1. Buffers and Acid/Base Titration

2. Common Ion Suppose we have a solution containing hydrofluoric acid (HF) and its salt sodium fluoride (NaF). Major species (HF, Na +, F -, and H 2 O) HF slightly dissociated, therefore common ion is F - How does this effect equilibrium of HF dissociation?

3. Common Ion Effect HF (aq)  H + (aq) + F - (aq) Added fluoride ions from NaF Shifts reaction left. Less H + ions present. This effect makes a solution of NaF and HF less acidic than a solution of HF alone.

4. Buffered Solutions A solution that resists a change in pH when either hydroxide ions or protons are added. Buffered solutions contain either: A weak acid and its salt A weak base and its salt

5. Acid/Salt Buffering Pairs The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)

6. Base/Salt Buffering Pairs The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO 3 )

7. Henderson-Hasselbalch Equation This is an exceptionally powerful tool, and it’s use will be emphasized in our problem solving.

8. Calculate the pH of the following mixtures: 0.75 M lactic acid (HC 3 H 5 O 3 ) and 0.25 M sodium lactate (Ka = 1.4 x 10 -4 ) 0.25 M NH 3 and 0.40 M NH 4 Cl (Kb = 1.8 x 10 -5 )

9. Buffer capacity The pH of a buffered solution is determined by the ratio [A - ]/[HA]. As long as this doesn’t change much the pH won’t change much. The more concentrated these two are the more H + and OH - the solution will be able to absorb. Larger concentrations bigger buffer capacity.

10. Buffer Capacity Calculate the change in pH that occurs when 0.010 mol of HCl(g) is added to 1.0L of each of the following: 5.00 M HC 2 H 3 O 2 and 5.00 M NaC 2 H 3 O 2 0.050 M HC 2 H 3 O 2 and 0.050 M NaC 2 H 3 O 2 K a = 1.8x10 -5

11. Weak Acid/Strong Base Titration A solution that is 0.10 M CH 3 COOH is titrated with 0.10 M NaOH Equivalence is above pH 7

12. Strong Acid/Strong Base Titration A solution that is 0.10 M HCl is titrated with 0.10 M NaOH Equivalance is at pH 7

13. Strong Acid/Strong Base Titration A solution that is 0.10 M NaOH is titrated with 0.10 M HCl Equivalence is at pH 7 It is important to recognize that titration curves are not always increasing from left to right.

14. Strong Acid/Weak Base Titration A solution that is 0.10 M HCl is titrated with 0.10 M NH 3 Equivalence is below pH 7

15. Titration of an Unbuffered Solution A solution that is 0.10 M CH 3 COOH is titrated with 0.10 M NaOH

16. Titration of a Buffered Solution A solution that is 0.10 M CH 3 COOH and 0.10 M NaCH 3 COO is titrated with 0.10 M NaOH

17. Comparing Results Buffered Unbuffered

18. Selection of Indicators The equivalence point of a titration, defined by the stoichiometry, is not necessarily the same as the end point, careful selection of the indicator will ensure that the error is negligible.

19. Indicator Transitions Source: Wikipedia

20. Indicators Weak acids that change color when they become bases. weak acid written HIn Weak base HIn H + + In - clear red Equilibrium is controlled by pH End point - when the indicator changes color.

21. Solubility Equilibria Lead (II) iodide precipitates when potassium iodide is mixed with lead (II) nitrate. Graphic: Wikimedia Commons user PRHaney

22. K sp Values for Some Salts at 25  C

23. Solving Solubility Problems For the salt AgI at 25  C, K sp = 1.5 x 10 -16 AgI(s)  Ag + (aq) + I - (aq) I C E O O +x x x 1.5 x 10 -16 = x 2 x = solubility of AgI in mol/L = 1.2 x 10 -8 M

24. Solving Solubility Problems For the salt PbCl 2 at 25  C, K sp = 1.6 x 10 -5 PbCl 2 (s)  Pb 2+ (aq) + 2Cl - (aq) I C E O O +x +2x x 2x 1.6 x 10 -5 = (x)(2x) 2 = 4x 3 x = solubility of PbCl 2 in mol/L = 1.6 x 10 -2 M

25. Solving Solubility with a Common Ion For the salt AgI at 25  C, K sp = 1.5 x 10 -16 What is its solubility in 0.05 M NaI? AgI(s)  Ag + (aq) + I - (aq) I C E 0.05 O +x 0.05+x x 1.5 x 10 -16 = (x)(0.05+x)  (x)(0.05) x = solubility of AgI in mol/L = 3.0 x 10 -15 M

26. Precipitation Ion Product, Q =[M + ] a [Nm - ] b If Q>Ksp a precipitate forms. If Q<Ksp No precipitate. If Q = Ksp equilibrium.

27. Precipitation A solution of 750.0 mL of 4.00 x 10 -3 M Ce(NO 3 ) 3 is added to 300.0 mL of 2.00 x 10 -2 M KIO 3. Will Ce(IO 3 ) 3 (Ksp= 1.9 x 10 -10 ) precipitate and if so, what is the concentration of the ions?

28. Precipitation and Qualitative Analysis

29. Complex Ions A Complex ion is a charged species composed of: 1. A metallic cation 2. Ligands – Lewis bases that have a lone electron pair that can form a covalent bond with an empty orbital belonging to the metallic cation

30. NH 3, CN -, and H 2 O are Common Ligands

31. Coordination Number  Coordination number refers to the number of ligands attached to the cation  2, 4, and 6 are the most common coordination numbers

32. Complex Ions and Solubility AgCl(s)  Ag + + Cl - K sp = 1.6 x 10 -10 Ag + + NH 3  Ag(NH 3 ) + K 1 = 2.1 x 10 3 Ag(NH 3 ) + NH 3  Ag(NH 3 ) 2 + K 2 = 8.2 x 10 3 AgCl + 2NH 3  Ag(NH 3 ) 2 + + Cl - K = K sp  K 1  K 2

33. Calculate the concentrations of Ag +, Ag(S 2 O 3 ) -, and Ag(S 2 O 3 ) 2 -3 in a solution made by mixing 150.0 mL of 1.00x10 -3 M AgNO3 with 200.0 mL of 5.00 M Na 2 S 2 O 3. Ag + + S 2 O 3 -2 Ag(S 2 O 3 ) - K 1 =7.4 x 10 8 Ag(S 2 O 3 ) - + S 2 O 3 -2 Ag(S 2 O 3 ) 2 -3 K 2 =3.9 x 10 4