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Chem 2 - Acid-Base Equilibria VII: Conjugate Acid/Base Pairs and Relationships Between Ka, Kb, and Kw

Chem 2 - Acid-Base Equilibria VII: Conjugate Acid/Base Pairs and Relationships Between Ka, Kb, and Kw

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Chem 2 - Acid-Base Equilibria VII: Conjugate Acid/Base Pairs and Relationships Between Ka, Kb, and Kw

  1. 1. Acid-Base Equilibria (Pt. 7) Conjugate Acid/Base Pairs and Relationships Between Ka, Kb, and Kw By Shawn P. Shields, Ph.D. This work is licensed by Dr. Shawn P. Shields-Maxwell under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
  2. 2. Recall: Conjugate Acid-Base Pairs Conjugate acid-base pairs are two related species differing only by a proton (H+). 𝐇𝐅 𝐚πͺ + 𝐇 𝟐 𝐎 π₯ β‡Œ 𝐇 πŸ‘ 𝐎+ 𝐚πͺ + π…βˆ’ (𝐚πͺ) 𝐍𝐇 πŸ‘ 𝐚πͺ + 𝐇 𝟐 𝐎 π₯ β‡Œ 𝐍𝐇 πŸ’ + 𝐚πͺ + πŽπ‡βˆ’ (𝐚πͺ) base conjugate acid for NH3 acid conjugate base for HF
  3. 3. The equilibrium constant K is β€œrenamed” for acids to Ka The Equilibrium Constant Ka for Weak Acids An equilibrium exists between the weak acid (HA) and its products. 𝐇𝐀 𝐚πͺ + 𝐇 𝟐 𝐎 π₯ β‡Œ 𝐇 πŸ‘ 𝐎+ 𝐚πͺ + π€βˆ’ (𝐚πͺ) 𝐊 𝐚 = 𝐇 πŸ‘ 𝐎+ π€βˆ’ 𝐇𝐀
  4. 4. Ka and Kb for Conjugate Acid-Base Pairs Suppose acetic acid (CH3COOH) is dissolved in water… 𝐂𝐇 πŸ‘ π‚πŽπŽπ‡ 𝐚πͺ + 𝐇 𝟐 𝐎 π₯ β‡Œ 𝐂𝐇 πŸ‘ π‚πŽπŽβˆ’ 𝐚πͺ + 𝐇 πŸ‘ 𝐎+ 𝐚πͺ acetic acid conjugate base for acetic acid β€œloseable” H+
  5. 5. Ka and Kb for Conjugate Acid-Base Pairs Suppose acetate (CH3COO-) is dissolved in water… 𝐂𝐇 πŸ‘ π‚πŽπŽβˆ’ + 𝐇 𝟐 𝐎 π₯ β‡Œ 𝐂𝐇 πŸ‘ π‚πŽπŽπ‡ 𝐚πͺ + πŽπ‡βˆ’ (𝐚πͺ) acetate conjugate acid for acetate acetate gained H+ from water
  6. 6. Ka and Kb for Conjugate Acid-Base Pairs CH3COOH and CH3COO- are conjugate acid-base pairs acetic acid acetate (base) Ka = 1.76 ο‚΄ 10-5 Kb = 5.68 ο‚΄ 10-10
  7. 7. Multiply Ka and Kb for Conjugate Acid-Base Pairs 𝐊 𝐚 βˆ™ 𝐊 𝐛= 𝟏. πŸ•πŸ” Γ— πŸπŸŽβˆ’πŸ“ [πŸ“. πŸ”πŸ– Γ— πŸπŸŽβˆ’πŸπŸŽ ] 𝐊 𝐚 βˆ™ 𝐊 𝐛= 𝟏 Γ— πŸπŸŽβˆ’πŸπŸ’ = 𝐊 𝐖 𝐊 𝐚 βˆ™ 𝐊 𝐛 = 𝐊 𝐖 Use this relationship to interconvert between Ka and Kb.
  8. 8. Calculating (Converting Between) Ka and Kb using Kw 𝐊 𝐚 βˆ™ 𝐊 𝐛 = 𝐊 𝐖 𝐊 𝐚 βˆ™ 𝐊 𝐛= 𝟏. 𝟎 Γ— πŸπŸŽβˆ’πŸπŸ’ = 𝐊 𝐖 𝐊 𝐚 = 𝐊 𝐖 𝐊 𝐛 𝐊 𝐛 = 𝐊 𝐖 𝐊 𝐚 Use KW to interconvert between Ka and Kb.
  9. 9. Calculating (Converting Between) Ka and Kb using Kw 𝐊 𝐚 βˆ™ 𝐊 𝐛= 𝟏. πŸ•πŸ” Γ— πŸπŸŽβˆ’πŸ“ [πŸ“. πŸ”πŸ– Γ— πŸπŸŽβˆ’πŸπŸŽ ] 𝐊 𝐚 βˆ™ 𝐊 𝐛= 𝟏. 𝟎 Γ— πŸπŸŽβˆ’πŸπŸ’ = 𝐊 𝐖 𝐊 𝐚 = 𝟏.πŸŽΓ—πŸπŸŽβˆ’πŸπŸ’ πŸ“.πŸ”πŸ–Γ—πŸπŸŽβˆ’πŸπŸŽ 𝐊 𝐛 = 𝟏.πŸŽΓ—πŸπŸŽβˆ’πŸπŸ’ 𝟏.πŸ•πŸ”Γ—πŸπŸŽβˆ’πŸ“ Use KW to interconvert between Ka and Kb.
  10. 10. pKa and pKb Calculate the pKa for an acid as pKa = ο€­ log Ka Calculate the pKb for a base as pKb = ο€­ log Kb
  11. 11. Calculating pKa and pKb Calculate the pKa pKa = ο€­ log Ka 𝐩𝐊 𝐚 = βˆ’π₯𝐨𝐠 𝟏. πŸ•πŸ” Γ— πŸπŸŽβˆ’πŸ“ = πŸ’. πŸ•πŸ“ Calculate the pKb pKb = ο€­ log Kb 𝐩𝐊 𝐛 = βˆ’π₯𝐨𝐠 πŸ“. πŸ”πŸ– Γ— πŸπŸŽβˆ’πŸπŸŽ = πŸ—. πŸπŸ“
  12. 12. Inverse logs for pKa, pKa, and pKw pKa = ο€­ log Ka 10ο€­pKa = Ka pKb = ο€­ log Kb 10ο€­pKb = Kb pKw = ο€­ log Kw = 14 10ο€­pKw = Kw Recall: Kw (and pKw) will have different values at temperatures other than 25ο‚°C
  13. 13. A Few More Relationships Between pKa, pKb, and pKw pKa + pKb = pKw pKw = ο€­ log [Kw] = 14 (at 25ο‚°C) pKa + pKb = 14 (at 25ο‚°C)
  14. 14. Next up, The Conjugate See-Saw and Analyzing Ka and Kb for Acid or Base Relative Strength (Pt. 8)

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