Acids

1. Chapter 15 Acids/Bases with common ion This is when you have a mixture like HF and NaF. The result is that you have both the acid and it�s conjugate base so you have to keep track of both initial concentrations. When doing a calculation make sure you include both initial concentrations

2. Calculate the pH for a aqueous solution containing 1.0 M HF and 1.0 M NaF Ka = 7.2 x 10-4 HF ? ? F- + H+ Chem. [init.] [equil.] HF 1.0 1.0 � x = 1.0 F- 1.0 1.0 + x = 1.0 assume H+ 0 + x 7.2 x 10-4 = (1.0)x / 1.0 x = 7.2 x 10-4 pH = - log (7.2 x 10-4) = 3.14 just like lab

3. Buffered Solutions Resists a change in pH by reacting with any H+ or OH- added to it. Done by having a weak acid and its conjugate base or a weak base and its conjugate acid in the same solution (common ion) Ex. HNO2 / NaNO2 (NO2-) Adding acid reacts with the NO2- to make HNO2, adding base reacts with the HNO2 to make NO2-

4. Henderson � Hasselbalch equation From the definition of Ka we can create this equation by taking the - log and simplifying pH = pKa + log ([A-] / [HA]) A- is the base HA is the acid This equation is an easy way to calculate the pH if you know the equilibrium concentrations of the acid and its conjugate base Buffering capacity

5. Calculate the pH for a solution of 0.75M lactic acid (HC3H5O3) Ka = 1.4 x 10-4 and 0.25M sodium lactate (conj. Base). Since the Ka and the conc. are not very close we can assume very little acid changes to base or base changes to acid So the equilibrium conc. are equal to the initial conc. and we can use the Hend-Hassel eq. pH = pKa + log ([A-] / [HA]) pH = -log(1.4 x 10-4) + log((0.25)/(0.75)) pH = 3.38

6. Titrations Systematic mixing of an acid with a base or vice versa to neutralize the solution. Moles H+ = moles OH- at the equiv. point #H+(VA)(MA) = #OH-(VB)(MB) pH curve The graph of the pH for the mixing of the acid/base combo. Forms a s shape with usually a fairly steep center section

7. Titration Examples We will use millimoles instead of mole since we usually do titrations using ml for volume M = mmol/ml so ml x M = mmol Strong Acid/Strong Base Before any base is added the pH is completely due to the strong acid concentration After adding the base we have to convert the indicated amount of acid and then calculate the pH

8. Calculate the pH for the titration of 50 ml of 0.2M HCl with 0.1M NaOH. At 0 ml NaOH pH = - log(0.2) = 0.699 At 20 ml we have to react the base with the acid H+ OH- init. 10 mmol 2 mmol Change - 2 mmol -2 mmol Final 8 mmol 0 mmol pH = - log (8mmol/70ml) = 0.942

9. At 100 ml we have H+ OH- init. 10 mmol 10 mmol Change -10 mmol -10 mmol Final 0 mmol 0 mmol pH = pH of water = 7

10. At 200 ml we have H+ OH- init. 10 mmol 20 mmol Change -10 mmol -10 mmol Final 0 mmol 10 mmol pH = 14 � pOH pOH = - log ( 10mmol/250ml ) = 1.4 pH = 14 � 1.4 = 12.6

11. Weak Acid/Strong Base Similar process but we end up with a weak acid equilibrium problem until we pass the equiv. pt. Calculate the pH for the titration of 50.0 ml of 0.10M acetic acid with 0.10M NaOH. Ka = 1.8 x 10-5 The equiv. pt is at 50.0 ml of NaOH so before that we have an equil. problem to solve

12. At 10.0 ml of 0.1 M NaOH HC2H3O2 OH- C2H3O2- init. 5 mmol 1 mmol 0 mmol Change - 1 mmol -1 mmol 1 mmol Final 4 mmol 0 mmol 1 mmol Chem [Init.] [Equil] HC2H3O2 4/60 4/60 � x = 4/60 C2H3O2- 1/60 1/60 + x = 1/60 H+ 0 + x

13. Chem [Init.] [Equil] HC2H3O2 4/60 4/60 � x = 4/60 C2H3O2- 1/60 1/60 + x = 1/60 H+ 0 + x pH = pKa + log [A-]/[HA] pH = -log(1.8 x 10-5) + log(1/60 / 4/60) pH = 4.14

14. At 25.0 ml of 0.1M NaOH HC2H3O2 OH- C2H3O2- init. 5 mmol 2.5 mmol 0 mmol Change - 2.5 mmol -2.5 mmol 2.5 mmol Final 2.5 mmol 0 mmol 2.5 mmol Chem [Init.] [Equil] HC2H3O2 2.5/75 2.5/75 � x = 2.5/75 C2H3O2- 2.5/75 2.5/75 + x = 2.5/75 H+ 0 + x

15. Chem [Init.] [Equil] HC2H3O2 2.5/75 2.5/75 � x = 2.5/75 C2H3O2- 2.5/75 2.5/75 + x = 2.5/75 H+ 0 + x pH = pKa + log [A-]/[HA] pH = -log(1.8 x 10-5) + log(2.5/75 / 2.5/75) pH = - log(1.8 x 10-5) + 0 = 4.74 So pH = pKa at the half-way pt.

16. At 50.0 ml of 0.1M NaOH (eq. pt.) HC2H3O2 OH- C2H3O2- init. 5 mmol 5 mmol 0 mmol Change - 5 mmol -5 mmol 5 mmol Final 0 mmol 0 mmol 5 mmol since all of the acid has been converted to the conjugate base we have to do a Kb problem C2H3O2- + H2O ? ? HC2H3O2 + OH-

17. Chem [Init.] [Equil] C2H3O2- 5/100 5/100 � x = 5/100 HC2H3O2 0 + x OH- 0 + x Kb = Kw/Ka = 1.0 x 10-14/1.8 x 10-5 Kb = 5.6 x 10-10 5.6 x 10-10 = x2/(5/100) x = 5.3 x 10-6 pOH = -log(5.3 x 10-6) = 5.28 pH = 14 � 5.28 = 8.72

18. Past the eq. pt. we have the weak base as well as the excess strong base so we can focus on the excess strong base to determine pH 60 ml 0.1MNaOH HC2H3O2 OH- C2H3O2- init. 5 mmol 6 mmol 0 mmol Change - 5 mmol -5 mmol 5 mmol Final 0 mmol 1 mmol 5 mmol

19. HC2H3O2 OH- C2H3O2- init. 5 mmol 6 mmol 0 mmol Change - 5 mmol -5 mmol 5 mmol Final 0 mmol 1 mmol 5 mmol So we focus on the new [OH-] [OH-] = 1 mmol / 110 ml = 9.1 x 10-3 pOH = - log(9.1 x 10-3) = 2.04 pH = 14 � 2.04 = 11.96

20. Weak Acid / Strong Base overview Acid only Ka problem Before eq. pt. Ka problem � way pt. pH = pKa Eq. pt. Kb problem Past eq. pt. strong base only

21. Calculating Ka We can use a titration set of data and the pH to determine an unknown Ka value. Just determine the final concentrations of the weak acid and conjugate base then insert into the equation pH = pKa + log [A-]/[HA]

22. Weak Base/Strong Acid Before any acid Kb problem Before eq. pt. Kb problem � way pt. pOH = pKb At eq. pt. Ka problem Past eq. pt. focus on strong acid conc.

23. Acid/Base Indicators Weak acids/bases that change color at specific pH points Usually use the ratio [I-]/[HI] = 1/10 to determine when an indicator changes color Phenolphthalein is most common indicator and is clear in acid solution and turns pink in base solution

24. Solubility Equilibria The equilibrium for a partially soluble ionic substance in water Solubility is the amount of solid that dissociates PbCl2 (s) ?? Pb2+ + 2Cl- Ksp Equilibrium constant for solubility Ksp = [Pb2+][Cl-]2

25. Calculate the Ksp for Bi2S3 that has a solubility of 1.0 x 10-15 M. Bi2S3 (S)?? 2Bi3+ + 3S2- Ksp = [Bi3+]2[S2-]3 1.0 x 10-15 M Bi2S3 = 2.0 x 10-15 M Bi3+ 1.0 x 10-15 M Bi2S3 = 3.0 x 10-15 M S2- Ksp = (2.0 x 10-15 )2(3.0 x 10-15 )3 Ksp = 1.1 x 10-73

26. Calculate the solubility for Cu(IO3)2 with a Ksp = 1.4 x 10-7 Cu(IO3)2 ?? Cu2+ + 2IO3- Ksp = [Cu2+][IO3-]2 1.4 x 10-7 = (x)(2x)2 = 4x3 x = 3.3 x 10-3 M

27. Common Ion Effect Presence of one of the ions that are formed by the ionic substance AgCl with Cl- (from NaCl) this would impact the solubility of the AgCl since some Cl- is already present

28. Calculate the solubility of CaF2 in a 0.025 M NaF solution if the Ksp = 4.0 x 10-11 Ksp = [Ca2+][F-]2 Chem [init.] [equil.] Ca2+ 0 + x F- 0.025 0.025 + 2x = 0.025 4.0 x 10-11 = (x)(0.025)2 x = 6.4 x 10-8 M Solubility is 6.4 x 10-8 M

29. pH effect A change in pH can effect some salts Sr(OH)2 ?? Sr2+ + 2OH- Adding OH- would shift the reaction left Adding H+ would shift the reaction right because it removes some of the OH-

30. Precipitation and Qualitative Analysis We can determine if a precipitate forms by comparing a set of concentrations to the Ksp by calculating Q Remember Q is the same formula as Ksp

31. Will Ce(IO3)3, Ksp = 1.9 x 10-10, precipitate from a solution made from 750 ml of 0.004 M Ce3+ and 300 ml of 0.02 M IO3-? Q = [Ce3+]o[IO3-]3o [Ce3+]o = (750ml)(0.004M)/1050ml = 2.86 x 10-3M [IO3-]o = (300ml)(0.02M)/1050ml = 5.71 x 10-3M Q = [Ce3+]o[IO3-]3o = (2.86 x 10-3)(5.71 x 10-3)3 Q = 5.32 x 10-10 Q>Ksp so it precipitates

32. Calculate the concentrations of Mg2+ and F- in a mixture of 150 ml of 0.01 M Mg2+ and 250 ml of 0.1 M F-. Ksp for MgF2 is 6.4 x 10-9 [Mg2+]o = (150ml)(0.01M)/400ml = 0.00375 M [F-]o = (250ml)(0.1M)/400ml = 0.0625 M Q = (0.00375)(0.0625)2 = 1.46 x 10-5 Since Q > Ksp a precipitate forms and we have to do an equilibrium problem

33. Chem. init. Change Mg2+ (150ml)0.01M 1.5 mmol - 1.5 2F- (250ml)0.1M 25 � 2(1.5) 25 mmol 22 mmol We have excess F- [F-] = 22 mmol/400ml = 0.055 M

34. Chem. [init.] [equil.] Mg2+ 0 + x F- 0.055 0.055 + 2x = 0.055 Ksp = 6.4 x 10-9 = [Mg2+][F-]2 6.4 x 10-9 = (x)(0.055)2 x = 2.1 x 10-6 M [Mg2+] = 2.1 x 10-6 M [F-] = 0.055 M

35. We can use Ksp values to determine which precipitate forms first since the smaller the Ksp the less ions that will exist in solution. By adding NaI to a mixture of Pb2+ and Cu+ which will precipitate first. PbI2 Ksp = 1.4 x 10-8 CuI Ksp = 5.3 x 10-12 Since CuI is much smaller we can predict that it precipitates first