Chemical Buffers and Titration in the Field

1. Welcome Back!!! • Bellwork: • Take a deep breath as you recollect your scientific minds. • Picture yourself as being successful the next couple of weeks as we finish off the semester. • Turn in your lab reports. • Turn to a partner and using 1-2 sentences, describe the main topic of each of the 5 units we have covered thus far. How might these topics be used in the chemical field?

2. Buffers and Titration Chapter 15

3. What is a buffer? • Buffered Solution: Resists change despite the addition of an acid or a base. • Made from a weak acid or base and its salt.

4. What causes buffers to work? • Common Ion Effect: a shift in the equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction. • Example: NH4Cl(s) +H2O(l) → NH4+(aq) + Cl-(aq) NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)

5. Acid/Salt Buffering Pairs • The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)

6. Base/Salt Buffering Pairs • The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3)

7. pH of a Buffer Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, and 0.50M sodium acetate, NaC2H3O2when the Ka= 1.8 x 10-5 ? Step #1: Write the dissociation equation: HC2H3O2 C2H3O2- + H+

8. pH of a Buffer Problem Step #2: ICE it! HC2H3O2 C2H3O2- + H+ 0.50 0 0.50 +x +x - x 0.50+x x 0.50 - x

9. pH of a Buffer Problem Step #3: Set up the law of mass action HC2H3O2 C2H3O2- + H+ E 0.50+x 0.50 - x x

10. pH of a Buffer Problem Step #4: Calculate x and pH: x = 1.8 x 10-5 pH = 4.74 This illustrates an important point: The buffer’s pH at equal concentrations of salt and weak acid (or base) equals the pKa (or pKb) of that substance, in other words, the –log of the Ka or Kb for that acid or base!

11. pH of Buffers GENERAL EQUATION: HA  A- + H+

12. Henderson-Hasselbalch Equation This is an exceptionally powerful tool, and it’s use will be emphasized in our problem solving.

13. Buffer Capacity • The pH of a buffered solution is determined by the ratio of [A-]/[HA]. • The capacity of a buffer solution is determined by the magnitudes of [HA] and [A-]. • Optimal buffering occurs when the [A-]/[HA] is equal to 1.

14. How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH is 9.00? (assume no change in volume) Analyze: Major species? NH4+, NH3, and Cl- NH4+ and NH3 will determine pH of the buffer solution: Kb = [NH4+] [OH-] = 1.8 x 10-5 [NH3]

15. pOH = 14.00-9.00 = 5.00 [OH-] = 1.0 x 10-5 [ NH3] = 0.10M Remember: Kb = [NH4+] [OH-] = 1.8 x 10-5 [NH3] 1.8 x 10-5 = [NH4+] (1.0 x 10-5) 0.10M [NH4+] = 0.18M So: (0.18 moles/liter)(2.0L) = 0.36 moles of NH4Cl

16. More Practice • A chemist needs a solution buffered at pH 4.30 and can choose from the following: • Chloroacetic acid (Ka = 1.35 x 10-3) • Propanoic acid (Ka = 1.3 x 10-5) • Benzoic acid (Ka = 6.4 x 10-5) • Hypochlorous acid (Ka = 3.5 x 10-8) • Which is best?

17. Homework • Pg. 740 #21, 23-25, 27-34, 37, 38, 41-46, 49, 50

18. Buffer Recap • A buffer contains a large concentration of both a conjugate acid and the conjugate base. • This is usually made with either a base or an acid and a salt of the conjugate species. • Weak acids and their conjugate bases make for good buffers, strong acids and bases do not! • When the pH > pKa, the base form has a higher concentration, but when pH < pKa, the acid form has a higher concentration.

19. Buffer Recap cont. • Ratio: [A-]/[HA] • The buffer capacity is related to the ratio of the conjugate acids and bases. • The closer the ratio is to 1, the better the buffer is. • The ratio of conjugate acid and base must change by a factor of 10 to change the pH level by 1.

20. Titrations • Adding a solution of a known concentration (titrant) to a solution of an unknown concentration (analyte or titrate) by a buret until the unknown solution has been consumed. • This is a common way to find the concentration of an acid or base • Reactions of acids and bases are called neutralization reactions, and these reactions generally have K>1, and thus can be considered to go to completion.

21. Neutralization Reactions • For a mixture of a strong acid with a strong base, the neutralization reaction is • Kw = 10-14 at 25◦C, so the reaction goes to completion. This allows the pH of mixtures of strong acids and bases to be determined from the limiting reactant, either the acid or the base. • Whatever is in excess (acid or base) will cause the pH to change.

22. Neutralization Reactions cont. • When a strong base is added to a solution of a weak acid, a neutralization reaction occurs: • conjugate acid + OH-→ conjugate base + H2O • When a strong acid is added to a solution of a weak base, a neutralization reaction occurs: • conjugate base + H3O+→ conjugate acid + H2O

23. Titration Curves (pH curves) • Graph displaying the pH changes with respect to the addition of the titrant. • As base is added to either a strong or weak acid solution, the hydronium concentration does not change much. The change in pH is less than ~1.5 for the region where 10-90 percent of the base needed to reach the equivalence point has been added.

24. Titration Curves (pH curves) • A titration technique exists for neutralization reactions. At the equivalence point, the moles of titrant and the moles of titrate are present in stoichiometric proportions. In the vicinity of the equivalence point, the pH rapidly changes. This can be used to determine the concentration of the titrate.

25. Equivalence Points • With both strong acids and bases titrated together the equivalence point will be at pH = 7. • With a weak acid titrated with a strong base the equivalence point will be at a pH > 7. • With a weak base titrated with a strong acid the equivalence point will be at a pH < 7.

26. Acid-Base Indicator • Indicators mark the end point of a titration by changing color. • The human eye will notice a color change when the ratio between the indicator and the conjugate base of the indicator are a factor of 10, such as 1/10.

27. Indicator Ranges

28. Homework • Pg. 741-742 • # 51-59, 61-63 • Pg. 743 • # 73-74

29. Bellwork • Write down two facts that you remember about solubility from honors chem. • Be prepared to share.

30. Solubility Chapter 15 cont.

31. Solubility as an Equilibrium Position • Solubility can represent the equilibrium position. CuBr (s)  Cu+ (aq) + Br- (aq) Ksp=[Cu+]m [Br-]n • m and n are the coefficients of the respective ions. • The “change” or “x” represents the “solubility” of the solid. Ksp is called the solubility-product constant • This is an equilibrium constant and is not in itself the solubility! • The solubility equilibrium position is not determined by pure liquids or solids.

32. Common Ions • With the presence of common ions, the solubility of a solid decreases. • What is the solubility of CaF2 (Ksp=4.0x10-11) in a 0.025M NaF solution? • CaF2(s)  Ca2+(aq) + 2F-(aq) • Ksp =4.0x 10-11= [Ca2+][F-]2

33. Common Ions/Practice cont.

34. Common Ions/Practice cont. • Ksp =4.0x 10-11= [x][0.025+2x]2 • We assume that x << 0.025M • Ksp =4.0x 10-11= [x][0.025]2 • Solve for x, x = 6.4 x 10-8M • Thus the solubility of the CaF2 in 0.025M NaF solution is 6.4 x 10-8mol/L

35. Solubility and Acidity • The solubility of a salt will be pH sensitive when one of the ions is an acid or base. • Example: • Mg(OH)2(s)  Mg2+(aq) + 2OH-(aq)

36. Equilibrium Position and Q • The dissolution of a substance in a solvent is a reversible reaction, and so has an associated equilibrium constant. For dissolution of a salt, the reaction quotient, Q, is referred to as the solubility product, and the equilibrium constant for this reaction is denoted as Ksp, the solubility product constant. • If Q > Kspthen precipitation will occur. • If Q < Ksp then no precipitation will occur.

37. Solubility Rules • All sodium, potassium, ammonium, and nitrate salts are soluble in water.

38. Homework • Pg.743 • #75-82, 91, 92, 95, 97, 98, 99-101