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Relationship Between K a and K b

NH 4 + (aq) NH 3 (aq) + H + (aq). (1). (2). NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH – (aq). [NH 3 ] [H + ]. [NH 4 + ] [OH – ]. K a =. K b =. [NH 4 + ]. [NH 3 ]. Relationship Between K a and K b. [NH 3 ] [H + ] [NH 4 + ] [OH – ]. [NH 4 + ] [NH 3 ].

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Relationship Between K a and K b

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  1. NH4+(aq) NH3(aq) + H+(aq) (1) (2) NH3(aq) + H2O(l) NH4+(aq) + OH–(aq) [NH3] [H+] [NH4+] [OH–] Ka = Kb = [NH4+] [NH3] Relationship Between Ka and Kb [NH3] [H+] [NH4+] [OH–] [NH4+] [NH3] Then Ka x Kb = = [H+] [OH–] For an acid and its conj. base: KaKb = Kw = 1.00 x 10–14

  2. Because they are easily calculated from Ka values, reference tables are often “short” on Kb values. Just like pH = –log [H+], pKa = –log Ka –log Kb pKb = If [H+] [OH–] = 1.00 x 10–14 and pH + pOH = 14 –log NUT –log CH then since Ka Kb = 1.00 x 10–14 –log SHOOTER pKa + pKb = 14 (@ 25oC) –log NOCHLE –log BRAIN –log SOUP

  3. HF H++ F– pKa = 3.17 pKb = 10.83 Kb = 1.5 x 10–11 F–+ H2OHF + OH– For hydrofluoric acid, Ka = 6.8 x 10–4. a. Write the formula of the conjugate base. F– b. Write the equation for which Kb applies. c. Write the equation for which Ka applies. d. Find pKa, pKb, and Kb. pKa = –log Ka = –log (6.8 x 10–4) 3.17 pKa + pKb = 14 Ka Kb = 1 x 10–14 (6.8 x 10–4)

  4. Acid-Base Properties of Salt Solutions Because ions can exhibit acid or base properties (i.e., donate or accept protons), salt solutions can be acidic or basic. -- hydrolysis: occurs when ions “react” w/H2O; the H2O is split into H+ and OH– Hydrolysis occurs when the acetate ion in sodium acetate reacts with water to produce acetic acid and hydroxide ion. OAc– + H2O HOAc + OH–

  5. Example: Consider the weak acid HX(aq) H+(aq) + X–(aq) X–+ H2OHX + OH– Anions of weak acids react with water to form OH–, and are thus basic. By the very definition of a weak acid, X– does NOT like to walk alone down the Boulevard of Broken Dreams. Now, say we have X–(aq) in solution from a salt. We’ll get... i.e., X– accepts protons, acting like a base

  6. Ka Ka ACID BASE Kb Kb Anions of strong acids do NOT influence pH. e.g., NO3– (from, say, KNO3) wants NOTHING to do with H+. Anions that still have ionizable protons (e.g., HSO3–) are amphoteric. -- For such substances, the larger of Ka or Kb determines the pH. HSO3– H+ + SO32– Equation for Ka: HSO3– + H2O H2SO3 + OH– Equation for Kb: For “amphoterics…”

  7. HPO42– H+ + PO43– HPO42– + H2O H2PO4– + OH– (a) H3PO4 H+ + H2PO4– (b) H2PO4– H+ + HPO42– (c) HPO42– H+ + PO43– Does Na2HPO4 form an acidic or a basic solution in water? Real question is: “How does HPO42– behave?” Like an acid... ...or a base? From a reference table, we find... Ka1 = 7.5 x 10–3 Ka2 = 6.2 x 10–8 Ka3 = 4.2 x 10–13

  8. (a) H3PO4 H+ + H2PO4– Ka1 = 7.5 x 10–3 (b) H2PO4– H+ + HPO42– Ka2 = 6.2 x 10–8 (c) HPO42– H+ + PO43– Ka3 = 4.2 x 10–13 Ka3 Ka3 Kb2 Compare (c) going forward with (b) going in reverse. “conjugatized” Kb2 4.2 x 10–13 1.6 x 10–7 BIGGER. smaller. Na2HPO4 acts as a base.

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