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Applications of Aqueous Equilibria

Applications of Aqueous Equilibria. Buffered Solutions. A solution that resists a change in pH when either hydroxide ions or protons are added. Buffered solutions contain either: A weak acid and its salt A weak base and its salt. Acid/Salt Buffering Pairs.

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Applications of Aqueous Equilibria

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  1. Applications of Aqueous Equilibria

  2. Buffered Solutions • A solution that resists a change in pH when either hydroxide ions or protons are added. • Buffered solutions contain either: • A weak acid and its salt • A weak base and its salt

  3. Acid/Salt Buffering Pairs The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)

  4. Base/Salt Buffering Pairs The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3)

  5. Weak Acid/Strong Base Titration A solution that is 0.10 M CH3COOH is titrated with 0.10 M NaOH Endpoint is above pH 7

  6. Strong Acid/Strong Base Titration Endpoint is at pH 7 A solution that is 0.10 M HCl is titrated with 0.10 M NaOH

  7. Strong Acid/Strong Base Titration A solution that is 0.10 M NaOH is titrated with 0.10 M HCl Endpoint is at pH 7 It is important to recognize that titration curves are not always increasing from left to right.

  8. Strong Acid/Weak Base Titration A solution that is 0.10 M HCl is titrated with 0.10 M NH3 Endpoint is below pH 7

  9. Titration of an Unbuffered Solution A solution that is 0.10 M HC2H3O2 is titrated with 0.10 M NaOH

  10. Titration of a Buffered Solution A solution that is 0.10 M HC2H3O2 and 0.10 M NaC2H3O2 is titrated with 0.10 M NaOH

  11. Comparing Results Unbuffered Buffered • In what ways are the graphs different? • In what ways are the graphs similar?

  12. Comparing Results Buffered Unbuffered

  13. Buffer capacity • The best buffers have a ratio [A-]/[HA] = 1 • This is most resistant to change • True when [A-] = [HA] • Make pH = pKa (since log1=0)

  14. General equation • Ka = [H+] [A-] [HA] • so [H+] = Ka [HA] [A-] • The [H+] depends on the ratio [HA]/[A-] • taking the negative log of both sides • pH = -log(Ka [HA]/[A-]) • pH = -log(Ka)-log([HA]/[A-]) • pH = pKa + log([A-]/[HA])

  15. This is called the Henderson-Hasselbalch Equation

  16. Using the Henderson-Hasselbalch Equation • pH = pKa + log([A-]/[HA]) • pH = pKa + log(base/acid) • Calculate the pH of the following mixtures • 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4) • 0.25 M NH3 and 0.40 M NH4Cl • (Kb = 1.8 x 10-5)

  17. Prove they’re buffers • What would the pH be if 0.020 mol of HCl is added to 1.0 L of both of the following solutions. • 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4) • 0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8 x 10-5) • What would the pH be if 0.050 mol of solid NaOH is added to each of the proceeding. To do this I must introduce the BAAM table!

  18. Buffer capacity • The pH of a buffered solution is determined by the ratio [A-]/[HA]. • As long as this doesn’t change much the pH won’t change much. • The more concentrated these two are the more H+ and OH- the solution will be able to absorb. • Larger concentrations bigger buffer capacity.

  19. Buffer Capacity • Calculate the change in pH that occurs when 0.010 mol of HCl(g) is added to 1.0L of each of the following: • 5.00 M HAc and 5.00 M NaAc • 0.050 M HAc and 0.050 M NaAc • Ka= 1.8x10-5

  20. Summary • Strong acid and base just stoichiometry. • Determine Ka, use for 0 mL base • Weak acid before equivalence point • Stoichiometry first • Then Henderson-Hasselbach • Weak acid at equivalence point Kb • Weak base after equivalence - leftover strong base.

  21. Summary • Determine Ka, use for 0 mL acid. • Weak base before equivalence point. • Stoichiometry first • Then Henderson-Hasselbach • Weak base at equivalence point Ka. • Weak base after equivalence - leftover strong acid.

  22. Selection of Indicators

  23. Solubility Equilibria Will it all dissolve, and if not, how much?

  24. All dissolving is an equilibrium. • If there is not much solid it will all dissolve. • As more solid is added the solution will become saturated. • Solid dissolved • The solid will precipitate as fast as it dissolves . • Equilibrium

  25. Watch out • Solubility is not the same as solubility product. • Solubility product is an equilibrium constant. • it doesn’t change except with temperature. • Solubility is an equilibrium position for how much can dissolve. • A common ion can change this.

  26. Ksp Values for Some Salts at25C

  27. Solving Solubility Problems For the salt AgI at 25C, Ksp = 1.5 x 10-16 AgI(s)  Ag+(aq) + I-(aq) O O +x +x x x 1.5 x 10-16 = x2 x = solubility of AgI in mol/L = 1.2 x 10-8 M

  28. Solving Solubility Problems For the salt PbCl2 at 25C, Ksp = 1.6 x 10-5 PbCl2(s)  Pb2+(aq) + 2Cl-(aq) O O +2x +x 2x x 1.6 x 10-5 = (x)(2x)2 = 4x3 x = solubility of PbCl2 in mol/L = 1.6 x 10-2 M

  29. Relative solubilities • Ksp will only allow us to compare the solubility of solids the that fall apart into the same number of ions. • The bigger the Ksp of those the more soluble. • If they fall apart into different number of pieces you have to do the math.

  30. The Common Ion Effect • When the salt with the anion of a weak acid is added to that acid, • It reverses the dissociation of the acid. • Lowers the percent dissociation of the acid. • The same principle applies to salts with the cation of a weak base.. • The calculations are the same as last chapter.

  31. Solving Solubility with a Common Ion For the salt AgI at 25C, Ksp = 1.5 x 10-16 What is its solubility in 0.05 M NaI? AgI(s)  Ag+(aq) + I-(aq) 0.05 O 0.05+x +x 0.05+x x 1.5 x 10-16 = (x)(0.05+x)  (x)(0.05) x = solubility of AgI in mol/L = 3.0 x 10-15 M

  32. Precipitation and Qualitative Analysis

  33. pH and solubility • OH- can be a common ion. • More soluble in acid. • For other anions if they come from a weak acid they are more soluble in a acidic solution than in water. • CaC2O4 Ca+2 + C2O4-2 • H+ + C2O4-2 HC2O4- • Reduces C2O4-2 in acidic solution.

  34. Precipitation • Ion Product, Q =[M+]a[Nm-]b • If Q>Ksp a precipitate forms. • If Q<Ksp No precipitate. • If Q = Ksp equilibrium. • A solution of 750.0 mL of 4.00 x 10-3M Ce(NO3)3 is added to 300.0 mL of 2.00 x 10-2M KIO3. Will Ce(IO3)3 (Ksp= 1.9 x 10-10M)precipitate and if so, what is the concentration of the ions?

  35. Selective Precipitations • Used to separate mixtures of metal ions in solutions. • Add anions that will only precipitate certain metals at a time. • Used to purify mixtures. • Often use H2S because in acidic solution Hg+2, Cd+2, Bi+3, Cu+2, Sn+4 will precipitate.

  36. Selective Precipitation • In Basic adding OH-solution S-2 will increase so more soluble sulfides will precipitate. • Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3, Al(OH)3

  37. Selective precipitation • Follow the steps first with insoluble chlorides (Ag, Pb, Ba) • Then sulfides in Acid. • Then sulfides in base. • Then insoluble carbonate (Ca, Ba, Mg) • Alkali metals and NH4+ remain in solution.

  38. Complex Ions A Complex ion is a charged species composed of: 1. A metallic cation 2. Ligands – Lewis bases that have a lone electron pair that can form a covalent bond with an empty orbital belonging to the metallic cation

  39. NH3, CN-, and H2O are Common Ligands

  40. The addition of each ligand has its own equilibrium • Usually the ligand is in large excess. • And the individual K’s will be large so we can treat them as if they go to equilibrium. • The complex ion will be the biggest ion in solution.

  41. Coordination Number • Coordination number refers to the number of ligands attached to the cation • 2, 4, and 6 are the most common coordination numbers

  42. Complex Ions and Solubility AgCl(s)  Ag+ + Cl- Ksp = 1.6 x 10-10 Ag+ + NH3 Ag(NH3)+ K1 = 1.6 x 10-10 Ag(NH3)+ NH3 Ag(NH3)2+ K2 = 1.6 x 10-10 K = KspK1K2 AgCl + 2NH3 Ag(NH3)2+ + Cl-

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