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 + HC 2 H 3 O 2 ● Addition of a salt containing the conjugate base. – HC 2 H 3 O 2 + NaC 2 H 3 O 2 ● 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. ● Consider the weak acid buffer of HC 2 H 3 O 2 and C 2 H 3 O 2 - (from the salt NaC 2 H 3 O 2 ). ● 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 ● K a = [H + ] [A - ] / [HA] ● Rearranges to: ● [H + ] = K a 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. ● H 2 CO 3 / HCO 3 - system – pK a = 6.1 – Requires [HA] = 0.0012M and [A - ] = 0.024M – Acid concentration controlled by shifting CO 2 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 = pK a.

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! PbCl 2 is insoluble per the rules, but you would NOT want to drink water that comes into contact with this solid. PbCl 2(s)  Pb +2 (aq) + 2 Cl - (aq) K sp = 1.7 E-5

20. Solubility Equilibria K sp 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 K sp from knowing “x”. Can calculate “x” from a K sp. 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 PbCl 2 equilibrium, we could add NaCl. PbCl 2(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 K sp reactions. Insoluble hydroxides depend entirely on the pH. Ex) Ca(OH) 2, K sp = 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: CaF 2(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 NH 3 will bond to Ag+ ion: Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) Unlike most of the equilibria in these chapters, this K is LARGE. K f = 1.7 E+7 (p. 747). Will need to solve these in two steps.

26. Factors in Solubility Can then combine a K sp with a K f reaction. Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 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 K sp. If Q > K sp, then a ppt will form. If Q < K sp, then no ppt forms. If Q = K sp, 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 K sp values. BaCrO 4, K sp = 1.2 E-10 SrCrO 4, K sp = 3.5 E-5 As K 2 CrO 4 is added drop wise, which will ppt first? How effective is the separation?