1. Chapter 17: Applications of Aqueous Equilibria By: Ms. Buroker

2. The Common Ion Effect Suppose we had a solution that contained the weak acid HF (K a = 7.2 x 10 -4 ) and its salt NaF. What would be the major species in solution? HF, Na +, F -, H 2 O The common ion is this solution is F - because it comes from both the HF and the NaF!!!

3. So … what effect does the presence of the NaF have on the dissociation of HF? Let’s compare: 1.0M HF with (1.0M HF + 1.0M NaF) According to LeChatelier’s principle we would expect the dissociation equilibrium for HF to be: HF (aq) ↔ H + (aq) + F - (aq) Equilibrium shifts away from the added component. Added F - ions from NaF

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) Weak AcidAcid FormulaExample of salt of the weak acid Hydrofluoric acid HFKF Hydrocyanic acid HCNNaCN Acetic Acid HC 2 H 3 O 2 Mg(C 2 H 3 O 2 ) 2

6. Base/Salt Buffering Pairs The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO 3 ) Weak BaseBase FormulaExample of salt of the weak base Ammonia NH 3 NH 4 Cl Methylamine CH 3 NH 2 CH 3 NH 2 Cl Ethylamine C 2 H 5 NH 2 C 2 H 5 NH 3 NO 3

7. Example #1 What is the pH of a buffer solution composed of.100M propanic acid (HC 3 H 5 O 2 ; K a =1.3x10 -5 ) and.100M sodium propanate? To decide pH, we need to figure out how the presence of the salt affects the acid’s equilibrium: HC 3 H 5 O 2 H + + C 3 H 5 O 2 -

8. Example #1 (continued) HC 3 H 5 O 2 H + + C 3 H 5 O 2 - ICEICE.100M 0 0.100M -x+x +x.100-x x 0.100 + x

9. Example #1 (continued) K = [HC 3 H 5 O 2 ] [H + ] [C 3 H 5 O 2 - ] K = [.100-x] [x] [.100+x] we will assume that the effect of “x” can be ignored (we will double check later) K =[x] = 1.3 x 10 -5 pH = -log[x] = 4.88

10. How do buffers work? The presence of the common ion from the conjugate helps moderate the equilibrium More of the common ion shifts the equilibrium toward the non-dissociated side of the equilibrium Adding more acid or base has a lesser effect on pH because of this

11. Example #2 How does the addition of 0.010 moles of solid NaOH affect the pH of 1 L of the buffer from the last example? HC 3 H 5 O 2 + OH -  C 3 H 5 O 2 - + H 2 O Stoichiometry Point of View 0. 010 moles of the base will react with the same amount of the acid, and produce the same amount of the conjugate Therefore we have: 0.100 -.010 =.090 moles of the acid remaining 0.100 +.010 =.110 moles of the conjugate present

12. Example #2 (continued) HC 3 H 5 O 2  H + + C 3 H 5 O 2 - ICEICE.090M 0.110M -x +x +x.090-x x.110 + x

13. Example #2 (continued) K = [HC 3 H 5 O 2 ] [H + ] [C 3 H 5 O 2 - ] K = [.090-x] [x] [.110+x] we will assume that the effect of “x” can be ignored (we will double check later) 1.3x10 -5 = [.090] [x] [.110] x = 1.06 x 10 -5 pH = -log[x] = 4.97

14. Another Way to Solve The stability of a buffer can be explained another way: K a = [HA] [H + ] [A - ] [H + ] = [HA] [A - ] KaKa can be re-arranged to:

15. Another Way: Henderson- Hasselbalch Taking negative log of both sides gives: -log[H + ] = -log(K a ) - log [HA] [A - ] pH = pK a + log [HA] [A - ] pH = pK a + log [acid] [base] Henderson-Hasselbalch Equation

16. Example #2 (again) pH = pK a + log [acid] [base] pH = p(1.3x10 -5 ) + log (.090) (.110) pH = 4.97

17. Example #3 A buffered solution contains.50M NH 3 (K b =1.8x10 -5 )and.30M NH 4 Cl. What is its pH? K w = (K a )(K b ) 1x10 -14 = (K a )(1.8x10 -5 ) 5.6x10 -10 = (K a ) To solve H-H:

18. Example #3 (continued) pH = pK a + log [acid] [base] pH = p(5.6x10 -10 ) + log [.30] [.50] pH = 9.26 + (-.22) = 9.04

19. Acid- Base Titrations Titration: Quantitatively determining the concentration of a solution by reacting it with another solution of precisely known concentration. Terminology – Titrant: The solution of known concentration used to titrate a volume of the solution of unknown concentration. – Standardized (standard) solution: Solution of precisely known concentration – Equivalence point: The point at which stoichiometrically equivalent amounts of reactants have reacted – Endpoint: The point where an indicator (such as pH) changes color.

20. Titration Curve The pH at the equivalence point of a strong acid-strong base titration is 7. The pH of the equivalence point is taken as the mid-point in the vertical portion of the pH verses volume of titrant curve.

21. Example What is the pH after 25.0mL of 0.100M NaOH has been added to 50.0mL of 0.100M HCl? What is the pH after 50.50mL of NaOH has been added? 1.) The first thing you need to do is find out how much of the acid will be left once the base has been added … you know all 25mL of the base will react.

22. Answer to First Part mol acid left: (0.0500L)(0.100M) – (.0250L)(0.100M) =.0025mol New [H 3 O + ] =.0025mol/.075L =.033M pH = -log(.033) = 1.48 Remember now that the base has been added to the acid … the volume has increased!! Draw a picture … it helps to keep things straight.

23. Now … back to the problem What is the pH after 25.0mL of 0.100M NaOH has been added to 50.0mL of 0.100M HCl? What is the pH after 50.50mL of NaOH has been added? 2.) Now you will notice that all of the acid has reacted and you have excess base remaining. So now you just have to find the hydroxide ion concentration and go from there.

24. Answer to Second Part At 50.50mL of NaOH all of the HCl is neutralized and an extra 0.50mL of.100M NaOH exists in solution. [OH - ] = (0.0005L)(0.100M) /.10050L = 4.975x10 -4 pOH = -log(4.975x10 -4 ) = 3.303 pH = 14.0 – 3.303 = 10.7 Remember now that the base has been added to the acid … the volume has increased!! Draw a picture … it helps to keep things straight.

25. Weak Acid- Strong Base Titrations * The pH before the titration begins is given by the equilibrium expression using K a of the weak acid. *The pH at the equivalence point is dominated by the conjugate base (in the salt that forms) from the weak acid that was titrated. *The pH at the halfway point (half-equivalence point) is given by: pH = pKa + log([A-]/[HA]) *at the halfway point [A-] = [HA] and the log(1) = 0 Therefore pH = pKa

26. Titration Curve

27. Example What is the pH of the solution when 35.0mL of 0.100M NaOH has been added to 100.0mL of 0.100M acetic acid? Answer: HC 2 H 3 O 2 + NaOH →NaC 2 H 3 O 2 + H 2 O Moles NaOH = (.0350L)(0.100M) =.0035 mol Moles HC 2 H 3 O 2 = (.1000L)(0.100M) =.0100 mol pH = pKa + log ([C 2 H 3 O 2 - ]/[HC 2 H 3 O 2 ]) pH = 4.74 + log (.0259M / 0.0481M)= 4.47 Moles of acid minus the moles of base it reacted with divided by liters of solution.

28. Example Calculate the pH after 75.0mL of 0.100M HCl has been added to 100.0mL of 0.100M NH 3 (K b of NH 3 = 1.8x10 -5 ) Answer: mol HCl = (.0750L)(0.100M) =.00750mol HCl initial moles NH 3 = (.100L)(0.100M) =.0100mol NH 3 Final: mol NH4 + =.00750mol mol NH 3 =.0100 -.00750 =.0025mol pOH = pK b + log([NH4 + ]/[NH 3 ]) pOH = 4.74 + log(.00750/.0025) = 5.22 pH = 14.00 – 5.22 = 8.78

29. Solubility of Salts For a given slightly soluble salt of the form: A x B y(s) ↔ xA y + (aq) + yB x - (aq) K sp = [A y + ] x [B x - ] y Where K sp is the solubility product constant (equilibrium constant) for the salt (in aqueous solution) Can be used to determine solubility (in mol/L or g/100mL H 2 O, etc.) Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a saturated solution. Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution.

30. Solubility Equilibria AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] K sp is the solubility product constant MgF 2 (s) Mg 2+ (aq) + 2F - (aq) K sp = [Mg 2+ ][F - ] 2 Ag 2 CO 3 (s) 2Ag + (aq) + CO 3 2 - (aq) K sp = [Ag + ] 2 [CO 3 2 - ] Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO 4 3 - (aq) K sp = [Ca 2+ ] 3 [PO 3 3 - ] 2 Dissolution of an ionic solid in aqueous solution: Q = K sp Saturated solution Q < K sp Unsaturated solution No precipitate Q > K sp Supersaturated solution Precipitate will form

31. What is the solubility of silver chloride in g/L ? AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] Initial (M) Change (M) Equilibrium (M) 0.00 +x 0.00 +x+x xx K sp = x 2 x = K sp  x = 1.3 x 10 -5 [Ag + ] = 1.3 x 10 -5 M [Cl - ] = 1.3 x 10 -5 M Solubility of AgCl = 1.3 x 10 -5 mol AgCl 1 L soln 143.35 g AgCl 1 mol AgCl x = 1.9 x 10 -3 g/L K sp = 1.6 x 10 -10

32. Example Knowing that the K sp value for MgF 2 is 5.2x10 -11, calculate the solubility of the salt in a)moles per liter and b)grams per liter Answer: 5.2x10 -11 = [X][2X] 2 X = 2.35x10 -4 = 2.4x10 -4 mol/L 2.4x10 -4 mol/L(62.3g/mol) = 0.015gMgF 2 /L Reaction: MgF 2 ↔ Mg +2 (aq) + 2F - (aq) K sp = [Mg +2 ][F - ] 2

33. If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl 2, will a precipitate form? The ions present in solution are Na +, OH -, Ca 2+, Cl -. Only possible precipitate is Ca(OH) 2 (solubility rules). Is Q > K sp for Ca(OH) 2 ? [Ca 2+ ] 0 = 0.100 M [OH - ] 0 = 4.0 x 10 -4 M K sp = [Ca 2+ ][OH - ] 2 = 8.0 x 10 -6 Q = [Ca 2+ ] 0 [OH - ] 0 2 = 0.10 x (4.0 x 10 -4 ) 2 = 1.6 x 10 -8 Q < K sp No precipitate will form

34. Solubility and the Common Ion Effect The effect of having an ion already in solution that is common to one of the ions in a substance being added to the solution (with a given K sp ) is to reduce the solubility of that substance. Example: Calculate the solubility of BaSO 4 a) in pure water and b) in the presence of 0.010M Ba(NO 3 ) 2. K sp for BaSO 4 is 1.1x10 -10. Think LeChatelier!!!

35. Answer a.) K sp = [Ba 2+ ][SO 4 2- ] so … 1.1x10 -10 = (x)(x) solving for x gives … x= 1.0x10 -5 mol/L b.) 1.1x10 -10 = (x +.010)(x) solving for x gives … x = 1.1x10 -8 mol/L This comes from the common ion that was added to the solution of BaSO 4.

36. Complex Ion Equilibria and Solubility A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. Co 2+ (aq) + 4Cl - (aq) CoCl 4 (aq) 2- Co(H 2 O) 6 2+ CoCl 4 2-

37. 16.10

38. Complex Ion Formation These are usually formed from a transition metal surrounded by ligands (polar molecules or negative ions). As a "rule of thumb" you place twice the number of ligands around an ion as the charge on the ion... example: the dark blue Cu(NH 3 ) 4 2+ (ammonia is used as a test for Cu 2+ ions), and Ag(NH 3 ) 2 +. Memorize the common ligands.

39. Common Ligands LigandsNames used in the ion H2OH2Oaqua NH 3 ammine OH-hydroxy Cl-chloro Br-bromo CN-cyano SCN-thiocyanato (bonded through sulfer) isothiocyanato (bonded through nitrogen)

40. Names Names: ligand first, then cation Examples: –tetraamminecopper(II) ion: Cu(NH 3 ) 4 2+ –diamminesilver(I) ion: Ag(NH 3 ) 2 +. –tetrahydroxyzinc(II) ion: Zn(OH) 4 2- The charge is the sum of the parts (2+) + 4(-1)= -2.

41. When Complexes Form Aluminum also forms complex ions as do some post transitions metals. Ex: Al(H 2 O) 6 3+ Transitional metals, such as Iron, Zinc and Chromium, can form complex ions. The odd complex ion, FeSCN 2+, shows up once in a while Acid-base reactions may change NH 3 into NH 4 + (or vice versa) which will alter its ability to act as a ligand. Visually, a precipitate may go back into solution as a complex ion is formed. For example, Cu 2+ + a little NH 4 OH will form the light blue precipitate, Cu(OH) 2. With excess ammonia, the complex, Cu(NH 3 ) 4 2+, forms. Keywords such as "excess" and "concentrated" of any solution may indicate complex ions. AgNO 3 + HCl forms the white precipitate, AgCl. With excess, concentrated HCl, the complex ion, AgCl 2 -, forms and the solution clears.

42. Coordination Number Total number of bonds from the ligands to the metal atom. Coordination numbers generally range between 2 and 12, with 4 (tetracoordinate) and 6 (hexacoordinate) being the most common.

43. Some Coordination Complexes molecular formula Lewis base/ligand Lewis aciddonor atom coordination number Ag(NH 3 ) 2 + NH 3 Ag + N2 [Zn(CN) 4 ] 2- CN-Zn 2+ C4 [Ni(CN) 4 ] 2- CN-Ni 2+ C4 [PtCl 6 ] 2- Cl-Pt 4+ Cl6 [Ni(NH 3 ) 6 ] 2+ NH 3 Ni 2+ N6