Chapter 15 Applications of Aqueous Equilibria

1. Chapter 15Applications of Aqueous Equilibria

2. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) THE COMMON-ION EFFECT • Consider a solution of acetic acid: • If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left.

3. THE COMMON-ION EFFECT “The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.” Add more of the same ion and there will be less ions of the weak one.

4. THE COMMON-ION EFFECT • The addition of concentrated HCl to a saturated solution of NaCl will cause some solid NaCl to precipitate out of solution. The NaCl has become less soluble because the addition of additional chloride ion. • NaCl+ HCl → more NaCl due to increased Cl-

5. THE COMMON-ION EFFECT pH • If a substance has a basic anion, it will be more soluble in an acidic solution. • Substances with acidic cations are more soluble in basic solutions.

6. THE COMMON-ION EFFECT • The addition of a common ion to a weak acid solution makes the solution LESS acidic. HC2H3O2(aq) ↔ H+(aq) + C2H3O2-(aq) • If NaC2H3O2 is added to the system, the equilibrium shifts to undissociated HC2H3O2 raising the pH. The new pH can be calculated by putting the concentration of the anion into the Ka equation and solving for the new [H+].

7. THE COMMON-ION EFFECT • Adding NaF to a solution of HF causes more HF to be produced. Major species are HF, Na+, F-, and H2O. Common ion is F-. • In a 1.0MNaF and 1.0M HF solution, there is more HF in the presence of NaF. HF(aq) ↔ H+(aq) + F-(aq) • Le Chatelier’s indicates that additional F- due to the NaF causes a shift to the left and thus generates more HF.

8. THE COMMON-ION EFFECT Finding the pH • Always determine the major species. • Write the equilibrium equation and expression. • Determine the initial concentrations. • Do ICE chart and solve for x. • Once [H+] has been found, find pH.

9. PRACTICE ONE The equilibrium concentration of H+ in a 1.0M HF solution is 2.7 x 10-2M and the percent dissociation of HF is 2.7%. Calculate [H+] and the percent dissociation of HF in a solution containing 1.0M HF (Ka = 7.2 x 10-4) and 1.0MNaF.

10. THE COMMON-ION EFFECT Notice: 1.0M HF % dissociation 2.7% versus 1.0M HF and 1.0MNaF % dissociation 0.072%

11. HF(aq) + H2O(l) H3O+(aq) + F−(aq) The Common-Ion Effect Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.

12. EQUATIONS QUIZ • For each of the following reactions, write an equation for the reaction. Write the net ionic equation for each. Omit formulas for spectator ions or molecules in the reaction. Put a box around your final answer.

13. DO NOW • Pick handout due tomorrow. • Turn in lab – make sure the you have: • One Title page, one Prelab, one Data Table, Calculations Table for everyone, and Calculations 1-8 for everyone in that order! • Get out notes.

14. BUFFERS • Solutions of a weak conjugate acid-base pair. • They are particularly resistant to pH changes, even when strong acid or base is added. • Just a case of the common ion effect.

15. BUFFERS • You are always adding a strong acid or strong base to a buffer solution. • Buffer system (conjugate acid-base pair) acts as a net – “catches” acid or base

16. BUFFERS If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.

17. BUFFERS If acid is added, the F− reacts to form HF and water.

18. BUFFERS Example: HC2H3O2/ C2H3O2- buffer system (the “net”) • Add a strong acid: H+ + C2H3O2- → HC2H3O2forms a weak acid • Add a strong base: OH-+ HC2H3O2 → C2H3O2- + H2O forms a weak base

19. BUFFERS Example: NH3/ NH4+ buffer system (the “net”) • Add a strong acid: H+ + NH3 → NH4+forms a weak acid • Add a strong base: OH-+ NH4+ → NH3 + H2O forms a weak base

20. PRACTICE TWO • A buffered solution contains 0.050Macetic acid (HC2H3O2,Ka = 1.8 x 10-5) and 0.50M sodium acetate (NaC2H3O2). Calculate the pH of the solution.

21. HELPFUL TIPS • Buffered solutions are simply solutions of weak acids and bases containing a common ion. • The pH calculations for buffered solutions require exactly the same procedures as determining the pH of weak acid or weak base solutions learned previously. • When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. After the stoichiometric calculations are completed, then consider the equilibrium calculations.

22. BUFFERS • Adding a strong acid or base to a buffered solution Requires moles

23. PRACTICE THREE Calculate the change in pH that occurs when 0.010mol solid NaOH is added to 1.0L of the buffered solution contains 0.050Macetic acid (HC2H3O2,Ka = 1.8 x 10-5) and 0.50M sodium acetate (NaC2H3O2). Compare this pH change with that which occurs when 0.010mol solid NaOH is added to 1.0L of water.

24. HOW BUFFERING WORKS • When hydroxide ions are added to the solution, the weak acid provides the source of protons. The OH- ions are not allowed to accumulate, but are replaced by A-. OH- + HA → A- + H2O • pH can be understood by looking at the EQ expression. Ka = or [H+] = Ka

25. HOW BUFFERING WORKS

26. HOW BUFFERING WORKS • So [H+] (and thus pH) is determined by the ratio of [HA]/[A-]. When OH- ions are added, HA converts to A- and the ratio decreases. However, is the amounts of [HA] and [A-] are LARGE, then the change in the ratio will be small. • If [HA]/[A-] = 0.50M / 0.50M = 1.0M initially. After adding 0.010M OH-, it becomes [[HA]/[A-] = 0.49M/ 0.51M = 0.96M. Not much of a change. [H+] and pH are essentially constant.

27. HOW BUFFERING WORKS

28. Ka = [H3O+] [A−] [HA] HA + H2O H3O+ + A− BUFFER CALCULATIONS Consider the equilibrium constant expression for the dissociation of a generic acid, HA:

29. [A−] [HA] [H3O+] Ka = base [A−] [HA] −log [H3O+] + −log −log Ka = pKa acid pH BUFFER CALCULATIONS Rearranging slightly, this becomes Taking the negative log of both sides, we get

30. [base] [acid] pKa = pH − log [base] [acid] pH = pKa + log BUFFER CALCULATIONS • So • Rearranging, this becomes • This is the Henderson–Hasselbalchequation.

31. BUFFER CALCULATIONS • For a particular buffering system, all solutions that have the same ratio of [A-] /[HA] have the same pH. • Optimum buffering occurs when [HA] = [A-] and the pKa of the weak acid used should be as close to possible to the desired pH of the buffer system.

32. HENDERSON-HASSELBACH • The equation needs to be used cautiously. • It is sometimes used as a quick, easy equation in which to plug in numbers. • A Ka or Kb problem requires a greater understanding of the factors involved and can ALWAYS be used instead of the HH equation. • However, at the halfway point (as in a titration), the HH is very useful.

33. PRACTICE FOUR What is the pH of a buffer that is 0.75 Mlactic acid, HC3H5O3, and 0.25 M in sodium lactate? Ka for lactic acid is 1.4  10−4.

34. [base] [acid] (0.25) (0.75) pH = pKa + log pH = −log (1.4  10−4) + log HENDERSON–HASSELBALCH EQUATION pH = 3.85 + (−.477) pH = 3.37

35. HINTS • Determine the major species involved. • If a chemical reaction occurs, write the equation and solve stoichiometry. • Write the EQ equation. • Set up the equilibrium expression (Ka or Kb) of the HH equation. • Solve. • Check the logic of the answer.

36. PRACTICE FIVE A buffered solution contains 0.25M NH3 (Kb = 1.8 x 10-5) and 0.40M NH4Cl. Calculate the pH of this solution.

37. PRACTICE SIX Calculate the pH of the solution that results when 0.10mol gaseous HCl is added to 1.0Lof the buffered solution of contains 0.25M NH3 (Kb = 1.8 x 10-5) and 0.40M NH4Cl.

38. BUFFERING CAPACITY • This is the amount of acid or base that can be absorbed by a buffer system without a significant change in pH. • In order to have a large buffer capacity, a solution should have large concentrations of both buffer components.

39. PRACTICE SEVEN Calculate the change in pH that occurs when 0.010mol gaseous HCl is added to 1.0L of each of the following solutions (Ka for acetic acid = 1.8 x 10-5): • Solution A: 5.00M HC2H3O2 and 5.00M NaC2H­3O2 • Solution B: 0.050M HC2H3O2 and 0.050M NaC2H­3O2

40. HINT • We see that the pH of a buffered solution depends on the ratio of the [base] to [acid] (or [acid] to [base]). • Big concentration difference = large pH change

41. PRACTICE EIGHT A chemist needs a solution buffered at pH 4.30 and can choose from the following list of acids and their soluble salt: a. chloroaceticacid Ka= 1.35 × 10-3 b. propanoic acid Ka= 1.3 × 10-5 c. benzoic acid Ka= 6.4 × 10-5 d. hypochlorusacid Ka= 3.5 × 10-8 Calculate the ratio of [HA] / [A-] required for each system to yield a pH of 4.30. Which system works best?

42. TITRATIONS and pH CURVES • Only when the acid AND base are both strong is the pH at the equivalence point 7. • Any other conditions and you get to do an equilibrium problem. It is really a stoichiometry problem with a limiting reactant. The “excess” is responsible for the pH • Weak acid + strong base equivalence pt. > pH 7 • Strong acid + weak base equivalence pt. < pH 7

43. END PT VS EQUIVALENCE PT • There is a distinction between the equivalence point and the end point. • The end point is when the indicator changes color. • If you’ve made a careful choice of indicators, the equivalence point, when the number of moles of acid = number of moles of base, will be achieved at the same time.

44. VOCABULARY • Titrant – solution of know concentration (usually in the buret). The titrant is added to a solution of unknown concentration until the substance being analyzed is just consumes (stoichiometric point or equivalence point). • Titration or pH Curve – plot of pH as a function of the amount of titrant added.

45. pH RANGE • The pH range is the range of pH values over which a buffer system works effectively. • It is best to choose an acid with a pKa close to the desired pH.

46. Titration A known concentration of base (or acid) is slowly added to a solution of acid (or base).

47. Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

48. STRONG ACID-STRONG BASE H+(aq) + OH-(aq) → H2O(l) • To compute H+, we have to know how much H+ remains at that point in the titration. • New unit: millimole, mmol – titrations usually involve small quantities. • This means Molarity = = . • So a 1.0M = =

49. STRONG ACID-STRONG BASE Example - For the titration of 50.0 mL of 0.200M HNO3 with 0.100M NaOH, calculate the pH of the solution at the following selected points of the titration: • 0.0 mL of 0.100MNaOH has been added. • 10.0 mL of 0.100MNaOH has been added. • 20.0 mL of 0.100MNaOH has been added. • 50.0 mL of 0.100MNaOH has been added.

50. STRONG ACID-STRONG BASE a. 0.0 mL of 0.100MNaOH has been added to 0.200M HNO3. • Major Species: H+, NO3-, H2O HNO3 = strong acid • pH = -log[H+] = -log (0.200) = 0.699