1. Buffers, Titrations, and Aqueous Equilibria

2. Common Ion Effect The shift in equilibrium that occurs because of the addition of an ion already involved in the equilibrium reaction. (Le Chatelier’s principle) AgCl(s)  Ag + (aq) + Cl  (aq) This affects the concentrations of other ions, notably H +

3. Calculations with ICE What is the [H + ] and % ionization of 1.0 M HF mixed with 1.0 M NaF. (1.0 M HF alone, [H + ]=2.7 x 10 -2, %ion=2.7%) HF(aq) H + (aq) + F - (aq) I 1.0 0 1.0 C -x +x+x E 1.0-x x1.0 + x K a = 7.2 x 10 -4 = x (1.0+x) / (1.0 - x) = x/1 x=[H+]= 7.2 x 10 -4 % ion = 7.2 x 10 -4 /1.0 =.072%

4. A Buffered Solution... resists change in its pH when either H + or OH  are added. 1.0 L of 0.50 M H 3 CCOOH + 0.50 M H 3 CCOONa pH = 4.74 Adding 0.010 mol solid NaOH raises the pH of the solution to 4.76, a very minor change.

5. Key Points on Buffered Solutions 1.They are weak acids or bases containing a common ion. 2.After addition of strong acid or base, deal with stoichiometry first, then equilibrium.

6. Demonstration of Buffer Action A buffered solution of 1.0 L of 0.5 M HC 2 H 3 O 2 K a =1.8 x 10 -5 and 0.5 M NaC 2 H 3 O 2 has 0.01 mol of solid NaOH added. What is the new pH? HC 2 H 3 O 2 (aq) C 2 H 3 O 2 - (aq) + H + (aq) I 0.50.5 0 C -x+x +x E 0.5 -x0.5 +x x K a =1.8 x 10 -5 = 0.5(x) / 0.5 x= 1.8 x 10 -5 pH= 4.74

7. When OH - is added, it takes away an equal amount of H + ions, and will affect also the acid concentration by the same amount. It will also add to the common ion. HC 2 H 3 O 2 (aq) C 2 H 3 O 2 - (aq) + H + (aq) I 0.50.5 1.8 x 10 -5 S -0.01 +0.01- 1.8 x 10 -5 due to adding base I 0.490.510 C -x+x+x E 0.49-x0.51 +xx K a = 1.8 x 10 -5 = x (.51)/.49 x= 1.73 x 10 -5 pH= 4.76 Practically no change in pH after base is added.

8. Henderson-Hasselbalch Equation -Useful for calculating pH when the [A  ]/[HA] ratios are known.

9. Base Buffers With H/H Equation pOH = pK b + log ( [HB + ] / [B] ) = pK b + log ([acid]/ [base]) Base buffers can be calculated in similar fashion to acid buffers. pH = 14 - pOH

10. Buffered Solution Characteristics -Buffers contain relatively large amounts of weak acid and corresponding base. -Added H + reacts to completion with the weak base. -Added OH  reacts to completion with the weak acid. -The pH is determined by the ratio of the concentrations of the weak acid and weak base.

11. H/H Calculations From the previous example, [HC 2 H 3 O 2 ] =.49 M after the addition of base, and [C 2 H 3 O 2 - ] =.51 M. Using the H/H equation, pH= pK a + log (.51)/(.49) = 4.74 + 0.02 = 4.76 same as using ICE, but easier A weak base buffer is made with 0.25 M NH 3 and 0.40 M NH 4 Cl. What is the pH of this buffer? K b = 1.8 x 10 -5 NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) pOH = pK b +log ([NH 4 + ] / [NH 3 ]) = 4.74 +log (.40/.25)= 4.94 pH = 14- 4.94 = 9.06

12. H/H Calculations The weak base buffer with pH of 9.06 from the previous example has 0.10 mol of HCl added to it. What is the new pH? NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) I.25.40 1.15 x 10 -5 S-.10 +.10 - 1.15 x 10 -5 due to acid I.15.50 0 Using H/H, pOH = 4.74 + log (.5)/(.15) = 5.26 pH= 14 - 5.26 = 8.74

13. Buffering Capacity... represents the amount of H + or OH  the buffer can absorb without a significant change in pH.

14. Titration (pH) Curve A plot of pH of the solution being analyzed as a function of the amount of titrant added. Equivalence (stoichiometric) point: Enough titrant has been added to react exactly with the solution being analyzed.

16. Strong Acid/Base Titration In titrations, it is easier to calculate millimoles(mmol), which would be Molarity x mL, in stoichiometry. What would be the pH of 50.0 mL of a 0.2 M solution of HCl after 20.0 mL of 0.1 M NaOH have been added? HCl---50.0 x 0.2 =10.0 mmol NaOH--- 20.0 x 0.1= 2.0 mmol H + + OH - ----> H 2 O volumes must be added I 10mmol 2 mmol 50 + 20 = 70 S -2 mmol -2mmol [H+] = 8 mmol/ 70 mL = 0.11 M End 8mmol 0 mmolpH= 0.95

17. Weak Acid - Strong Base Titration Step 1 -A stoichiometry problem - reaction is assumed to run to completion - then determine remaining species. Step 2 -An equilibrium problem - determine position of weak acid equilibrium and calculate pH.

19. Weak Acid/Strong Base First do a stoichiometry, then equilibrium using H/H equation What is the pH of 50.0 mL of a 0.100M HCN solution after 8.00 mL of 0.100 M NaOH has been added? K a = 6.2 x 10 -10 HCN + OH - -----> H 2 O + CN - Volume I 5 mmol 0.8 mmol 0 mmol 50 + 8 = 58 S - 0.8 mmol -0.8 mmol +0.8 mmol I 4.2 mmol/58mL 0.8 mmol /58 mL 0.072M0.0138 M pH = pK a + log (0.0138/0.072) = 8.49

24. Acid-Base Indicator... marks the end point of a titration by changing color. The equivalence point is not necessarily the same as the end point.

26. 15_334 012345678910111213 The pH ranges shown are approximate. Specific transition ranges depend on the indicator solvent chosen. pH Crystal Violet Cresol Red Thymol Blue Erythrosin B 2,4-Dinitrophenol Bromphenol Blue Methyl Orange Bromcresol Green Methyl Red Eriochrome* Black T Bromcresol Purple Alizarin Bromthymol Blue Phenol Red m - Nitrophenol o -Cresolphthalein Phenolphthalein Thymolphthalein Alizarin Yellow R * Trademark CIBA GEIGY CORP.

28. Homework !! p. 761 33, 35, 45 (for 35 only)

29. Solubility Product For solids dissolving to form aqueous solutions. Bi 2 S 3 (s)  2Bi 3+ (aq) + 3S 2  (aq) K sp = solubility product constant and K sp = [Bi 3+ ] 2 [S 2  ] 3 K sp = [Bi 3+ ] 2 [S 2  ] 3

30. Solubility Product “Solubility” = s = concentration of Bi 2 S 3 that dissolves, which equals 1/2[Bi 3+ ] and 1/3[S 2  ]. Bi 2 S 3 (s) 2 Bi 3+ (aq)+ 3 S 2   aq) Bi 2 S 3 (s) 2 Bi 3+ (aq)+ 3 S 2   aq) 2 s 3 s 2 s 3 s Note:K sp is constant (at a given temperature) s is variable (especially with a common ion present) ion present)

31. Relation of K sp to s Ionic compounds will dissociate according to set equations, which will give set K sp to s relation. Use ICE to find it. A 2 B 2A + B 2s s K sp = (2s) 2 (s)=4s 3 2s s K sp = (2s) 2 (s)=4s 3 AB 3 A + 3B s 3s K sp = (s)(3s) 3 =27s 4

32. K sp from s If 4.8 x 10 -5 mol of CaC 2 O 4 dissolve in 1.0 L of solution, what is its K sp ? [CaC 2 O 4 ]= mol / L = 4.8 x 10 -5 / 1.0 L = 4.8 x 10 -5 M CaC 2 O 4 (s) Ca +2 (aq) + C 2 O 4 -2 (aq) s s K sp = (s)(s)= s 2 = (4.8 x 10 -5 ) 2 = 2.3 x 10 -9 If the molar solubility (s) of BiI 3 is 1.32 x 10 -5, find its K sp. BiI 3 (s) Bi +3 + 3I - s 3s K sp = (s)(3s) 3 = 27 s 4 K sp = 27(1.32 x 10 -5 ) 4= 8.20 x 10 -19

33. s from K sp If K sp for Ag 2 CO 3 is 8.1 x 10 -12, find its molar solubility. Ag 2 CO 3 (s) 2 Ag + (aq) + CO 3 -2 (aq) 2 s s K sp = (2s) 2 (s) = 4 s 3 = 8.1 x 10 -12 s= 1.27 x 10 -4 If K sp for Al(OH) 3 is 2 x 10 -32, find its molar solubility. Al(OH) 3 (s) Al +3 (aq) + 3 OH - (aq) s 3 s K sp = (s)(3s) 3 = 27s 4 = 2 x 10 -32 s = 5.22 x 10 -9

34. Equilibria Involving Complex Ions Complex Ion: A charged species consisting of a metal ion surrounded by ligands (Lewis bases). Coordination Number: Number of ligands attached to a metal ion. (Most common are 6 and 4.) Formation (Stability) Constants: The equilibrium constants characterizing the stepwise addition of ligands to metal ions.

36. Complex Ion Examples Coordination number=2 Ag(NH 3 ) 2 + Coordination number=4 Al(OH) 4 -1 Cu(NH 3 ) 4 +2 Coordination number=6 Fe(CN) 6 -4 Fe(SCN) 6 -3

37. Homework and XCR p. 762ff 50, 51, 59, 60 XCR 72, 76, 86 Super XCR 90