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Buffers, Titrations, and Aqueous Equilibria

Buffers, Titrations, and Aqueous Equilibria. 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 )

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Buffers, Titrations, and Aqueous Equilibria

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  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 x 1.0 + x Ka = 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 H3CCOOH + 0.50 M H3CCOONa • 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 HC2H3O2 Ka=1.8 x 10-5 and 0.5 M NaC2H3O2 has 0.01 mol of solid NaOH added. What is the new pH? HC2H3O2 (aq)<====> C2H3O2-(aq) + H+ (aq) I 0.5 0.5 0 C -x +x +x E 0.5 -x 0.5 +x x Ka=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. HC2H3O2 (aq)<====> C2H3O2-(aq) + H+ (aq) I 0.5 0.5 1.8 x 10-5 S -0.01 +0.01 - 1.8 x 10-5 due to adding base I 0.49 0.51 0 C -x +x +x E 0.49-x 0.51 +x x Ka= 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 = pKb + log ( [HB+] / [B] ) = pKb + 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, [HC2H3O2] = .49 M after the addition of base, and [C2H3O2-] = .51 M. Using the H/H equation, pH= pKa + 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 NH3 and 0.40 M NH4Cl. What is the pH of this buffer? Kb = 1.8 x 10-5 NH3 (aq) + H2O (l) <===> NH4+ (aq) + OH- (aq) pOH = pKb +log ([NH4+] / [NH3]) = 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? NH3 (aq) + H2O (l) <===> NH4+ (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.

  15. 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- ----> H2O 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 mmol pH= 0.95

  16. 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.

  17. 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? Ka= 6.2 x 10-10 HCN + OH- -----> H2O + 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 0.072M 0.0138 M pH = pKa + log (0.0138/0.072) = 8.49

  18. 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.

  19. m - Nitrophenol * Trademark CIBA GEIGY CORP. pH 15_334 0 1 2 3 4 5 6 7 8 9 10 11 12 13 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 o -Cresolphthalein Phenolphthalein Thymolphthalein Alizarin Yellow R The pH ranges shown are approximate. Specific transition ranges depend on the indicator solvent chosen.

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

  21. Solubility Product • For solids dissolving to form aqueous solutions. • Bi2S3(s)  2Bi3+(aq) + 3S2(aq) • Ksp = solubility product constant • and Ksp = [Bi3+]2[S2]3

  22. Solubility Product • “Solubility” = s = concentration of Bi2S3 that dissolves, which equals 1/2[Bi3+] and 1/3[S2]. • Bi2S3 (s) <==> 2 Bi3+ (aq)+ 3 S2(aq) • 2 s 3 s • Note: Ksp is constant (at a given temperature) • s is variable (especially with a common ion present)

  23. Relation of Ksp to s • Ionic compounds will dissociate according to set equations, which will give set Ksp to s relation. Use ICE to find it. • A2B <==> 2A + B • 2ss Ksp = (2s)2(s)=4s3 • AB3 <==> A + 3B • s 3s Ksp= (s)(3s)3=27s4

  24. Ksp from s If 4.8 x 10-5 mol of CaC2O4 dissolve in 1.0 L of solution, what is its Ksp? [CaC2O4]= mol / L = 4.8 x 10-5 / 1.0 L = 4.8 x 10-5 M CaC2O4 (s) <==> Ca+2 (aq) + C2O4-2 (aq) s s Ksp= (s)(s)= s2 = (4.8 x 10-5 )2 = 2.3 x 10-9 If the molar solubility (s) of BiI3 is 1.32 x 10-5, find its Ksp. BiI3 (s) <==> Bi+3 + 3I- s 3s Ksp= (s)(3s)3= 27 s4 Ksp = 27(1.32 x 10-5)4= 8.20 x 10-19

  25. s from Ksp If Ksp for Ag2CO3 is 8.1 x 10-12, find its molar solubility. Ag2CO3 (s) <==> 2 Ag+ (aq) + CO3-2 (aq) 2 ss Ksp = (2s)2(s) = 4 s3 = 8.1 x 10-12 s= 1.27 x 10-4 If Ksp 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 Ksp = (s)(3s)3 = 27s4 = 2 x 10-32s = 5.22 x 10-9

  26. 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.

  27. Complex Ion Examples • Coordination number=2 • Ag(NH3)2+ • Coordination number=4 • Al(OH)4-1 Cu(NH3)4+2 • Coordination number=6 • Fe(CN)6-4 Fe(SCN)6-3

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

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