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Chapter 15

Chapter 15. Applications of Equilibrium. Common Ions. An ion that is present in both An acid and its conjugate base HNO 2 and NaNO 2 A base and its conjugate acid NH 3 and NH 4 Cl This common ion effects the position of the equilibrium and the pH.

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Chapter 15

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  1. Chapter 15 Applications of Equilibrium

  2. Common Ions • An ion that is present in both • An acid and its conjugate base • HNO2 and NaNO2 • A base and its conjugate acid • NH3 and NH4Cl • This common ion effects the position of the equilibrium and the pH

  3. How will the pH of a solution containing 0.10M HNO2 compare to a solution containing 0.10M HNO2 and 0.10M NaNO2? Why?

  4. Points To Remember • In common ion situations you MUST consider the initial concentrations of any common ions in solution before you begin the problem • The larger Ka or Kb will govern the equilibrium

  5. Calculate the pH of a solution containing 0.10M HNO2 and 0.10M NaNO2? (Ka = 4.0x10-4)

  6. Homework • P. 774 #’s 21,23a&d, 34,36

  7. Buffered Solutions • Solutions that resists changes in pH when an acid or base is added • Made of • Weak acid and its conjugate base • Weak base and its conjugate acid • Made by adding both chemicals or reacting

  8. Adding NaF .1M HF

  9. Reacting .1M NaOH .1M HF

  10. Determine the pH of a solution made by dissolving NaHCO3 to 0.25M in 0.0233M H2CO3 (Ka = 4.3x10-7)

  11. Henderson-Hasselbalch Equ. • A way of calculating buffer problems • ONLY USED 4 BUFFERS!!!!!!

  12. HA  H+ + A-

  13. Determine the pH of a solution made by dissolving NaHCO3 to 0.25M in 0.0233M H2CO3 (Ka = 4.3x10-7) Use H-H Equ.

  14. How Do Buffers Work • Strong acid or base is not allowed to build up in solution • Strong acid is replaced by weak acid • Strong base is replaced by weak base • The strong acid or base reacts completely!!! • Then equilibrium kicks in

  15. Acidic Buffers HF  H+ + F- Add NaOH Add HCl

  16. Basic Buffer NH3+ H2O  NH4+ + OH- Add NaOH Add HCl

  17. Determine the pH after 0.20 grams of NaOH is added to 500.mL a solution made by dissolving NaHCO3 to 0.25M in 0.0233M H2CO3 (Ka = 4.3x10-7)

  18. What Makes a Good Buffer? • The best buffers have acids and bases at the same, large concentration • Added acid or base has little effect on pH • When the concentrations are the same pH=pKa

  19. Homework • P. 774 #’s 23d,31,37ab, 40, 42,46,48,49

  20. Titrations • The controlled addition of a chemical of known concentration to a chemical of unknown concentration • Known Chemical – Standard or titrant • Unknown Chemical - Analyte

  21. Titrations are carried out to the equivalence point • Point when mole ratio of chemicals is reached • Often an invisible point • Marked by and indicator • Indicator is a chemical that changes color in different pH’s • Indicator color change marks the endpoint • Hopefully endpoint and equivalence point are the same

  22. Titration Curves • Plotting of the pH of the titration solution as solution is added • Used as a way of determining equivalence and Ka or Kb values • Specific ranges for specific acid/base combinations • Strong/Strong or Strong/Weak

  23. Titration Curve HCl vs NaOH http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html

  24. Titration Curve Problems • Remember the process. Do NOT memorize! • Treat each step as a separate problem • For Strong/Strong systems just stoichiometry • For Strong/Weak First stoichiometry then equilibrium

  25. Strong Acid / Strong Base • Beginning – Just the pH of the chemical you have • Before Equivalence – Stoichiometry • Equivalence – pH is 7.00 • After Equivalence - Stoichiometry

  26. Strong Acid / Strong Base Titration #54 page 776 80.0 mL of 0.100M Ba(OH)2 is titrated with 0.400M HCl 0.00 mL 20.0 mL 30.0 mL 40.0 mL 80.0 mL

  27. Weak / Strong Titration • Beginning – Just the pH of the chemical you have (Might need equilibrium) • Before Equivalence – Stoichiometry then equilibrium • Equivalence – pH is NOT 7 – Stoichiometry then equilibrium • After Equivalence - Stoichiometry

  28. Weak / Strong Combos • Weak Acid / Strong Base • Equivalence Point is ABOVE seven • Conjugate base of a weak acid controls pH • Weak Base / Strong Acid • Equivalence Point is BELOW seven • Conjugate acid of a weak base controls pH • At halfway pH = pKa or pOH = pKb!!!!!!!!

  29. Weak Base / Strong Acid Titration http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html

  30. Weak Acid / Strong Base Titration http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html

  31. Strong Base/ Weak Acid Titration #55 page 776 100.0 mL of 0.200M Acetic Acid is titrated with 0.100M KOH 0.00 mL 50.0 mL 100.0 mL 200.0 mL 250.0 mL

  32. Homework • P. 775 #’s 52,53,56

  33. Which Indicator to Use? • Choosing the appropriate indicator is important when doing titrations • There are so many good ones to use • So how does a chemist know???????

  34. Which Indicator to Use? • Indicators are weak acids and change color with varying pH • Pick the indicator that changes a little above your calculated equivalence point. • Pick indicator with pKa value close to equivalence point • Change color +/- 1 from pKa

  35. Example • Choose an indicator for the titration of HCl with NH3 pH=5.36

  36. Polyprotic Acids • These acids have multiple equivalence points • Carbonic acid has two • Phosphoric acid has three • Multiple step points on a titration curve

  37. Diprotic Acid Titration

  38. Molar Mass of an Acid • You can find the molar mass of an acid by titration too • Titrate a acid of known mass with a standard to determine number of moles • Divided grams by moles • Molar Mass

  39. Solubility Equilibrium • Dissolving is an equilibrium process • Chemicals that we say are insoluble are actually very slightly soluble • Meaning only a small amount of the solute dissolves • Concentrations are very small • 10-5M to 10-20M are common

  40. Solubility Equilibrium • Iron (II) Sulfide is “insoluble” by solubility rules • However, it still dissolved to a small extent • Consider the reaction FeS(s) Fe+2 + S-2 • Since dissolving is an equilibrium process we can write an equilibrium expression • Keq = [Fe+2] [S-2] • FeS is omitted because it is a solid

  41. Solubility Product Constant • The previous equilibriums expression is a “special” type of equilibrium expression • Ksp – solubility product constant • Ksp = [Fe+2] [S-2] • Only used for slightly soluble salts • Never includes the reactants • Ksp’s are very small • Check out the table 15.4 page 753

  42. Ksp Values • Lead (II) Chloride 2.4X10-4 • Strontium Carbonate 3.8X10-9 • Nickel (II) Hydroxide 5.5X10-16 • Copper (II) Hydroxide 2.2X10-20 • Cadmium Sulfide 8.0X10-28 • Silver Sulfide 6.0X10-51 • And my personal favorite • Bismuth Sulfide 1.6x10-72

  43. Ksp Values • Q) What does a small Ksp value mean? • A) Low concentration of ions • Small solubility • Q) Is the compound with the smallest Ksp the least soluble? • A) Not necessarily • There are different numbers of ions that changes the expression

  44. Solubility of Salts • We can easily compare solubility for salts that have the same # of ions • When there are the same # of ions the salt with the smallest Ksp is least soluble • Which salt is least soluble, most soluble AgCl, FeS, BaSO4 • FeS 6.0x10-19 Least soluble • AgCl 1.6x10-10 Most soluble • BaSO4 1.1x10-10 Middle

  45. Calculating Solubility • We can calculate solubility of salts and the concentration of the ions in solutions from Ksp • Deal with saturated solutions • The salt has dissolved as much as it can • Ksp has been reached • In saturated solutions the concentrations of the ions are related to the mole ratio

  46. Example • Calculate the solubility of silver chloride Ksp=1.6x10-10

  47. Example • Calculate the solubility of mercury (II) sulfide and the concentration of each ion in a saturated solution Ksp=1.6x10-54

  48. Example • Calculate the solubility of calcium phosphate Ksp=1.3x10-32

  49. Example • A saturated solution of Iron (III) Hydroxide has a concentration of Fe+3 of 1.8x10-8. What is the Ksp of Iron (III) Hydroxide?

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