Buffer Solutions. Chapter 16 Additional Aqueous Equilibria. Buffer Action. Buffer Action. The ph of Buffer Solutions.

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1 Buffer Solutions John W. Moore Conrad L. Stanitski Peter C. Jurs Chapter 16 Additional Aqueous Equilibria Buffer = chemical system that resists changes in ph. Add mol of HCl or NaOH to: ph 1 L Solution Initial after HCl after NaOH pure H 2 O [CH 3 COOH] = 0.5 M [CH 3 COONa] = 0.5 M Stephen C. Foster Mississippi State University A buffered solution Buffer Action Buffers must contain: A weak acid to react with any added base. A weak base to react with any added acid. These components must not react with each other. Buffers are made with ~equal quantities of a conjugate acid-base pair. (e.g. CH 3 COOH + CH 3 COONa) Buffer Action Added OH - removed by the acid: CH 3 COOH + OH - CH 3 COO - + H 2 O Added H 3 O + removed by the conjugate base: CH 3 COO - + H 3 O + CH 3 COOH + H 2 O The equilibrium maintains the acid/base ratio. CH 3 COOH + H 2 O CH 3 COO - + H 3 O + ph remains stable. Buffer Action Blood ph Blood is buffered at ph = 7.40 ± ph too low: acidosis. ph too high: alkalosis. CO 2 generates the most important blood buffer. CO 2 + H 2 O(l) H 2 CO 3 + H 2 O(l) H 2 CO 3 H 3 O + + HCO 3- The ph of Buffer Solutions Depends on the [acid]/[base] not absolute amounts. Henderson-Hasselbalch Hasselbalch equation [H 3 O + ] = K a log [H 3 O + ] = log K a + log ph = pk a + log With pk a = -log K a Note: ph = pk a when = 1

2 The ph of Buffer Solutions What is the ph of a M (monoprotic) pyruvic acid M sodium pyruvate buffer? K a = 3.2 x ph = log(3.2 x 10-3 ) + log(0.060/0.050) = = 2.57 What is the HPO 4 /H 2 PO 4- ratio in blood at ph =7.40. K a (H 2 PO 4- ) = K a,2 (H 3 PO 4 ) = 6.2 x = log(6.2 x 10-8 ) + log([hpo 4 ]/[H 2 PO 4- ]) log([hpo 4 ]/[H 2 PO 4- ]) = [HPO 4 ]/[H 2 PO 4- ] = 1.5 Common Buffers ph weak acid weak base K a (weak acid) pk a 4 lactic acid lactate ion 1.4 x acetic acid acetate ion 1.8 x carbonic acid hydrogen carbonate ion 4.2 x dihydrogen phosphate hydrogen phosphate ion 6.2 x hypochlorous acid hypochlorite ion 3.5 x ammonium ion ammonia 5.6 x hydrogen carbonate carbonate ion 4.8 x Useful buffer range: ph = pk a ±1 (10:1 or 1:10 ratio of /). Addition of Acid or Base to a Buffer 1.0 L of buffer is prepared with [NaH 2 PO 4 ] = 0.40 M and [Na 2 HPO 4 ] = 0.25 M. Calculate the ph of: (a) the buffer (b) after 0.10 mol of NaOH is added. K a (H 2 PO 4- ) = 6.2 x 10-8 pk a = -log(6.2 x 10-8 ) = 7.21 (a) No base added: ph = pk a + log ph = log 0.25 = Addition of Acid or Base to a Buffer 1.00 L Buffer: [NaH 2 PO 4 ]= 0.40 M ; [Na 2 HPO 4 ] = 0.25 M. (b) calculate ph after 0.10 mol of NaOH is added. K a (H 2 PO 4- ) = 6.2 x 10-8 (b) 0.10 mol of NaOH, converts conj. acid to base: ph = pk a + log H 2 PO OH - ph = log 0.35 = HPO 4 + H 2 O n initial n added 0.10 n after ( ) 0 ( ) Use n directly. [ ] = n/ V and V is the same for both (cancels) Buffer Capacity The amount of acid (or base) that can be added without large ph changes. / determines the buffer ph. Magnitude of and determine buffer capacity The buffer capacity for: Base addition = n conjugate acid Acid addition = n conjugate base Acid-Base Titrations Standard solution (titrant) is added from a buret. The equivalence point occurs when a stoichiometric amount of titrant has been added. Use a ph meter. An indicator is used to find an end point. Color change observed. End point equivalence point (should be close...). n titrant = n analyte n titrant = V titrant [titrant] n analyte = V analyte [analyte] 2

3 Detection of the Equivalence Point Acid-Base Indicator = weak acid that changes color with changes in ph. HIn + H 2 O(l) H 3 O + + In - color 1 color 2 Observed color will vary (depends on and [In - ] in solution). K a = [H 3 O + ][In - ] Detection of the Equivalence Point The acidic form dominates when the >> [In - ] If - pk a = - ph - 1 ph = pk a -1 Basic color shows when [In - ] >> If [In - ] = 10 [In - ] = 10 K a = [H 3 O + ][In - ] - pk a = 1 - ph ph = pk a +1 = [H 3 O + ] 10 [H 3 O + ][In - ] K a = = 10 [H 3 O + ] Detection of the Equivalence Point Acid-base titration curve = plot of ph vs V titrant added. Titrate 50.0 ml of M HCl with M NaOH Red ph 4 Yellow ph 6 Initial ph = -log(0.100) = (fully ionized acid) Moles of acid = L mol L Methyl Red Bromothymol blue Phenolphthalein = 5.00 x 10-3 mol After 40.0 ml of base added (before equivalence) Base removes H 3 O + : H 3 O + + OH - 2 H 2 O [H 3 O + ] = original n acid total n base added V acid (l) + V base added (l) [H 3 O + ] = (5.00 x x 10-3 ) mol = M ( ) L ph = -log(0.0111) = 1.95 At Equivalence (50.0 ml added) All acid and base react. Neutral salt. ph = After Equivalence (50.2 ml added) All acid consumed. n OH- added = L original n acid = L mol L mol L = 5.02 x 10-3 mol = 5.00 x 10-3 mol n OH- remaining = 0.02 x 10-3 mol V total = ( ) L = L poh = - log(0.02 x10-3 / ) = 3.70 ph = =

4 V titrant /ml V excess /ml V total /ml [OH - ]/mol L -1 ph ml of base increased the ph by 3 units! More complicated. Weak acid is in equilibrium with its conjugate base. Titrate 50.0 ml of M acetic acid with M NaOH. What is the ph after the following titrant additions: 0 ml, 40.0 ml, 50.0 ml, and 50.2 ml? 0 ml added HA + H 2 O H 3 O + + A - [H x 2 3 O + ] K a = 1.8 x 10-5 = x= (0.100) 40.0 ml added 50 ml of M HA + OH - A - + H 2 O(l) n initial n added n left Each OH - removes 1 HA n left = ml of M and makes 1 A - V total = ( ) ml = 90.0 ml = L ph = -log(0.0013) = ml added Henderson-Hasselbalch: ( /0.0900) ph = -log(1.8 x 10-5 ) + log ( /0.0900) ph = log = = n A- / V = n HA / V Note: V cancels (could be omitted) (c) 50.0 ml added Equivalence. All HA is converted to A -. HA + OH - A - + H 2 O(l) n initial n added n left V total = ml, so = mol/0.100 L = M A - is basic! 4

5 Use K b to solve for the [OH - ] generated by : K b = [OH - ] A - + H 2 O HA + OH - ( x) x x K b = 5.6 x x poh = 5.28 ph = 14 - poh = 8.72 K b = K w / K a = 5.6 x x = 5.3 x 10-6 M (c) 50.2 ml added 0.2 ml of extra OH - dominates ph. Ignore any contribution from A -. [OH - ] = ( L x mol/l) / L = 2.0 x 10-4 M poh = 3.7 ph = = 10.3 Titration of Weak Base with Strong Acid Titrate of 50.0 ml of M acetic acid with M NaOH. Volume of M HCl added (ml) ph = pk a at 50% titration (25 ml) pk a (acetic acid) = Short vertical section compared to strong acid/strong base ml of M ammonia titrated with 0.100M HCl. ph = pk a at 50% to equivalence (pk a = 9.25). Solubility Equilibria and K SP Some ionic compounds are slightly water soluble. Saturation occurs at low concentration. AgCl(s) The solubility product constant is: K sp = [ Ag + ] [ Cl - ] Ag + + Cl - As always, [ ] of solid is omitted Solubility Equilibria and K SP K sp (AgCl)=1.8 x Calculate the solubility (mol/l). At equilibrium, K sp = [Ag + ][Cl - ] = 1.8 x [Ag + ] = [Cl - ] = S Then: 1.8 x = (S)(S) = S 2 S = 1.3 x 10-5 M Solubility = 1.3 x 10-5 mol/l. 5

6 Factors Affecting Solubility ph and Dissolving Slightly Soluble Salts Using Acids Insoluble salts containing anions of Bronsted-Lowry bases dissolve in acidic solutions. Pure water: CaCO 3 (s) Ca 2+ + CO 3 K sp = 8.7 x 10-9 Acid: CO 3 + H 3 O + HCO 3- + H 3 O + HCO 3- + H 2 O(l) H 2 CO 3 + H 2 O(l) H 2 CO 3 CO 2 (g) + H 2 O(l) K 10 5 Common Ion Effect AgCl is slightly soluble in water: AgCl(s) Called the common ion effect. Ag + + Cl - Add more Ag + or Cl -. Equilibrium will move left. e.g. AgCl is less soluble in a NaCl solution than in water. The common ion is Cl -. Solubility and the Common Ion Effect Calculate the solubility of PbI 2 in (a) water (b) M NaI. K sp (PbI 2 ) = 8.7 x (a) Pure water: PbI 2 Pb I - K [Pb 2+ ][I - ] 2 sp = = (S)(2S) 2 = 4S 3 = 8.7 x 10-9 S S = 1.3 x 10-3 M 2S Solubility and the Common Ion Effect (b) NaI supplies I - lowering the PbI 2 solubility. PbI 2 Pb I - [ ] initial (from NaBr) [ ] change (from PbBr 2 ) S 2S [ ] equil S S K sp = 8.7 x 10-9 = [Pb 2+ ][I - ] 2 = S(2S ) 2 Assume S << x 10-9 S(0.010) 2 S = 8.7 x 10-5 M 15 times lower than in (a) Complex Ion Formation Metal ions (Lewis acids) can react with Lewis bases. AgBr(s) + 2S 2 O 3 Ag(S 2 O 3 ) Br - The complex is much more soluble than AgBr. Black & white photography light AgBr in film Ag(s) Fix a negative: complex and dissolve unexposed Ag salts. Complex Ion Formation Complex ion solubility can be evaluated using the formation constant, K f. AgBr(s) Ag S 2 O 3 net AgBr(s) + 2 S 2 O 3 Ag + + Br - Ag(S 2 O 3 ) 2 3- [Ag(S K net = K sp K f = [Ag + ][Br - ] 2 O 3 ) 3-2 ] [Ag(S = 2 O 3 ) 3-2 ][Br - ] [Ag + ][S 2 O 3 ] 2 [S 2 O 3 ] 2 K sp = 3.3 x10-13, K f = 2 x 10 13, so K net = 7 The reaction is favored. Ag(S 2 O 3 ) Br - 6

7 Amphoterism Precipitation: Will It Occur? Calculate Q, compare it to K sp. If Q > K sp Q must decrease. remove ions, precipitate solid. If Q = K sp at equilibrium (saturated solution). If Q < K sp Q must increase. dissolve more solid (if present). Form more ions. Precipitation: Will It Occur? Mix 25.0 ml of M HCl and 10.0 ml of M AgNO 3. Will AgCl precipitate? K sp (AgCl) = 1.8 x n Cl- = L ( mol/l) = 6.25 x 10-5 mol n Ag+ = L (0.010 mol/l) = 1.0 x 10-4 mol V total = ( ) = L. [Cl - ] = 6.25 x 10-5 mol/ L = 1.79 x 10-3 mol/l [Ag + ] = 1.0 x 10-4 mol / L = 2.86 x 10-3 mol/l Q = [Ag + ][Cl - ] = (1.79 x 10-3 )(2.86 x 10-3 ) = 5.1 x 10-6 Precipitation: Will It Occur? Slowly add HCl to a solution that is M in Cu + and M in Pb 2+. Which salt will precipitate first? K sp for PbCl 2 and CuCl are 1.7 x 10-5 and 1.9 x Find the minimum [Cl - ] that will cause precipitation: CuCl: K SP = 1.9 x 10-7 = [Cu + ][Cl - ] = 0.010[Cl - ] [Cl - ] = 1.9 x 10-5 M PbCl 2 : K SP = 1.7 x 10-5 = [Pb 2+ ][Cl - ] 2 = 0.500[Cl - ] 2 [Cl - ] = 5.8 x 10-3 M Least soluble, precipitates 1st Q > K SP Precipitation will occur 7