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UNIT IV

UNIT IV. pH. Ionization of Water. Pure distilled water still has a small conductivity. Why? 2H 2 O (l) + 59kJ H 3 O + (aq) + OH - (aq). Ionization of Water.

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UNIT IV

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  1. UNIT IV pH

  2. Ionization of Water Pure distilled water still has a small conductivity. Why? 2H2O(l) + 59kJ H3O+(aq) + OH-(aq)

  3. Ionization of Water 2H2O(l) + 59kJ H3O+(aq) + OH-(aq) • In neutral water… [H3O+] = [OH-] • In acidic solutions… [H3O+] > [OH-] • In basic solutions… [OH-] > [H3O+] Keq = [H3O+] [OH] or... Kw = [H3O+] [OH-]

  4. Ionization of Water 2H2O(l) + 59kJ H3O+(aq) + OH-(aq) • Since reaction is endothermic: • At higher temperatures ______________are favoured and Kw is _____________er. •  At lower temperatures ______________ are favoured and Kw is _____________er.

  5. Ionization of Water Always: [H3O+] [OH-] = Kw At 250C only: [H3O+] [OH-] = 1.00 x 10-14

  6. [H3O+] & [OH-] in Neutral Water At 25oC:(NOTE: Assume T = 25oC unless otherwise noted) • [H3O+] [OH-] = 1.00 x 10-14 • and [H3O+] = [OH-] if water is neutral Substitute [H3O+] for [OH-]: •  [H3O+] [H3O+] = 1.00 x 10-14 •  [H3O+]2 = 1.00 x 10-14 •  √[H3O+] = √1.00 x 10-14 = 1.00 x 10-7 M • Also [OH-] = [H3O+] = 1.00 x 10-7 M

  7. [H3O+] & [OH-] in Neutral Water Example Given: Kw at 600C = 9.55 x 10-14. Calculate [H3O+] & [OH-] at 600C.

  8. [H3O+] & [OH-] in Acids and Bases 2H2O(l) H3O+(aq) + OH-(aq) • Add acid, H3O+increases, so equilibrium shifts LEFT and [OH-] decreases. • Add base, [OH-] increases, so the equilibrium shifts LEFT and [H3O+] decreases.

  9. [H3O+] & [OH-] in Acids and Bases Ex. Find the [OH-] in 0.0100 M HCl. Ex.Find [H3O+] in 0.300 M NaOH.

  10. [H3O+] & [OH-] in Acids and Bases Ex. Find [H3O+] in 0.020 M Ba(OH)2 . Ex. Calculate [OH-] in 0.00600 M HNO3 at 600C. Kw at 600C = 9.55 x 10-14 • Hebden Textbook Page 127 Questions #28-30

  11. pH • Shorthand method of showing acidity (or basicity/alkalinity) pH= -log[H3O+] • If [H3O+] = 1.0 x 10-7 • pH = -log (1.0 x 10-7 ) = 7

  12. Converting [H3O+] to pH Ex.Find the pH of 0.030 M HCl. **In pH or pOH notation, only the digits after the decimal are significant digits. The digits before the decimal come from the “non-significant” power of 10 in the original [H3O+].

  13. Converting [H3O+] to pH Ex.Find the pH of 0.00100 M NaOH at 250 C. Ex. Find the pH of neutral water at 250 C.

  14. pH Scale • In neutral water pH =7.0 • In acid solution pH < 7.0 • In basic solution pH > 7.0

  15. Converting pH to [H3O+] [H3O+] = antilog (-pH) Ex.) If pH = 11.612 , find [H3O+]. Ex.) If pH = 3.924 calculate [H3O+].

  16. Logarithmic Nature of pH • A change of 1 pH unit  a factor of 10 in [H3O+] (or acidity). • How many times more acidic is pH 3 than pH 7? • Natural rainwater pH ~ 6 • Extremely acidic acid rain pH ~ 3 • So, acid rain is 1000 times more acidic than natural rain water!

  17. pOH pOH = -log [OH-] [OH-] = antilog (-pOH)

  18. pOH Ex.Calculate the pOH of 0.0020M KOH. Ex. Find the pH of the same solution. Notice: pH + pOH = 14.00

  19. Relation of pH to pOH Since... [H3O+] [OH-] = Kw -log[H3O+ ] + -log [OH- ] = -log Kw pH + pOH = pKw where pKw = -log Kw (definition of pKw) TRUE AT ALL TEMPERATURES!

  20. Relation of pH to pOH Specifically at 250C: Kw = 1.00 x 10-14 pKw = -log (1.00 x 10-14) pKw = 14.000 pH + pOH = 14.000 TRUE AT 25°C!

  21. pH and pOH Ex.Find the pH of 5.00 x 10-4 M LiOH (250C). Ex.Find the pOH of 0.0300 M HBr (250C). See pOH scale & pH scale on page 140 of Hebden Textbook.

  22. pH and pOH When not at 250C: Ex. At 600C Kw = 9.55 x 10-14. Find the pH of neutral water at 600C. 

  23. pH and pOH • Is pH always 7.00 in neutral water?_________ • At higher temperatures: 2H2O + heat  H3O+ + OH- • [H3O+] > 1.0 x 10-7 so pH < 7 • [OH-] > 1.0 x 10-7 so pOH < 7

  24. Summary of pH and pOH These are very important? Make sure you study these! • In neutral water pH = pOH at any temperature • pH & pOH = 7.00 at 250C only • At lower temps pH and pOH are > 7 • At higher temps pH and pOH are < 7 • At any temperature: pH + pOH = pKw • At 250C: pH + pOH = 14.000 • Hebden Textbook Pages 139-141 Questions #49-53, 55-57

  25. Summary of pH and pOH Kw = [H3O+][OH-] [H3O+] [OH-] pH =-log[H3O+] pOH = -log[OH-] [H3O+] = antilog(-pH) [OH-] = antilog(-pH) pH pOH pH + pOH = 14

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