1. Chapter 16: Aqueous Ionic Equilibria CHE 124: General Chemistry II Dr. Jerome Williams, Ph.D. Saint Leo University

2. Overview Basic Buffers & Hendersen-Hasselbalch Buffering Effectiveness (Buffer Capacity) Titration Fundamentals Indicators Solubility Equilibria

3. Basic Buffers B: (aq) + H 2 O (l)  H:B + (aq) + OH − (aq) Buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B + Cl − H 2 O (l) + NH 3 (aq)  NH 4 + (aq) + OH − (aq) 3 Tro: Chemistry: A Molecular Approach, 2/e

4. The Henderson-Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product B: + H 2 O  H:B + + OH − To apply the Henderson-Hasselbalch equation, the chemical equation of the basic buffer must be looked at like an acid reaction H:B + + H 2 O  B: + H 3 O + this does not affect the concentrations, just the way we are looking at the reaction Henderson-Hasselbalch Equation for Basic Buffers 4 Tro: Chemistry: A Molecular Approach, 2/e

5. Relationship between pK a and pK b Just as there is a relationship between the K a of a weak acid and K b of its conjugate base, there is also a relationship between the pK a of a weak acid and the pK b of its conjugate base K a  K b = K w = 1.0 x 10 −14 −log(K a  K b ) = −log(K w ) = 14 −log(K a ) + −log(K b ) = 14 pK a + pK b = 14 5 Tro: Chemistry: A Molecular Approach, 2/e

6. Example 16.4: What is the pH of a buffer that is 0.50 M NH 3 (pK b = 4.75) and 0.20 M NH 4 Cl? 6 find the pK a of the conjugate acid (NH 4 + ) from the given K b assume the [B] and [HB + ] equilibrium concentrations are the same as the initial substitute into the Henderson- Hasselbalch equation check the “x is small” approximation NH 3 + H 2 O  NH 4 + + OH − Tro: Chemistry: A Molecular Approach, 2/e

7. The Henderson-Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product B: + H 2 O  H:B + + OH − We can rewrite the Henderson-Hasselbalch equation for the chemical equation of the basic buffer in terms of pOH Henderson-Hasselbalch Equation for Basic Buffers 7 Tro: Chemistry: A Molecular Approach, 2/e

8. Example 16.4: What is the pH of a buffer that is 0.50 M NH 3 (pK b = 4.75) and 0.20 M NH 4 Cl? 8 find the pK b if given K b assume the [B] and [HB + ] equilibrium concentrations are the same as the initial substitute into the Henderson-Hasselbalch equation base form, find pOH check the “x is small” approximation calculate pH from pOH NH 3 + H 2 O  NH 4 + + OH − Tro: Chemistry: A Molecular Approach, 2/e

9. Buffering Effectiveness A good buffer should be able to neutralize moderate amounts of added acid or base However, there is a limit to how much can be added before the pH changes significantly The buffering capacity is the amount of acid or base a buffer can neutralize The buffering range is the pH range the buffer can be effective The effectiveness of a buffer depends on two factors (1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base 9 Tro: Chemistry: A Molecular Approach, 2/e

10. HAA−A− OH − mols before0.180.0200 mols added ── 0.010 mols after 0.170.030≈ 0 Effect of Relative Amounts of Acid and Conjugate Base Buffer 1 0.100 mol HA & 0.100 mol A − Initial pH = 5.00 Buffer 2 0.18 mol HA & 0.020 mol A − Initial pH = 4.05 pK a (HA) = 5.00 HA + OH −  A  + H 2 O HAA−A− OH − mols before0.100 0 mols added ── 0.010 mols after 0.0900.110≈ 0 A buffer is most effective with equal concentrations of acid and base after adding 0.010 mol NaOH pH = 5.09 after adding 0.010 mol NaOH pH = 4.25

11. HAA−A− OH − mols before0.500.5000 mols added ── 0.010 mols after 0.490.51≈ 0 HAA−A− OH − mols before0.050 0 mols added ── 0.010 mols after 0.0400.060≈ 0 Effect of Absolute Concentrations of Acid and Conjugate Base Buffer 1 0.50 mol HA & 0.50 mol A − Initial pH = 5.00 Buffer 2 0.050 mol HA & 0.050 mol A − Initial pH = 5.00 pK a (HA) = 5.00 HA + OH −  A  + H 2 O A buffer is most effective when the concentrations of acid and base are largest after adding 0.010 mol NaOH pH = 5.02 after adding 0.010 mol NaOH pH = 5.18

12. Buffering Capacity a concentrated buffer can neutralize more added acid or base than a dilute buffer 12 Tro: Chemistry: A Molecular Approach, 2/e

13. Effectiveness of Buffers A buffer will be most effective when the [base]:[acid] = 1 – equal concentrations of acid and base A buffer will be effective when 0.1 < [base]:[acid] < 10 A buffer will be most effective when the [acid] and the [base] are large 13 Tro: Chemistry: A Molecular Approach, 2/e

14. Buffering Range We have said that a buffer will be effective when 0.1 < [base]:[acid] < 10 Substituting into the Henderson-Hasselbalch equation we can calculate the maximum and minimum pH at which the buffer will be effective Lowest pHHighest pH Therefore, the effective pH range of a buffer is pK a ± 1 When choosing an acid to make a buffer, choose one whose is pK a closest to the pH of the buffer 14 Tro: Chemistry: A Molecular Approach, 2/e

15. Formic acid, HCHO 2 pK a = 3.74 Example 16.5a: Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with pH 4.25? Chlorous acid, HClO 2 pK a = 1.95 Nitrous acid, HNO 2 pK a = 3.34 Hypochlorous acid, HClOpK a = 7.54 The pK a of HCHO 2 is closest to the desired pH of the buffer, so it would give the most effective buffering range 15 Tro: Chemistry: A Molecular Approach, 2/e

16. Example 16.5b: What ratio of NaCHO 2 : HCHO 2 would be required to make a buffer with pH 4.25? 16 Formic acid, HCHO 2, pK a = 3.74 To make a buffer with pH 4.25, you would use 3.24 times as much NaCHO 2 as HCHO 2 Tro: Chemistry: A Molecular Approach, 2/e

17. Practice – What ratio of NaC 7 H 5 O 2 : HC 7 H 5 O 2 would be required to make a buffer with pH 3.75? Benzoic acid, HC 7 H 5 O 2, pK a = 4.19 17 Tro: Chemistry: A Molecular Approach, 2/e

18. Practice – What ratio of NaC 7 H 5 O 2 : HC 7 H 5 O 2 would be required to make a buffer with pH 3.75? Benzoic acid, HC 7 H 5 O 2, pK a = 4.19 to make a buffer with pH 3.75, you would use 0.363 times as much NaC 7 H 5 O 2 as HC 7 H 5 O 2 18 Tro: Chemistry: A Molecular Approach, 2/e

19. Buffering Capacity Buffering capacity is the amount of acid or base that can be added to a buffer without causing a large change in pH The buffering capacity increases with increasing absolute concentration of the buffer components As the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves Buffers that need to work mainly with added acid generally have [base] > [acid] Buffers that need to work mainly with added base generally have [acid] > [base] 19 Tro: Chemistry: A Molecular Approach, 2/e

20. Titration In an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete – when the reaction is complete we have reached the endpoint of the titration An indicator may be added to determine the endpoint – an indicator is a chemical that changes color when the pH changes When the moles of H 3 O + = moles of OH −, the titration has reached its equivalence point 20 Tro: Chemistry: A Molecular Approach, 2/e

21. Titration 21 Tro: Chemistry: A Molecular Approach, 2/e

22. Titration Curve A plot of pH vs. amount of added titrant The inflection point of the curve is the equivalence point of the titration Prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH The pH of the equivalence point depends on the pH of the salt solution – equivalence point of neutral salt, pH = 7 – equivalence point of acidic salt, pH < 7 – equivalence point of basic salt, pH > 7 Beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH 22 Tro: Chemistry: A Molecular Approach, 2/e

23. Titration Curve: Unknown Strong Base Added to Strong Acid 23 Tro: Chemistry: A Molecular Approach, 2/e

24. Before Equivalence (excess acid) After Equivalence (excess base) Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH Equivalence Point equal moles of HCl and NaOH pH = 7.00  Because the solutions are equal concentration, and 1:1 stoichiometry, the equivalence point is at equal volumes 24 Tro: Chemistry: A Molecular Approach, 2/e

25. Titration of a Strong Base with a Strong Acid If the titration is run so that the acid is in the burette and the base is in the flask, the titration curve will be the reflection of the one just shown 25 Tro: Chemistry: A Molecular Approach, 2/e

26. Titration of a Weak Acid with a Strong Base Titrating a weak acid with a strong base results in differences in the titration curve at the equivalence point and excess acid region The initial pH is determined using the K a of the weak acid The pH in the excess acid region is determined as you would determine the pH of a buffer The pH at the equivalence point is determined using the K b of the conjugate base of the weak acid The pH after equivalence is dominated by the excess strong base – the basicity from the conjugate base anion is negligible 26 Tro: Chemistry: A Molecular Approach, 2/e

27. Titrating Weak Acid with a Strong Base The initial pH is that of the weak acid solution – calculate like a weak acid equilibrium problem e.g., 15.5 and 15.6 Before the equivalence point, the solution becomes a buffer – calculate mol HA init and mol A − init using reaction stoichiometry – calculate pH with Henderson-Hasselbalch using mol HA init and mol A − init Half-neutralization pH = pK a 27 Tro: Chemistry: A Molecular Approach, 2/e

28. Titrating Weak Acid with a Strong Base At the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established – mol A − = original mole HA calculate the volume of added base as you did in Example 4.8 – [A − ] init = mol A − /total liters – calculate like a weak base equilibrium problem e.g., 15.14 Beyond equivalence point, the OH is in excess – [OH − ] = mol MOH xs/total liters – [H 3 O + ][OH − ]=1 x 10 −14 28 Tro: Chemistry: A Molecular Approach, 2/e

29. Indicators Many dyes change color depending on the pH of the solution These dyes are weak acids, establishing an equilibrium with the H 2 O and H 3 O + in the solution HInd (aq) + H 2 O (l)  Ind  (aq) + H 3 O + (aq) The color of the solution depends on the relative concentrations of Ind  :HInd – when Ind  :HInd ≈ 1, the color will be mix of the colors of Ind  and HInd – when Ind  :HInd > 10, the color will be mix of the colors of Ind  – when Ind  :HInd < 0.1, the color will be mix of the colors of HInd 29 Tro: Chemistry: A Molecular Approach, 2/e

30. Phenolphthalein 30 Tro: Chemistry: A Molecular Approach, 2/e

31. Methyl Red 31 Tro: Chemistry: A Molecular Approach, 2/e

32. Monitoring a Titration with an Indicator For most titrations, the titration curve shows a very large change in pH for very small additions of titrant near the equivalence point An indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH – pK a of HInd ≈ pH at equivalence point 32 Tro: Chemistry: A Molecular Approach, 2/e

33. Acid-Base Indicators 33 Tro: Chemistry: A Molecular Approach, 2/e

34. Solubility Equilibria All ionic compounds dissolve in water to some degree – however, many compounds have such low solubility in water that we classify them as insoluble We can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water 34 Tro: Chemistry: A Molecular Approach, 2/e

35. Solubility Product The equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, K sp For an ionic solid M n X m, the dissociation reaction is: M n X m (s)  nM m+ (aq) + mX n− (aq) The solubility product would be K sp = [M m+ ] n [X n− ] m For example, the dissociation reaction for PbCl 2 is PbCl 2 (s)  Pb 2+ (aq) + 2 Cl − (aq) And its equilibrium constant is K sp = [Pb 2+ ][Cl − ] 2 35 Tro: Chemistry: A Molecular Approach, 2/e

36. 36 Tro: Chemistry: A Molecular Approach, 2/e

37. Molar Solubility Solubility is the amount of solute that will dissolve in a given amount of solution – at a particular temperature The molar solubility is the number of moles of solute that will dissolve in a liter of solution – the molarity of the dissolved solute in a saturated solution for the general reaction M n X m (s)  nM m+ (aq) + mX n− (aq) 37 Tro: Chemistry: A Molecular Approach, 2/e

38. Example 16.8: Calculate the molar solubility of PbCl 2 in pure water at 25  C 38 write the dissociation reaction and K sp expression create an ICE table defining the change in terms of the solubility of the solid [Pb 2+ ][Cl − ] Initial00 Change+S+2S EquilibriumS2S PbCl 2 (s)  Pb 2+ (aq) + 2 Cl − (aq) K sp = [Pb 2+ ][Cl − ] 2 Tro: Chemistry: A Molecular Approach, 2/e

39. Example 16.8: Calculate the molar solubility of PbCl 2 in pure water at 25  C 39 substitute into the K sp expression find the value of K sp from Table 16.2, plug into the equation, and solve for S [Pb 2+ ][Cl − ] Initial00 Change+S+2S EquilibriumS2S K sp = [Pb 2+ ][Cl − ] 2 K sp = (S)(2S) 2 Tro: Chemistry: A Molecular Approach, 2/e

40. Example 16.10: Calculate the molar solubility of CaF 2 in 0.100 M NaF at 25  C 40 substitute into the K sp expression, assume S is small find the value of K sp from Table 16.2, plug into the equation, and solve for S [Ca 2+ ][F − ] initial00.100 change+S+2S equilibriumS0.100 + 2S K sp = [Ca 2+ ][F − ] 2 K sp = (S)(0.100 + 2S) 2 K sp = (S)(0.100) 2 Tro: Chemistry: A Molecular Approach, 2/e