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Buffers

Buffers. solutions that resist pH changes. addition of. acid. base. contain. acidic component. H A. + OH -. H 2 O + A -. . . basic component. HA. A -. + H +. conjugate pair. weak acid + conjugate base. weak base + conjugate acid. [ A - ]. [ H A]. [C - ]. [HC].

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Buffers

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  1. Buffers solutions that resist pH changes addition of acid base contain acidic component HA + OH- H2O + A-   basic component HA A- + H+ conjugate pair weak acid + conjugate base weak base + conjugate acid

  2. [A-] [HA] [C-] [HC] Henderson-Hasselbalch Equation pH = pKa + log [A-] best buffering  [A-] [HA] [HA] 10:1 maximum [A-] = [HA] pH = pKa  pH = 1 pKa adjust pH by changing H+ + C- HC  If you want a buffer at pH = 8.60 10-8.6 = 2.5 x 10-9 8.60 = 8.59 + log [C-] a) HA Ka = 2.7 x 10-3 [HC] Ka = 4.4 x 10-6 b) HB = 1.02 Ka = 2.6 x 10-9 8.59 c) HC

  3. HA Calculate the pH of a buffer 1.0 M CHOOH 1.0 M KCHOO A- Ka = 1.8 x 10-4 [A-] (0.9) pH = pKa + log = 3.74 + = 3.74 + log = 3.65 log [1.00] = 3.74 [HA] [1.00] (1.1) after addition 0.1 mole of HCl to 1.0 L What is the pH  HCl H+ CHOO- base H+ + Cl- acid reacts with  + H+ CHOO- CHOOH [CHOOH] [CHOO-] [H+] in H2O I 1.1 0.9 0.0 pH = 1 -x +x +x C E 1.1 - x 0.9 + x x

  4. Preparation of a buffer of specific pH [A-] pH = pKa + log “phosphate buffer” pH = 7.4 [HA] 10-7.4 = 3.98 x 10-8 H3PO4 H2PO4- + H+ Ka1 = 7.5 x 10-3  pKa = 7.21 H2PO4- HPO42- + H+ Ka2 = 6.2 x 10-8  HPO42- PO43- + H+ Ka3 = 4.8 x 10-13  dissolve NaH2PO4 and Na2HPO4 in water 7.4 = -log (6.2 x 10-8) + log [HPO42-] [H2PO4-] 0.19 = log [HPO42-] [HPO42-] [H2PO4-] 1.55 = [H2PO4-]

  5. Buffers + salt of conjugate base 1. Mix weak acid Mix weak base + salt of conjugate acid Partial neutralization of weak acid with strong base 2. weak base with strong acid there must be an excess of weak acid (weak base) assume that all of the strong base reacts forming stoichiometric amount of A- leaving unreacted HA

  6. x 1.0 mol x 0.1 mol L L Calculate the pH of a buffer prepared by mixing: 40.0 mL of 1.0 M C2H5OOH Ka = 1.3 x 10-5 60.0 mL of 0.1 M NaOH mol C2H5OOH = 0.04 L = 0.04 mol mol OH- = 0.06 L = 0.006 mol C2H5OOH + OH-  C2H5OO- + H2O mol C2H5OOH = 0.040 = 0.034 - 0.006 mol C2H5OO- = volume = 0.006 0.100 L [C2H5OOH] = 0.34 M [C2H5OO-] = 0.06 M

  7. Calculate the pH of a buffer prepared by mixing: 40.0 mL of 1.0 M C2H5OOH Ka = 1.3 x 10-5 60.0 mL of 0.1 M NaOH pKa = - log (1.3 x 10-5) pH = + log [C2H5COO-] pKa [C2H5OOH] pH = + log = 4.14 4.89 0.06 0.34 3.89 < 0.16 (0.1 – 10) < 5.89 [C2H5OOH] = 0.34 M [C2H5OO-] = 0.06 M

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