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Chapter 6 Problems

Chapter 6 Problems. 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,. Outline. Equilibrium of Acids and Bases Bronsted-Lowry Acids/Bases Define strong Define weak pH of pure water at 25 o C Define K a and K b Relationship b/w K a and K b Chapter 8 – Activity Relationship with K.

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Chapter 6 Problems

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  1. Chapter 6 Problems • 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

  2. Outline • Equilibrium of Acids and Bases • Bronsted-Lowry Acids/Bases • Define strong • Define weak • pH of pure water at 25oC • Define Ka and Kb • Relationship b/w Ka and Kb • Chapter 8 – Activity • Relationship with K

  3. Acids and Bases & Equilibrium Section 6-7

  4. Strong Bronsted-Lowry Acid • A strong Bronsted-Lowry Acid is one that donates all of its acidic protons to water molecules in aqueous solution. (Water is base – electron donor or the proton acceptor).

  5. Strong Bronsted-Lowry Base • Accepts protons from water molecules to form an amount of hydroxide ion, OH-, equivalent to the amount of base added.

  6. Question • Can you think of a salt that when dissolved in water is not an acid nor a base?

  7. Weak Bronsted-Lowry acid • One that DOES not donate all of its acidic protons to water molecules in aqueous solution. • Example?

  8. Weak Bronsted-Lowry base • Does NOT accept an amount of protons equivalent to the amount of base added, so the hydroxide ion in a weak base solution is not equivalent to the concentration of base added. • example:

  9. Common Classes of Weak Acids and Bases Weak Acids Weak Bases

  10. Equilibrium and Water Question: Calculate the Concentration of H+ and OH- in Pure water at 250C.

  11. EXAMPLE: Calculate the Concentration of H+ and OH- in Pure water at 250C. H2O H+ + OH- Kw = KW=

  12. EXAMPLE: Calculate the Concentration of H+ and OH- in Pure water at 250C. H2O H+ + OH- Kw = KW=

  13. Example • What is the concentration of OH- in a solution of water that is 1.0 x 10-3 M in [H+] (@ 25 oC)? “From now on, assume the temperature to be 25oC unless otherwise stated.” Kw = [H+][OH-] 1.0 x 10-14 = [1 x 10-3][OH-] 1.0 x 10-11 = [OH-]

  14. pH ~ -3 -----> ~ +16 pH + pOH = - log Kw = pKw = 14.00

  15. HA + H2O(l) H3O+ + A- Weak Acids and Bases Ka HA H+ + A-

  16. Weak Acids and Bases Kb B + H2O BH+ + OH-

  17. Relation Between Ka and Kb

  18. Kb NH3 + H2O NH4+ + OH- Ka NH4+ + H2O NH3 + H3O+ Relation between Ka and Kb • Consider Ammonia and its conjugate acid.

  19. Example The Ka for acetic acid is 1.75 x 10-5. Find Kb for its conjugate base.

  20. Example • Calculate the hydroxide ion concentration in a 0.0100 M sodium hypochlorite solution. OCl- + H2O  HOCl + OH- The acid dissociation constant = 3.0 x 10-8

  21. 1st Insurance Problem Challenge on page 120

  22. Chapter 8 Activity

  23. Write out the equilibrium constant for the following expression Fe3+ + SCN-D Fe(SCN)2+ Q: What happens to K when we add, say KNO3 ?

  24. Keq K decreases when an inert salt is added!!! Why?

  25. 8-1 Effect of Ionic Strength on Solubility of Salts • Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions. Hg2(IO3)2(s)D Hg22+ + 2IO3- Ksp=1.3x10-18 some - - -x +x +2x some-x +x +2x I C E A seemingly strange effect is observed when a salt such as KNO3 is added. As more KNO3 is added to the solution, more solid dissolves until [Hg22+] increases to 1.0 x 10-6 M. Why?

  26. Increased solubility • Why? • LeChatelier’s Principle? • Complex Ion? • ?

  27. The potassium hydrogen tartrate example

  28. Alright, what do we mean by Ionic strength? • Ionic strength is dependent on the number of ions in solution and their charge. Ionic strength (m) = ½ (c1z12+ c2z22 + …) Or Ionic strength (m) = ½ S cizi2

  29. Examples • Calculate the ionic strength of (a) 0.1 M solution of KNO3and (b) a 0.1 M solution of Na2SO4 (c) a mixture containing 0.1 M KNO3 and 0.1 M Na2SO4. (m) = ½ (c1z12+ c2z22 + …)

  30. Alright, that’s great but how does it affect the equilibrium constant? • Activity = Ac = [C]gc • AND

  31. Relationship between activity and ionic strength Debye-Huckel Equation m = ionic strength of solution g = activity coefficient Z = Charge on the species x a = effective diameter of ion (nm) 2 comments • What happens to g when m approaches zero? • Most singly charged ions have an effective radius of about 0.3 nm Anyway … we generally don’t need to calculate g – can get it from a table

  32. Activity coefficients are related to the hydrated radius of atoms in molecules

  33. Relationship between m and g

  34. Back to our original problem • Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions. Hg2(IO3)2(s)D Hg22+ + 2IO3- Ksp=1.3x10-18

  35. Back to our original problem • Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions. Hg2(IO3)2(s)D Hg22+ + 2IO3- Ksp=1.3x10-18 In 0.1 M KNO3 - how much Hg22+ will be dissolved?

  36. Back to our original problem • Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions. Hg2(IO3)2(s)D Hg22+ + 2IO3- Ksp=1.3x10-18

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