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Chapter 16

Chapter 16. Aqueous Ionic Equilibria. Common Ion Effect. Water dissolves many substances and often many of these interact with each other. A weak acid, HA, dissolves in water and produces a few H + and A - ions. What happens if another source of H + or A - is present also?

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Chapter 16

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  1. Chapter 16 Aqueous Ionic Equilibria

  2. Common Ion Effect Water dissolves many substances and often many of these interact with each other. A weak acid, HA, dissolves in water and produces a few H+ and A- ions. What happens if another source of H+ or A- is present also? What does LeChatelier's Principle suggest will happen?

  3. Common Ion Effect Addition of a strong acid to a weak acid. HCl + HC2H3O2 Addition of a salt containing the conjugate base. HC2H3O2 + NaC2H3O2 Can also do the same with weak bases.

  4. Common Ion Effect

  5. Buffers Solutions that contain amounts of both a weak acid and a salt containing its conjugate base are said to be buffered solutions. A buffer is a solution that resists changes in pH. pH of blood buffered at 7.4 pH of seawater buffered at 8.2

  6. Buffers

  7. Buffers Consider the weak acid buffer of HC2H3O2 and C2H3O2- (from the salt NaC2H3O2). What happens when a strong base is added? What happens when a strong acid is added? Buffers have a built-in acid/base to counteract any substance.

  8. Henderson-Hasselbach Ka = [H+] [A-] / [HA] Rearranges to: [H+] = Ka x [HA] / [A-] Taking the negative natural log of both sides yields:

  9. Buffer Capacity Capacity is the amount of acid or base the buffer can neutralize. Buffers are most effective when HA:A- ratio is 1:1. The quantity of HA and/or A- cannot be exceeded = absolute capacity. When ratio becomes 1:10 or 10:1 = practical capacity.

  10. Buffers pH of a buffer can be calculated from H-H equation. If strong acid or base is added, then a reaction needs to be written and an s.c.e. table is constructed. Do not need to re-calculate the concentrations – just plug in millimole amounts.

  11. Blood Buffers The pH of blood must be maintained at 7.40 +/- 0.05. H2CO3 / HCO3- system pKa = 6.1 Requires [HA] = 0.0012M and [A-] = 0.024M Acid concentration controlled by shifting CO2 amounts

  12. Titrations In first semester, we titrated a known base against an unknown acid using an indicator. What if we had monitored the solution with a pH meter? Graphing the results would yield a titration curve.

  13. Titration Curves Strong acid – strong base pH at equivalence is 7.00.

  14. Titration Curves Weak acid – strong base. pH at equivalence is NOT 7.00.

  15. Titration Curve Can perform a derivative test, which shows a single peak. At halfway point, pH = pKa.

  16. Polyprotic Acids A polyprotic acid will show multiple peaks. Measure from end of the first to second for the second proton.

  17. Titration Curves Curve appearance depends on how weak the acid is. Weaker the acid, the lesser the inflection point and higher equivalence pH.

  18. Titration Curves • Suitable indicator depends on what pH the equivalence occurs.

  19. Solubility Equilibria • The solubility rules in Ch. 4 predict whether or not a compound is soluble. • Even an insoluble compound will dissolve slightly! • PbCl2 is insoluble per the rules, but you would NOT want to drink water that comes into contact with this solid. • PbCl2(s) Pb+2(aq) + 2 Cl-(aq) • Ksp= 1.7 E-5

  20. Solubility Equilibria • Ksp is called the solubility product constant. • Molar solubility is the number of moles of per liter of the insoluble compound that is present in solution. • Molar solubility always = “x”. • Can calculate a Ksp from knowing “x”. • Can calculate “x” from a Ksp. • MUST be able to write the reaction of the break-up correctly!

  21. Factors in Solubility • Many other factors may affect the molar solubility of a compound. • Common-Ion • pH • Complex-Ion formation • Amphoterism

  22. Factors in Solubility • The presence of a common ion will reduce the molar solubility. • For the PbCl2 equilibrium, we could add NaCl. • PbCl2(s) Pb+2(aq) + 2 Cl-(aq) • The NaCl(s) completely dissociates to Na+(aq) and Cl-(aq). • What does LeChatelier’s Principle tell us will happen?

  23. Factors in Solubility • pH of a solution can affect many Ksp reactions. • Insoluble hydroxides depend entirely on the pH. • Ex) Ca(OH)2, Ksp = 6.5 E-6. • A saturated solution will contain ___________ M and have a pH of __________. • What is the molar solubility if the pH is 10.00?

  24. Factors in Solubility • pH can also affect the solubility if the anion is the conjugate base of a weak acid. • Consider: CaF2(s) Ca+2(aq) + 2 F-(aq) • Because F- is the conjugate of HF, the following can happen: F-(aq) + H+(aq)  HF(aq) • Multiply this by 2 and add to the first, yields:

  25. Factors in Solubility • Formation of a complex ion occurs when several small molecules or anions bond to the metal ion – 2, 4, or 6 are common. • For example, two NH3 will bond to Ag+ ion: • Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq) • Unlike most of the equilibria in these chapters, this K is LARGE. Kf = 1.7 E+7 (p. 747). • Will need to solve these in two steps.

  26. Factors in Solubility • Can then combine a Ksp with a Kf reaction. • Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq) • AgCl(s)  Ag+(aq) + Cl-(aq) • Net = • The resulting K value for the net reaction = • Two possibilities for the molar solubility.

  27. Factors in Solubility • Several hydroxides are amphoteric. • Most notably, Al(OH)3. • Al(OH)3(aq) + OH-(aq)  Al(OH)4-(aq). • Others are: Cr+3, Zn+2, and Sn+2.

  28. Amphoterism

  29. Precipitation • Calculation of Q given amounts of ions, then compare to Ksp. • If Q > Ksp, then a ppt will form. • If Q < Ksp, then no ppt forms. • If Q = Ksp, then the solution is saturated. • Reminder: if adding two solutions, then you MUST account for the dilution effect!

  30. Precipitation • Ions can be selectively precipitated based on the differences in their Ksp values. • BaCrO4, Ksp = 1.2 E-10 • SrCrO4, Ksp = 3.5 E-5 • As K2CrO4 is added drop wise, which will ppt first? • How effective is the separation?

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