Chapter 15

1. Chapter 15 Applications of Aqueous Equilibria

2. Common Ion Effect When a salt with the anion of a weak acid is added to that acid (or the cation of a weak base is added to that base), it reverses the dissociation. HF(aq) H+(aq) + F-(aq) NH3(aq) NH4+(aq) + OH-(aq)

3. What is a Buffer? • A weak acid and its conjugate base is present in some form (usually within a salt) OR • A weak base and its conjugate acid is present • in some form (usually within a salt)

4. pH of Buffers Practice Calculate the pH of a solution containing a mixture of 0.100 M HC3H5O2 (Ka = 1.3 x 10-5) and 0.100 M NaC3H5O2 Calculate the pH of a solution containing a mixture of 0.25 M NH3 (Kb = 1.8 x 10-5) and 0.10 M NH4Br.

5. Henderson-Hasselbalch Equation Useful for calculating pH when the [A-]/[HA] Ratios are known. pH = pKa + log [A-]/[HA] pH = pKa + log [base]/[acid] Only can be used if you are dealing with a buffer system!!! [H+] pH Ka pKa

6. Calculate the pH of the following: 0.75 M lactic Acid (HC3H5O3, Ka = 1.4 x 10-4) with 0.25 M sodium lactate (NaC3H5O3)

7. A Buffered Solution …….resists change in its pH when either H+ or OH- are added. 1.0 L of 0.50 M H3CCOOH + 0.50 M H3CCOONa pH = 4.74 Adding 0.010 mol solid NaOH raises the pH of the solution to 4.76, a very minor change

8. Adding a Strong Acid……. ….to a weak acid buffer system:

9. Adding a Strong Acid……. ….to a weak acid buffer system: A strong acid will add its proton to the conjugate base …..to a weak base buffer system:

10. Adding a Strong Base……. ….to a weak acid buffer system:

11. Adding a Strong Base……. ….to a weak acid buffer system: A strong base will add grab protons from the weak acid …..to a weak base buffer system:

12. Buffer Practice Write the reaction that occurs when NaOH is added to the following buffer solutions: 1.00 M HNO2/1.00M NaNO2 Write the reaction that occurs when HCl is added to the following buffer solutions: 0.050 M NH3/0.15 M NH4Cl

13. Buffering Capacity ........represents the amount of H+ or OH- the buffer can absorb without a significant change in pH The pH of a buffered solution is determined by the ratio [A-]/[HA] The capacity of a buffered solution is determined by the magnitudes of [A-]/[HA]

14. Buffering Capacity Practice Calculate the pH of each buffer system below: 5.00 M HAc and 5.00 M NaAc 0.05 M Hac and 0.050 M NaAc Ka = 1.8 x 10-5 After HCl(g) is added to each buffer system, the first one has a resulting pH of 4.74, and the second one has a resulting pH of 4.57. Which one has the better buffering capacity?

15. Review Session #2 Are You Ready?

16. How to Choose a Buffer • The most effective buffers (most resistant to • change) have a ratio [A-]/[HA] = 1 • True when [A-] = [HA] • Make pH = pKa (since log 1 = 0) When choosing a buffer for a desired pH, choose a buffer system whose pKa is close to the desired pH.

17. Review Session #3 Spring Break

18. TITRATION CURVES We can determine the amount of a certain substance by performing a technique called “titration.” Analyte

19. 3 Criteria Must Be Met for a Successful Titration: • Exact reaction between titrant and analyte MUST be known. • Equivalence point MUST be marked accurately. • Volume of titrant required to reach the equivalence point MUST be accurate. Note: When analyte is an acid or base…titrant is a strong base or acid, respectively.

20. Titration (pH) Curve A plot of the solution being analyzed as a function of the amount of titrant added. Equivalence (stoichiometric) point: Enough titrant has been added to react exactly with the solution being analyzed. Moles acid = moles of base

21. Strong Acid-Strong Base Titration WMX : W = Water; M = Molarity; X = add or sub. NmN : N = Neutralization; m = moles; N = numbers

22. Strong Acid-Strong Base Titration • They both dissociate completely • Do the stoichiometry • There is no equilibrium The titration of 40.0 mL of 0.200 M HCl04 with 0.100 M KOH Calculate the pH after the following volumes of KOH have been added: 0,10.0, 40.0, 80.0 and 100.0 mL

23. Strong Acid-Strong Base Titration The titration of 40.0 mL of 0.200 M HCl04 with 0.100 M KOH Calculate the pH after the following volumes of KOH have been added: 0,10.0, 40.0, 80.0 and 100.0 mL

24. Strong Acid-Base Titration

25. Weak Acid-Strong Base Titration Consider the titration of 100.0 mL of 0.200 M acetic acid (Ka = 1.8 x 10-5) by 0.100 M KOH. Calculate the pH of the resulting solution after The following volumes of KOH have been added: 0,50.0, 100.0, 200.0 and 250.0 mL

26. Weak Acid-Strong Base Titration Consider the titration of 100.0 mL of 0.200 M acetic acid (Ka = 1.8 x 10-5) by 0.100 M KOH. Calculate the pH of the resulting solution after The following volumes of KOH have been added: 0,50.0, 100.0, 200.0 and 250.0 mL

27. Titration Practice A beaker contains 100.0 mL of a solution of HOCl (Ka = 3.5 x 10-8) of unknown concentration. • The solution was titrated with 0.100 M NaOH, and the equivalence point was reached when 40.0 mL of NaOH was added. What was the original (initial) concentration of the HOCl solution? • What is the concentration of OCl- ions at the equivalence point? • What is the pH of the solution at the equivalence point?

28. Weak Acid-Strong Base Titration The pH at the equivalence point of a titration of a weak acid with a strong base is always greater than 7

29. Weak Base-Strong Acid TitrationSummary Consider the titration of 100.0 mL of 0.100 M NH3 (Kb = 1.8 x 10-5) by 0.200 M HCl. Calculate the pH of the resulting solution after the following volumes of HCl have been added: 0, 20.0, 25.0, 50.0 and 100.0 mL

30. Weak Base-Strong Acid Titration The pH at the equivalence point of a titration of a weak base with a strong acid is always less than 7

31. Acid-Base Indicators Used in titrations to indicate when we have passed the equivalence point. • Weak acids that change color when they become bases HIn H+ + In- yellow Red • Equilibrium is controlled by pH • End point – when the indicator changes • color….tells us we have reached the eq. point.

32. Acid-Base Indicators Phenolphthalein Ka = 1.9 x 10-9 What is the pKa? 9 Bottom Line…….Take Home Message pH of color change = pKa +/- 1

33. Solubility Product For solids dissolving to from aqueous solutions: Bi2S3(s) 2Bi3+(aq)+ + 3S2-(aq) Ksp = solubility product constant and Ksp = [Bi3+]2 [S2-]3

34. Solubility Product “Solubility” = S = concentration of Bi2S3 that dissolves , which equals ½[Bi3+] and 1/3[S2-]. Note: Ksp is constant (at a given temperature) S is variable (especially with a common ion present)

35. Solubility Product Practice The solubility of BiI3 is 1.32 x 10-5mol/L. Calculate the Ksp value. The solubility of Ca3(PO4)2 is 2.05 x 10-7 mol/L Calculate the Ksp value The Ksp value for PbBr2 is 4.6 x 10-6 Calculate the solubility at 25oC. The Ksp value for Ag2CO3 is 8.1 x 10-12 Calculate the solubility at 25oC.

36. Relative Solubilities • Ksp will only allow us to compare the solubility of solids that fall apart into the same number of ions. • The bigger the Ksp of those the more soluble PbBr2 Ag2CO3 If they fall apart into different number of pieces you have to do the math! (see #3 & 4) from previous slide)

37. Common Ion Effect If we try to dissolve the solid in a solution with either the cation or anion already present, less solid will dissolve. 1. Calculate the solubility of solid SrSO4 (Ksp = 3.2 x 10-7) in a solution of 0.01M Na2SO4. 2. Calculate the solubility of solid Ca3(PO4)2 (Ksp = 1.3 x 10-32) in a solution of 0.20M Na3PO4.

38. pH and Solubility If the anion X- is an effective base – that is, HX is a weak acid – salt MX will show increased solubility in an acidic solution. CaC2O4 Ca2+ + C2O42- H+ + C2O42- HC2O4-