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Chapter 15

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.

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Chapter 15

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  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 • 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. 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 • HC2H3O20 + 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 • ½ 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 • 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 • 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

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