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Acid Base Review

Acid Base Review. 2H 2 O  H 3 O + + OH -. K w = 1 x 10 -14 (at 25 ˚C) = [H 3 O + ][OH - ]. pH = - log [H 3 O + ]. pOH = - log [OH - ]. pH + pOH = 14. K a = [H 3 O + ][A - ] [HA]. Weak Monoprotic acids. Generic: HA + H 2 O  H 3 O + + A -.

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Acid Base Review

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  1. Acid Base Review 2H2O  H3O+ + OH- Kw = 1 x 10-14 (at 25˚C) = [H3O+][OH-] pH = - log [H3O+] pOH = - log [OH-] pH + pOH = 14

  2. Ka = [H3O+][A-] [HA] Weak Monoprotic acids Generic: HA + H2O  H3O+ + A- pKa = -log Ka Acid (HA)Conjugate base (A-) Ka HF F- 7.2 x 10-4 HNO2 NO2- 4.5 x 10-4 CH3COOH CH3COO- 1.8 x 10-5 HOCl OCl- 3.5 x 10-8 HCN CN- 4.0 x 10-10 pH and [ ]s of all species in 0.23M HOCl?

  3. Types of problems ….. 1) Ka/Kb and [acid]/[base] to pH and other [ ]s. 2) Ka/Kb and pH to [acid]/[base] + other [ ]s. 3) pH and [acid]/[base] to Ka and other [ ]s. 4) pH of solution containing salts of weak acids (e.g. sodium acetate) or salts of weak bases (e.g. ammonium chloride). Or vice versa.

  4. Kb = [HB+][OH-] [B] Weak bases B + H2O HB+ + OH- Base (B)Conjugate acid (HB+) Kb NH3 NH4+ 1.8 x 10-5 CH3NH2 CH3NH3+ 5.0 x 10-4 C5H5N C5H5NH+ 1.5 x 10-9 C6H5NH2 C6H5NH3+ 4.2 x 10-10 [methylamine] if pH = 9.7?

  5. x = -b ± (b2 – 4ac)1/2 2a What is the pH of a 0.280 M solution of HF? HF + H2O↔ H3O+ + F- Ka = 7.2 x 10-4 = ([H3O+] [F-])/[HF] Let x = [H3O+] …. 7.2 x 10-4 = x2/(0.280 – x) x ~ (0.280 • 7.2 x 10-4)1/2 = (2.016 x 10-4 M)1/2 = 1.42 x 10-2 M = 0.0142 M Is this acceptable error? 0.0142/0.280 • 100 = 5.07% Using quadratic equation x2 = (2.016 x 10-4 – 7.2 x 10-4 x x2 + 7.2 x 10-4x - 2.016 x 10-4 =0 x = 0.0138 M pH = 1.860 (1.848)

  6. Salts of weak acids are bases referred to as the conjugate base 0.15 M NaCH3COO (sodium acetate) Na(CH3COO-)  Na+ + CH3COO- CH3COO- + H2O CH3COOH+ OH- Kb = [OH-][CH3COOH] = ? [CH3COO-] Ka = 1.8 x 10-5 & Kw = KaKb so Kb = Kw/Ka = 5.6 x 10-10 5.6 x 10-10 = x2/(0.15 – x) where x = [OH-] x = 9.16 x 10-6 M (0.6% error) pOH = 5.04 pH = 8.96

  7. Polyprotic acids – An acid with 2 or more acidic protons e.g. Phosphoric acid (H3PO4) H3PO4+ H2O H2PO4- + H3O+ Ka1 = 7.5 x 10-3 H2PO4-+ H2O HPO42- + H3O+ Ka2 = 6.2 x 10-8 HPO42-+ H2O PO43- + H3O+ Ka3 = 3.6 x 10-13 What is pH and [ ]s of all species in 0.270 M H3PO4? Use Equilibrium #1 Ka expression to find [H2PO4-] & [H3O+]. Use Equilibrium #2 Ka expression & [H2PO4-] from #1 to find additional [H3O+] and [HPO4-]. Use Equilibrium #3 Ka expression & [HPO4=] to find [PO43-] & additional [H3O+].

  8. 0.270M Phosphoric acid (H3PO4) H3PO4+ H2O H2PO4- + H3O+ Ka1 = 7.5 x 10-3 H2PO4-+ H2O HPO42- + H3O+ Ka2 = 6.2 x 10-8 HPO42-+ H2O PO43- + H3O+ Ka3 = 3.6 x 10-13 Use Equilibrium #1 Ka expression to find [H2PO4-] & [H3O+]. x12 = (.270 • 7.5 x 10-3 - 7.5 x 10-3x) = [H2PO4-] x1 = -7.5 x 10-3 ± {(7.5 x 10-3)2 – 4 • -2.025 x 10-3}1/2 ÷ 2 =0.0414 M [H3PO4] = 0.23M Use Equilibrium #2 Ka expression & [H2PO4-] from #1 to find additional [H3O+] and [HPO4-]. 6.2 x 10-8 =(0.0414 + x2) • x2÷ (0.0414 – x2) x2 = 6.2 x 10-8 M = [HPO42-] [H3O+] = 0.0414 + 0.0000062 = 0.0414 M Use Equilibrium #3 Ka expression & [HPO4=] to find [PO43-] & additional [H3O+]. 3.6 x 10-13 = (0.0414 + x3) • x3 ÷ (6.2 x 10-8 – x3) x3 = [PO43-] = 5.39 x 10-19 M pH = - log [H3O+] = - log 0.0414 = 1.38

  9. H3PO4+ H2O H2PO4- + H3O+ Ka1 = 7.5 x 10-3 H2PO4-+ H2O HPO42- + H3O+ Ka2 = 6.2 x 10-8 HPO42-+ H2O PO43- + H3O+ Ka3 = 3.6 x 10-13 Colas contain ~ 0.106 g L-1 of H3PO4 (98.0 g mol-1). (for flavor?) Estimate its pH and [ ]s of all species if this is the only acid present. (note that carbonated beverages contain CO2/H2CO3 that is also acidic) 7.5 x 10-3 = [H2PO4-][H3O+]/[H3PO4] = x2/(0.0011 – x) x is not small compared to 0.00110 M: x ~ 0.000974M [H3O+] = [H2PO4-] = 9. 74 x 10-4 M pH = 3.01 [HPO42-] = 6.2 x 10-8 M [PO43-] = 2.3 x 10-17 M

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