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Applications of Aqueous Equilibria

This document explores the principles of aqueous equilibria and buffered solutions, emphasizing their role in resisting pH changes upon the addition of acids or bases. It outlines the composition of buffered solutions, including weak acids with their salts and weak bases with their salts, and explains the theoretical basis using the Henderson-Hasselbalch equation. Additionally, it presents the concepts of titration curves for unbuffered and buffered solutions, discusses buffer capacity, and provides calculations related to pH changes in various scenarios.

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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. Titration of an Unbuffered Solution A solution that is 0.10 M CH3COOH is titrated with 0.10 M NaOH

  6. Titration of a Buffered Solution A solution that is 0.10 M CH3COOH and 0.10 M NaCH3COO is titrated with 0.10 M NaOH

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

  8. Comparing Results Buffered Unbuffered

  9. 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)

  10. 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])

  11. This is called the Henderson-Hasselbalch Equation

  12. 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)

  13. Prove they’re buffers • What would the pH be if .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.

  14. 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.

  15. 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

  16. 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

  17. 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

  18. Strong Base/Strong Acid 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.

  19. Weak Base/Strong Acid Titration

  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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