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A. Polyprotic Acids Acids which can donate more than one H + . Each “donation” has its own Ka

A. Polyprotic Acids Acids which can donate more than one H + . Each “donation” has its own Ka H 2 SO 4  H + + HSO 4 - HSO 4 -  H + + SO 4 -2 Ka 2 is always much lower than Ka 1 . Much harder to pull a second H + off of an ion.

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A. Polyprotic Acids Acids which can donate more than one H + . Each “donation” has its own Ka

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  1. A. Polyprotic Acids Acids which can donate more than one H+. Each “donation” has its own Ka H2SO4 H+ + HSO4- HSO4-  H+ + SO4-2 Ka2 is always much lower than Ka1. Much harder to pull a second H+ off of an ion. In pH calculation, Ka2 and Ka3 can be ignored. (Calculate pH based on first H+ being removed.) Example – Determine the pH of a 0.1 M carbonic acid (H2CO3) solution. (Ka1 = 4.3 x 10-7, Ka2 = 5.6 x 10-11) H2CO3  H+ + HCO3- Ka1 = very large Ka2 1 x 10-2 Only 1st reaction is important! 0.0 M 0.1 M 0.0 M -x +x +x 0.1 x x

  2. Ka = [H+][HCO3-] [H2CO3] 4.3 x 10-7 = x2 0.1 X = 2.07 x 10-4 pH = -log 2.07 x 10-4 pH = 3.68 B. Weak Bases Bases we will deal with pull H+ off of H2O B + H2O  HB+ + OH- NH3 + H2O  NH4+ + OH- Types 1. Neutral, but have a nonbonding pair of electrons can attract H+ Most important = nitrogen compounds such as amines Equilib constant = Kb

  3. H | X – N – H H H H | | | H - C – C – N – H | | H H Can accept H+ 2. Anions of weak acids NaNO2 Since HNO2 is a weak acid, NO2- in solution absorbs some H+ NO2- + H2O  HNO2 + OH-

  4. Writing reactions of weak acids and bases a. Weak acids HX  H+ + X- or HX + H2O  H3O+ + X- b. Weak bases X + H2O  HX+ + OH- or X- + H2O  HX + OH- Examples – Write out the reaction when the following react with water NH3 (base) HS- (base) H2S (acid) NH3 + H2O  NH4+ + OH- HS- + H2O  H2S + OH- H2S + H2O  HS- + H3O+

  5. Kb Like Ka, Kb describes how far the base reaction runs B + H2O  HB+ + OH- Kb = [HB+] [OH-] [B] Like Ka a high Kb indicates that the substance has acted like a stronger base. Kb can be used in the same manner as Ka Just remember to write the proper reaction!

  6. What is the pH of a 0.015 M CH3NH2 solution, if the chemical has a Kb of 4.4x10-4? CH3NH2 + H2O  CH3NH3+ + OH- Kb = [CH3NH3+] [OH-] [CH3NH2] 4.4 x 10-4 = x2 0.015 X = 2.569 x 10-3 This is the [OH-] 0.015 0 0 -x +x +x x x 0.015 pOH = - log (2.569 x 10-3) pOH = 2.59 pH = 11.41

  7. Example – If the pH of a 2.0 L solution of ammonia is 10.5, how many grams of NH3 were added? (Kb = 1.8 x 10-5)? • Use pH to get [OH-] • 2. Set up chart to determine [NH3] from Keq • 3. Use molarity to get moles of NH3 • 4. Convert moles to g • pH = 10.5 • pOH = 3.5 • [OH-] = 10-3.5 • NH3 + H2O  NH4+ + OH- 3.16 x 10-4 M x M 0.0 M 0.0 M -y +y +y x 3.16 x 10-4 M 3.16 x 10-4 M

  8. 1.85 x 10-5 = (3.16 x 10-4)2 x X = 0.0056 M 0.0056 = x 2 L X = 0.012 moles G = gmm x moles G = 17.04 x 0.012 g = 0.2045 g D. Ka and Kb For a conjugate acid base pair, the Ka and the Kb are related Ka x Kb = Kw = 1 x 10-14 This means that pKa + pKb = 14 (pX = -log x)

  9. Example Calculate the Kb of F- if the Ka of HF is 6.8 x 10-4 Ka x Kb = Kw 6.8 x 10-4 (Kb) = 1 x 10-14 Kb = 1.47 x 10-11

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