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ACID BASE TITRATIONS

ACID BASE TITRATIONS. Chapter 15. Titration Curves. A plot of pH versus amount of acid or base added. At the EQUIVALENCE POINT on a titration curve, the amount of acid = the amount of base. The END POINT of a titration is determined by a color change of an indicator.

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ACID BASE TITRATIONS

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  1. ACID BASE TITRATIONS Chapter 15

  2. Titration Curves • A plot of pH versus amount of acid or base added. • At the EQUIVALENCE POINT on a titration curve, the amount of acid = the amount of base. • The END POINT of a titration is determined by a color change of an indicator. • Ideally, the end point and equivalence point will be within 1 drop of each other.

  3. INDICATORS • Select an indicator based on the pH range of the equivalence point. pKa of indicator within +/- 1 pH unit of the equivalence point. • Indicators are organic dyes whose colors depend on the [H3O+] or pH of a solution. • Most are produced synthetically. • Ex. • Phenolphthalein • Universal indicator – mix of organic acids that indicate over different ranges • Many are vegetable dyes. • Ex. • Litmus • Purple cabbage indicator

  4. INDICATORS

  5. Generally they are WEAK ORGANIC ACIDS • Symbolized HIn • Dissociation reaction: HIn+ H2O  H30+ + In-

  6. Ex. Bromthymol blue • HIn – yellow • In- - blue HIn+ H20  In- + H30+ • Ka = [H30+] [In-1] [HIn] When the ratio goes to 1/10, a color change will occur. • Adding an acid shifts the equilibrium left. • Adding a base shifts the equilibrium right.

  7. Indicator sample problem • An indicator, HIn, has a Ka = 1.0 x 10-7 . Determine the pH at which a color change will occur given the following scenarios: • Acid titration • Base titration

  8. INDICATOR EXAMPLE Titration of an acid Titration of a base The solution is initially basic, [In] is dominant. The color change occurs when [In]/[HIn] = 10/1 Ka = 1.0 x 10-7 = [H+] (10/1) [H+] = (1.0 x 10-7)/10 = 1.0 x 10-8 pH = 8.0 • The solution is initially acidic, [HIn] is dominant. • The color change occurs when [In]/[HIn] = 1/10 Ka = 1.0 x 10-7 = [H+] (1/10) [H+] = 10 (1.0 x 10-7) = 1.0 x 10-6 pH = 6.0

  9. STRONG ACID/ STRONG BASE • Consider the titration of 50.0 ml of 0.2 M nitric acid with 0.1 M NaOH

  10. Calculate the pH @ • 0.0 ml base added • 10.0 ml base added • 20.0 ml base added Major species in soln?

  11. SA/SB • 50.0 ml base added • 100.0 ml base added • 200.0 ml base added Major species in soln?

  12. STRONG BASE/ STRONG ACID • The pH Curve for the titration of 100.0 mL of 0.50 MNaOH with 1.0 M HCI

  13. WEAK ACID/ STRONG BASE • The pH Curve for the Titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 MNaOH

  14. WEAK ACID/ STRONG BASE • Before any BASE is added, the pH depends only on the weak acid. • After some base is added, but before the EQUIVALENCE POINT, a series of weak acid/ salt buffer solutions determine the pH. • At the EQUIVALNECE POINT, hydrolysis of the anion of the weak acid determines the pH. • Beyond the equivalence point, EXCESS STRONG BASE determines the pH.

  15. WA/SB Sample problem: • 30.0 ml of 0.10 M NaOH is added to 50.0 ml of 0.10 M HF. What is the pH after all 30.0 ml are added? Ka = 7.2 x 10-4 • Major species in soln? • Rxn? • HF initial? (Use mmol) • OH- added? • HF consumed? • F- formed?

  16. WA/SB After equilibrium • Rxn? • Ka expression • Calculate concentrations using mmoles and volumes M = mmol/ml • ICE • [H+] and pH

  17. WEAK BASE/ STRONG ACID • Calculate the pH at each of the following points in the titration of 50.00 ml of a 0.01000M sodium phenolate (NaOC6H5) solution with 1.000 M HClsoltuion. Ka for HOC6H5 = 1.05 x 10-10. • Initial • Midpoint • Equivalence point

  18. WB/SA • Initial – weak base Kb • pH = pKa at the midpoint, so pOH = pKb since [BH+]/[B] = 1

  19. WB/SA • At equivalence. • How many moles of HCl were needed to neutralize? MaVa = MbVbHCl + OC6H5-  • New volume? • Weak acid dissociation reaction. HOC6H5- + H20  • Ka expression. • [H+] • pH

  20. POLYPROTIC ACIDS • Consider 20.00ml or 0.100 M polyprotic acid H2A titrated with 0.100 M NaOH. • Ka1 = 1x10-3 • Ka2 = 1x10-7 • 2 equivalence points are expected.

  21. POLY pH calculations @ • 0 ml base added • H2A dissociation • H2A  H+ + HA- • 10.0 ml base added • H2A/ HA- buffer H2A + OH- HA- + H20

  22. POLY pH calculations @ • 20.0 ml base added – first equivalence point • 30.0 ml base added – ½ way between 1st and 2nd equivalence point. H2A + OH- HA- + OH-  A2-

  23. POLY pH calculations @ • 40.0 ml base added – 2nd equivalence point H2A + OH- HA- + OH-  A2- A2- + H20  HA- + OH- • 50.0 ml NaOH added – excess OH H2A + OH- HA- + OH-  A2-

  24. SOLUBILITY EQUILIBRIUM Ksp

  25. KspSOLUBILITY PRODUCT EQUILIBRIA • Problems dealing with solubility of PARTIALLY soluble ionic compounds (in other words, salts that barely dissociate/dissolve in water) • General Form, called the solubility product: MX(s) n M+ (aq) + p X – (aq) • Ksp = [M+]n[X -]p • ExBa(OH)2 (s)

  26. Solubility • Is NOT the solubility product. • Uses an equilibrium problem to determine how much can dissolve at a certain temperature. • Is related stoichiometrically to the initial formula. Ex. Ba(OH)2 yields twice as many OH-1 ions.

  27. Calculate Ksp • Given the solubility of FeC2O4at eq. = 65.9mg/L • Given the solubility of Li2CO3 is 5.48 g/L.

  28. Calculate solubility • Of SrSO4 with a Ksp of 3.2 x 10-7 in M and g/L • Of Ag2CrO4 with Ksp of 9.0 x 10-12 in M and g/L

  29. Relative solubilities • Ksp can be used to compare the solubility of solids that break apart into the same number of ions. • The bigger the Ksp, the more soluble. • An ICE table is necessary if different numbers of ions are produced.

  30. Common Ion Effect • If we try to dissolve the solid in solution with either the cation or anion present, less will dissolve. • Calculate the solubility of strontium sulfate in a 0.100M solution of Na2SO4

  31. pH and solubility • OH- can be a common ion. • More soluble in acid, since OH- will be removed from the reaction. • For other anions, if they come from a weak acid they are more soluble in acid than in water. Ex. CaC2O4 Ca+2 + C2O4-2 H+ + C2O4-2  HC2O4- Reduces the C2O4-2 in acidic solution

  32. Precipitation • Ion product, Q = [M+]n[X -]p • If Q > Ksp, the precipitate forms • Q< Ksp, no precipitate • Q = Ksp at equilibrium

  33. Sample PPT problem • A solution of 750.0 ml of 4.00 x 10-3 M cerium (III) nitrate is added to 300.0 ml of 2.00 x 10-2 M potassium iodate. Will cerium (III) iodate, Ksp = 1.9 x 10-10 ppt. and if so what is the concentration of ions in solution?

  34. Selective precipitation • Used to separate mixtures of metal ions in solution. • Add anions that will only ppt. certain metals at a time. • Used to purify mixtures. • Often use H2S because in acidic solution, Hg+2, Cd+2, Bi+3, Cu+2, and Sn+4 will ppt.

  35. Selective precipitation • In basic solution, adding OH- solution, S-2 will increase so more soluble sulfides will ppt. • Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3, Al(OH)3

  36. Selective precipitation • Follow the steps with first insoluble chlorides (Ag, Pb, Ba) • Then sulfides in acid • Then sulfides in base • Then insoluble carbonates (Ca, Ba, Mg) • Alkali metals and NH4+ remain in solution • Flame test, NH4+ yields an ammonia smell when heated.

  37. Complex ion equilibria • A charged ion surrounded by LIGANDS. • Ligands are LEWIS BASES using their lone pair to stabilize the charged metal ions. • Common ligands are NH3, H2O, CN-, Cl- • COORDINATION NUMBER is the number of attached ligands – usually twice the cations charge. • Ex. Cu(NH3)4+2 has a coordination # = 4

  38. LIGAND attachment • The addition of each ligand has its own equilibrium. • Usually the ligand is in large EXCESS. • The complex ion will be the biggest ion in solution. • Complex formation helps dissolve otherwise insoluble compounds.

  39. Complex ion equilibrium • Calculate the concentrations of Ag+ and Ag(CN-)2-1 in a solution prepared by mixing 100.0 ml of 5.0 x 10-3 M AgNO3 with 100.0 ml of 2.00 M KCN. Ag+ + 2 CN- Ag(CN)2-1 K1 = 1.3 x10-21

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