0.100M HBr a) 10.0 ml NaOH NaOH  Na+ + OH- in water So ALL these will react with H+ until one or the other runs out!

  17. So 10 ml 0.001 mol NaOH yields 0.001 mol OH- in H2O. OH- + H+ H2O Consumed by OH- Started with 0.002 mol H+ That leaves 0.001 mol H+ pH = 3

  18. B) 20.0 ml NaOH total So, is our pH 14? NO Autoprotolysis H2O  H+ + OH- pH would be that of the salt + autoprotolysis  7

  19. C) 30.0 ml NaOH Consumed by 0.002 mol H+ in initial solution pOH = 3 pH = 11

  20. Buffered Solutions Solutions that resist change in pH when small amounts of acid or base are added. Many neutral systems are buffered systems.

  21. Buffers contain both an acidic species and a basic species. A weak acid/base conjugate pair is common. NaC2H3O2 and HC2H3O2 NH4Cl and NH3

  22. Choose appropriate components and adjust concentrations to get buffer at any pH.

  23. Ratio of conjugate acid/base pair Na HCl NaCl OH- + HX  H2O + X- [HX]  [X-]  But if [HX] and [X-] are large enough, a small amount of [OH-] won’t matter.

  24. H+ + X- HX Same goes for a small amount of acid. When [HX] and [X-] are approximately equal, buffer is most effective for pH change in either direction. Choose buffer whose acid has a pKa close to desired pH.

  25. Buffer Capacity and pH In General: Henderson-Hasselbach equation Use starting concentration of acid and base components of the buffer.

  26. Ka2 Ka1 ? pH of Buffer 0.15M NaHCO3 0.10M Na2CO3

  27. Strong Acid-Weak Base and Weak Acid-Strong Base Titrations

  28. * pH > 7 at stoichiometric endpoint for weak acid/strong base titration. * pH < 7 at stoichiometric endpoint for strong acid/weak base titration.

  29. Example 30.0ml sample of 0.20M C6H5COOH is titrated with 0.30M KOH. Ka C6H5COOH = 6.5x10-5 Calculate pH at stoichiometric endpoint. C6H5COOH weak acid C6H5COOH  C6H5COO- + H+ Ka = 6.5x10-5in water But what about with a strong base?

  30. OH- will deprotonate all of the C6H5COOH So, at the endpoint, enough OH- has been added to react with all the C6H5COOH. Up to this point, the problem looks just like a strong acid/strong base titration!

  31. What’s different? Weak acid yields a strong conjugate base. So, C6H5COO- + H2O  C6H5COOH + OH- That’s the logic.

  32. Must have same moles of OH- Total volume = 0.03L + 0.02L = 0.05L = 50ml

  33. What is happening in solution? Relatively strong base in water: C6H5COO- + H2O  C6H5COOH + OH-

  34. x2 = 1.85x10-11 x = 4.3x10-6 x = [C6H5COOH] = [OH-] pOH = 5.37 pH = 8.6

  35. ? pH 10.0g KCH3CO2 dissolved in 250ml solution. Ka(CH3CO2) = 1.8x10-5 10.0g KCH3CO2 = 0.093mol KCH3CO2 0.093mol KCH3CO2 yields 0.093mol CH3CO2-

  36. Solubility Equilibria Ksp = the degree to which a solid is soluble in water.

  37. Table for Ksp at 25ºC Appendix D Ksp BaSO4 = 1.1x10-10 The smaller Ksp is means less will dissolve in H2O Ksp Ca3(PO4)2 = [Ca2+]3 [PO43-]2 = 2.0x10-29 Ca3(PO4)2 3Ca2+ + 2PO43-

  38. Solubility Molar solubility Molar of compound of compound [ions] Ksp (g/L) (mol/L) Converting between Solubility and Ksp

  39. 17.34) PbBr2 molar solubility = 1.0x10-2 mol/L ? Ksp Ksp = [Pb2+][Br-]2 Ksp = (1.0x10-2)(2.0x10-2)2 Ksp = 4.0x10-6

  40. Common Ion Effect Remember Le Châtlier! 17.36) ? solution of CaF2 in g/L in 0.15M KF solution Ksp = [Ca2+][F-]2 molar solution CaF2 = [Ca2+] = X [F-] = 2X another source of F- = 0.15M F-

  41. So [Ca2+] = X [F-] = 2X + 0.15 Ksp = (X)(2X + 0.15)2 * assume X is small compared to 0.15 Ksp = 3.9x10-11 = (X)(0.15)2 = 0.0225X X = 1.7x10-9 M Ca

  42. Precipitation will happen when Q > Ksp Q = ion product if Q = Ksp, equilibrium exists, saturated solution if Q < Ksp, solid dissolves until Q = Ksp

  43. Solubility and pH Hg2+ Very important in environmental systems Consider Hg(OH)2 Ksp = 3.0x10-26 Hg(OH)2 Hg2+ + 2OH- Ksp = [Hg2+][OH-]2 = 3.0x10-26 at pH = 3, pOH = 11  [OH] = 1.0x10-11 [Hg2+][1.0x10-11]2 = 3.0x10-26 [Hg2+] = 3x10-4M = 6.0x10-2g/L = 60mg/L = 60ppm

  44. Selective Precipitation Say you have Cu2+ and Zn2+ in solution You can ppt one out of solution without the other use S2- by adding H2S(g) CuS Ksp = 6.3x10-36 ZnS Ksp = 1.1x10-21 Must adjust pH to accomplish this.

  45. Say the solution in 0.10M in Zn2+ and 0.10 in Cu2+ Now look at how H2S affects pH [H+]2[S2-] = 7x10-22

More Related