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 K a = 7.2 x 10 -4 –HF   F - + H + –Chem.[init.][equil.] –HF1.01.0 – x = 1.0 –F - 1.01.0 + x = 1.0 –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

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. HNO 2 / NaNO 2 (NO 2 -) –Adding acid reacts with the NO 2 - to make HNO 2, adding base reacts with the HNO 2 to make NO 2 -

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 (HC 3 H 5 O 3 ) K a = 1.4 x 10 -4 and 0.25M sodium lactate (conj. Base). –Since the K a 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+(V A )(M A ) = #OH-(V B )(M B ) 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 mmol2 mmol –Change- 2 mmol-2 mmol –Final8 mmol0 mmol –pH = - log (8mmol/70ml) = 0.942

9. –At 100 ml we have H+OH- – init.10 mmol10 mmol –Change- 10 mmol-10 mmol –Final0 mmol0 mmol –pH = pH of water = 7

10. –At 200 ml we have H+OH- – init.10 mmol20 mmol –Change- 10 mmol-10 mmol –Final0 mmol10 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 –HC 2 H 3 O 2 OH-C 2 H 3 O 2 - – init.5 mmol1 mmol0 mmol –Change- 1 mmol-1 mmol1 mmol –Final4 mmol0 mmol1 mmol –Chem[Init.][Equil] –HC 2 H 3 O 2 4/604/60 – x = 4/60 –C 2 H 3 O 2 - 1/601/60 + x = 1/60 –H+0+ x

13. –Chem[Init.][Equil] –HC 2 H 3 O 2 4/604/60 – x = 4/60 –C 2 H 3 O 2 - 1/601/60 + 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 –HC 2 H 3 O 2 OH-C 2 H 3 O 2 - – init.5 mmol2.5 mmol0 mmol –Change- 2.5 mmol-2.5 mmol2.5 mmol –Final2.5 mmol0 mmol2.5 mmol –Chem[Init.][Equil] –HC 2 H 3 O 2 2.5/752.5/75 – x = 2.5/75 –C 2 H 3 O 2 - 2.5/752.5/75 + x = 2.5/75 –H+0+ x

15. –Chem[Init.][Equil] –HC 2 H 3 O 2 2.5/752.5/75 – x = 2.5/75 –C 2 H 3 O 2 - 2.5/752.5/75 + 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.) –HC 2 H 3 O 2 OH-C 2 H 3 O 2 - – init.5 mmol5 mmol0 mmol –Change- 5 mmol-5 mmol5 mmol –Final0 mmol0 mmol5 mmol – since all of the acid has been converted to the conjugate base we have to do a K b problem –C 2 H 3 O 2 - + H 2 O   HC 2 H 3 O 2 + OH-

17. –Chem[Init.][Equil] –C 2 H 3 O 2 - 5/1005/100 – x = 5/100 –HC 2 H 3 O 2 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 = x 2 /(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 –HC 2 H 3 O 2 OH-C 2 H 3 O 2 - – init.5 mmol6 mmol0 mmol –Change- 5 mmol-5 mmol5 mmol –Final0 mmol1 mmol5 mmol

19. –HC 2 H 3 O 2 OH-C 2 H 3 O 2 - – init.5 mmol6 mmol0 mmol –Change- 5 mmol-5 mmol5 mmol –Final0 mmol1 mmol5 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 K a problem –Before eq. pt. K a problem –½ way pt. pH = pK a –Eq. pt. K b problem –Past eq. pt. strong base only

21. Calculating K a –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 –½ 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 –PbCl 2 (s)  Pb 2+ + 2Cl - Ksp –Equilibrium constant for solubility –Ksp = [Pb 2+ ][Cl - ] 2

25. Calculate the K sp for Bi 2 S 3 that has a solubility of 1.0 x 10 -15 M. –Bi 2 S 3 (S)  2Bi 3+ + 3S 2- –Ksp = [Bi 3+ ] 2 [S 2- ] 3 –1.0 x 10 -15 M Bi 2 S 3 = 2.0 x 10 -15 M Bi 3+ –1.0 x 10 -15 M Bi 2 S 3 = 3.0 x 10 -15 M S 2- –K sp = (2.0 x 10 -15 ) 2 (3.0 x 10 -15 ) 3 –K sp = 1.1 x 10 -73

26. Calculate the solubility for Cu(IO 3 ) 2 with a K sp = 1.4 x 10 -7 –Cu(IO 3 ) 2  Cu 2+ + 2IO 3 - –K sp = [Cu 2+ ][IO 3 - ] 2 –1.4 x 10 -7 = (x)(2x) 2 = 4x 3 –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 CaF 2 in a 0.025 M NaF solution if the K sp = 4.0 x 10 -11 –Ksp = [Ca 2+ ][F - ] 2 –Chem[init.][equil.] –Ca 2+ 0 + x –F - 0.0250.025 + 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  Sr 2+ + 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 K sp by calculating Q –Remember Q is the same formula as K sp

31. –Will Ce(IO 3 ) 3, K sp = 1.9 x 10 -10, precipitate from a solution made from 750 ml of 0.004 M Ce 3+ and 300 ml of 0.02 M IO 3 - ? –Q = [Ce 3+ ] o [IO 3 - ] 3 o –[Ce 3+ ] o = (750ml)(0.004M)/1050ml = 2.86 x 10 -3 M –[IO 3 - ] o = (300ml)(0.02M)/1050ml = 5.71 x 10 -3 M Q = [Ce 3+ ] o [IO 3 - ] 3 o = (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 Mg 2+ and F - in a mixture of 150 ml of 0.01 M Mg 2+ and 250 ml of 0.1 M F -. K sp for MgF 2 is 6.4 x 10 -9 –[Mg 2+ ] 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 –Mg 2+ (150ml)0.01M 1.5 mmol- 1.5 2F - (250ml)0.1M25 – 2(1.5) 25 mmol22 mmol We have excess F- [F-] = 22 mmol/400ml = 0.055 M

34. –Chem.[init.][equil.] –Mg 2+ 0+ x –F - 0.0550.055 + 2x = 0.055 –Ksp = 6.4 x 10 -9 = [Mg 2+ ][F - ] 2 –6.4 x 10 -9 = (x)(0.055) 2 –x = 2.1 x 10 -6 M –[Mg 2+ ] = 2.1 x 10 -6 M –[F - ] = 0.055 M

35. We can use K sp values to determine which precipitate forms first since the smaller the K sp the less ions that will exist in solution. –By adding NaI to a mixture of Pb 2+ and Cu + which will precipitate first. –PbI 2 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