Ap chem unit 14 presentation part 2


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Ap chem unit 14 presentation part 2

  1. 1. STRONG BASESAll the hydroxides of the of the Group 1 and 2 elements (LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Ba( OH)2, Sr(OH)2 ) are strong bases, but only NaOH and KOH are common laboratory reagents.• The alkaline earth hydroxides are not very soluble and are used only when the solubility factor is not important.• Low solubility of bases can sometimes be an advantage.
  2. 2. PRACTICE PROBLEM 12Calculate the pH of a 5.0x10-2 M NaOH solution.Major species: Na+, OH-, and H2ONarrow down contributing species and its concentration. Write balanced equation. [OH-] = 5.0x10-2 MpOH 1.3 pH=12.70
  3. 3. BASESMany types of proton acceptors do not contain the hydroxide ion. However, when dissolved in water, these substances increase the concentration of hydroxide ion because of their reaction with water. + - NH 3(aq) + H 2O(l) Û NH 4(aq) + OH (aq)
  4. 4. BASESKb refers to the equilibrium expression of the reaction of a base with water to form the conjugate acid and the hydroxide ion.• pH calculations for weak bases are similar to weak acids.
  5. 5. BASESOther bases that produce the hydroxide ion by reaction with water:
  6. 6. PRACTICE PROBLEM 13Calculate the pH for a 15.0 M solution of NH3 (Kb=1.8 x 10-5)pH=12.20
  7. 7. PRACTICE PROBLEM 14Calculate the pH of a 1.0 M solution of methylamine (Kb= 4.38 x 10-4)pH = 12.32
  8. 8. POLYPROTIC ACIDSAn acid that can furnish more than one proton are called polyprotic acids.• A polyprotic acid dissociates in a stepwise manner, one proton at a time.• Each dissociation of a proton, gives its own Kavalue.• The conjugate base in the first step is the acide in the second step.
  9. 9. POLYPROTIC EXAMPLETriprotic example:H3PO4  H+ + H2PO4- ; Ka1 = 7.5 x 10-3H2PO4-  H+ + HPO42- ; Ka2 = 6.2 x 10-8HPO42-  H+ + HPO42- ; Ka3 = 4.8 x 10-13• Ka1 > Ka2 > Ka3• Each acid involve in dissociation steps is successively weaker.
  11. 11. POLYPROTIC ACIDSMost polyprotic acids have very different successive Ka values.• Typically, the first dissociation step is the only step to make an important contribution to [H+]
  12. 12. POLYPROTIC ACIDSSulfuric acid is unique among the common polyprotic acids.• Sulfuric acid is a strong acid in its first dissociation step and a weak acid in the second step.• When acid concentration is below 1.0M, both steps contribute to the overall contribution of [H+]
  13. 13. PRACTICE PROBLEM 15Calculate the pH of a 5.0 M H3PO4 solution and the equilibrium concentrations of the species H3PO4, H2PO4-, HPO42-, and PO43-pH=0.72, [H+]=[H2PO4- ]=0.19M, [H3PO4]=4.8M,[HPO42-]=6.2x10-8M, [PO43-]=1.6x10-19M
  14. 14. PRACTICE PROBLEM 16Calculate the pH of a 1.0 M H2SO4 solution.[H+]=1.0M, pH=0.00
  15. 15. POLYPROTIC SUMMARY1. Typically, successive Ka values are so much smaller than the first value that only the first dissociation step makes a significant contribution.2. Sulfuric acid is unique. At 1.0M and higher the large concentration of H+ from the first dissociation step represses the second step and the second step is negligible. For dilute solutions, the second step does make a significant contribution and the quadratic equation must be used to obtain the total H+ concentration.
  16. 16. PROPERTIES OF SALTSIonic compounds, also known as salts, can dissolve in water and under certain conditions, these ions can behave as acids or bases.• Respective partners of strong acids and strong bases, do not combine with H+ or OH- and therefore have no effect on pH in an aqueous solution.• examples: K+, Na+, Cl-, NO3-,
  17. 17. BASIC SALTSIons that are the conjugate base of a weak acid are strong bases. These ions produce hydroxide in an aqueous solution.• example: C2H3O2-, F- - - C2 H3O2(aq) + H 2O(l) Û HC2 H3O(aq) + OH(aq)
  18. 18. KB AND KAIf Ka is known for a weak acid the Kb for its conjugate base can be found.• example: C2H3O2 [H + ][[C2 H 3O2 ] [OH - ][[HC2 H 3O2 ] - - Ka x Kb = x - [HC2 H 3O2 ] [C2 H 3O2 ] = [H + ][OH - ] = K w K a x K b = K w
  19. 19. BASIC SALTS SUMMARYFor any salt whose cation has neutral properties (such as Na+ or K+) and whose anion is the conjugate base of a weak acid, the aqueous solution will be basic.
  20. 20. PRACTICE PROBLEM 18Calculate the pH of a 0.30 M NaF solution. The Ka value for HF is 7.2x10-4.pH=8.31
  21. 21. BASIC SALTSIn the first reaction CN- is competing with water forthe proton, Ka=6.2x10-10 : + - HCN(aq) + H 2O(l) Û H3O (aq) + CN (aq)In the second reaction CN- is competing with OH-for the proton, Kb=1.6x10-5: - - CN (aq) + H 2O(l) Û OH (aq) + HCN(aq) Generally, OH- > CN- > H2O
  22. 22. ACIDIC SALTSFor any salt whose anion has neutral properties (such as Cl- or NO3-) and whose cation is the conjugate acid of a weak base, the aqueous solution will be acidic.• example: NH4+, CH3NH3+
  23. 23. ACIDIC SALTSA second type of salt that produces an acidic solution is one that contains a highly charged metal ion.• Example: Al3+, Al(H2O)63+Al(H2O)5OH2++ H+• The high charge on the metal ion polarizes the O-H bonds in the water molecule, leaving an acidic solution.• Typically, the higher the charge on the metal ion, the stronger the acidity of the hydrated ion.
  24. 24. PRACTICE PROBLEM 19Calculate the pH of a 0.10 M NH4Cl solution. The Kb value for NH3 is 1.8x10-5.pH = 5.13
  25. 25. PRACTICE PROBLEM 20Calculate the pH of a 0.010 M AlCl3 solution. The Ka value for Al(H2O)63+ is 1.4x10-5.pH= 3.43
  26. 26. CONFLICTING SALTSFor many salts both ions can affect the pH of the aqueous solution.• We can predict whether the solution will be basic, acidic, or neutral by comparing the Ka value for the acidic ion with the Kb value for the basic ion.
  27. 27. CONFLICTING SALTSFor many salts both ions can affect the pH of the aqueous solution.• The larger of the two constants determines the characteristic of the solution: acidic, basic or neutral.• Ka > Kb =acidic• Ka < Kb = basic• Ka = Kb = neutral
  28. 28. PRACTICE PROBLEM 21Predict whether an aqueous solution of each of the following salts will be acidic, basic or neutral.a. NH4C2H3O2b. NH4CNc. Al2(SO4)3a. neutral b. basic c. acidic
  30. 30. STRUCTURE CONSIDERATIONSAny molecule containing a hydrogen atom is potentially an acid. Therefore, there are two main factors that determine if the substance will act like an acid and if so, what relative strength it will have:1. Bond polarities (electronegativity)2. Number of oxygen atoms in the molecule
  31. 31. STRUCTURE CONSIDERATIONSWhen looking at relative bond polarities of binary acids we find: H – F > H – Cl > H – Br > H – I Electronegativity goes down the group, therefore HF is extremely polar and very strong. HF is the weakest of the acids.
  33. 33. STRUCTURE CONSIDERATIONSGenerally, oxyacids increase with strength with an increase in the number of oxygen atoms attached to the central atom.• Example: HClO is a weak acid, but HClO4 is strong.• This happens because the very electronegative oxygen atoms are able to pull electrons away from the O-H bond and weaken it.
  35. 35. STRUCTURE CONSIDERATIONSOxyacids with the H – O – X grouping, The higher the electronegativity of X, the greater the acidity of the molecule.
  37. 37. STRUCTURE CONSIDERATIONSOxyacids behave similarly to hydrated metal ions.• The acidity of the water molecules attached to the metal ion is increased by the attraction of electrons to the positive metal ion.• The greater the charge on the metal ion, the more acidic the hydrated ion becomes.
  38. 38. OXIDE PROPERTIESWhy is NaOH not an acid?• The strenth of the OH bond is stronger than Na+ ability to bond to O.
  39. 39. OXIDE PROPERTIESIf an oxide with an H – O – X group has a highly electronegative X, the H is lost before the OH.If an oxide with an H – O – X group has a low electronegative X, OH- can be formed instead of H+
  40. 40. OXIDE PROPERTIESWhen a covalent oxide dissolves in water, an acidic solution forms. These oxides are called acidic oxides.• SO3(g) + H2O(l) H2SO4(aq)• CO2(g) + H2O(l) H2CO3(aq)
  41. 41. OXIDE PROPERTIESWhen an ionic oxide dissolves in water, a basic solution results. The most ionic oxides, such as those of the Group 1 and 2, produce basic solutions when they are dissolved in water. These oxides are called basic oxides.• CaO(s) + H2O(l) Ca(OH)2(aq)• K2O(s) + H2O(l) 2KOH(aq)
  42. 42. THREE MODELS FOR ACIDS AND BASESArrhenius is the most limiting model and was replaced with a more general (Bronsted- Lowry) model. An even more general model was suggested in the 1920’s.
  43. 43. LEWIS ACID-BASE MODELA Lewis acid is an electron-pair acceptor, and a Lewis base is an electron-pair donor. In other words, a Lewis acid has an empty atomic orbital that can accept (share) an electron pair from a molecule with a lone pair.
  44. 44. LEWIS THEORY AND COMPLEX IONSThe Al3+ ion accepts one electron pair from each of six water molecules to form Al(H2O)63+
  45. 45. LEWIS THEORY AND COVALENT OXIDESSulfur trioxide gains lone pairs from a water molecule.SO3(g) + H 2O(l) Û H 2 SO4(aq)
  46. 46. PRACTICE PROBLEM 22For each reaction, identify the Lewis acid and base.a. Ni2+(aq) + 6NH3(aq)  Ni(NH3)62+(aq)b. H+(aq) + H2O(aq)  H3O+(aq)a. Ni2+ = acid, NH3 = base b. H+ = acid, H2O =base
  47. 47. THE END