1 / 9

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 -.

wiley
Download Presentation

Acid Base Review

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related