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Unit 6: pH Calculations

Unit 6: pH Calculations. Homework: Read pages: 353-357 WS: Concentration and pH. pH. What is Log 10 ? Power to which 10 must be raised to equal that number Log 1000 = ? 3 pH = -log[H + ]. pH. power of [H + ] pH = – log [H + ] pH = power to which 10 must be raised to equal the [H + ]

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Unit 6: pH Calculations

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  1. Unit 6: pH Calculations Homework: Read pages: 353-357 WS: Concentration and pH

  2. pH • What is Log10? • Power to which 10 must be raised to equal that number • Log 1000 = ? • 3 • pH = -log[H+]

  3. pH • power of [H+] • pH = – log [H+] • pH = power to which 10 must be raised to equal the [H+] example: [H+] = 1x10–7 M pH = – log (1x10–7) = 7

  4. Self ionization of water • Because water molecules are polar, they are in continuous random motion H2O +H2O  H3O+ + OH- H3O+ = hydronium ion OH- = hydroxide ion

  5. Simple dissociation of water H2O  H+ + OH- Where [H+] = [OH-] [ ] = concentration (Molarity)

  6. Kw = ion product constant for water At 25˚C the concentration of OH- = concentration of H+ = neutral solution [H+] = 1x10-7 M [OH-] = 1x10-7 M [H+] = [OH-]

  7. Kw = ion product constant for water Kw = [H+] x [OH-] Kw = 1x10-7 x 1x10-7 = 1x10-14

  8. pH H+ pH OH– 7

  9. pH calculation # 1 What is the exact pH of a solution if the [H+] = 1.26 x 10-6? Is the solution acidic, basic or neutral? pH = 5.900 Sig. Figs. only decimal places count

  10. pH calculation # 2 What is the exact pH of a solution if the [OH-] = 4.62 x 10-11? Is the solution acidic, basic or neutral? pH = 3.665 [H+] = 2.16 x 10-4

  11. antilogarithm • Determining the number (exponent) that corresponds to a certain logarithm of base 10. • Antilog or inverse log function (10 must be raised to what power to give that number) • Inverse log of 1.371= ? • 23.5

  12. pH calculation # 3 What is the [H+] of a solution if the pH = 3.7? Is the solution acidic, basic or neutral? [H+] = 2 x 10-4 M

  13. Concentration and pH answer key 1. pH of the following: a. 3.058 b. 12.866 c. 2.063 d. 9.795 2. Concentrations of the following solutions: a. H2SO4= 7.2 x 10-4 M b. HNO3= 5.1 x 10-4 M c. NaOH= 5.3 x 10-3 M d. CsOH= 2.2 x 10-5 M

  14. 3. 12.7 = pH of CsOH 4. 0.764 = pH (AgNO3 = limiting reagent) 5. HCl is limiting reagent ; pH=12.108

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