CHEMISTRY Chapter 15 Applications of Aqueous Equilibria

1. CHEMISTRYChapter 15Applications of Aqueous Equilibria

2. Two important points: • Reactions with strong acids or strong bases go to completion. • Reactions with only weak acids and bases reach an equilibrium.

3. The pH scale • The pH scale ranges from 0 to 14. • Acids have a pH less than 7. • A base has a pH greater than 7. • Pure water has a pH equal to 7.

6. Acids and bases in your body • Many reactions, such as the ones that occur in your body, work best at specific pH values.

7. pH and blood • The pH of your blood is normally within the range of 7.3–7.5. • Holding your breath causes blood pH to drop. • High blood pH can be caused by hyperventilating.

8. Acids & Bases • Strong acids: • Know the names and formulas of the 7 common strong acids: • HCl (aq) hydrochloric acid • HBr (aq) hydrobromic acid • HI (aq) hydroiodic acid • HClO3 chloric acid • HClO4 perchloric acid • HNO3 nitric acid • H2SO4 sulfuric acid

9. Acids & Bases • Examples of Weak Acids HF (aq) hydrofluoric acid H3PO4 phosphoric acid CH3COOH acetic acid

10. Acids & Bases • Strong Bases: Know the names and formulas of the strong bases • Alkali metal (1A) hydroxides • LiOH lithium hydroxide • NaOH sodium hydroxide • KOH potassium hydroxide • RbOH rubidium hydroxide • CsOH cesium hydroxide

11. Acids & Bases • Strong bases to know (con’t): • Heavy alkaline earth metal (2A) hydroxides • Ca(OH)2 calcium hydroxide • Sr(OH)2 strontium hydroxide • Ba(OH)2 barium hydroxide

12. Acids & Bases • Examples of Weak Bases: ammonia (NH3) sodium bicarbonate (NaHCO3) • baking soda • a component of Alka-Seltzer

13. Generally, when solutions of an acid and a base are combined, the products are a salt and water. CH3COOH(aq) + NaOH(aq)CH3COONa (aq) + H2O(l) HCl (aq) + NaOH(aq) NaCl (aq) + H2O(l) All neutralization reactions are double displacement reactions Acids + Bases Neutralization Reaction

14. Neutralization Reactions When a strong acid reacts with a strong base, the net ionic equation is… HCl (aq) + NaOH(aq)  NaCl (aq) + H2O(l) H+(aq) + Cl- (aq) + Na+ (aq) + OH-(aq) Na+ (aq) + Cl- (aq) + H2O (l) H+(aq) + OH-(aq) H2O(l)

15. Strong Acids and Bases Chapter 16

16. Weak Acids and Strong Bases Chapter 16

17. Strong Acid Weak Base Chapter 16

18. Weak Acid Weak Base Chapter 16

19. COMMON-ION EFFECT A shift in equilibrium due to addition of same ion salt to an aqueous weak acid or weak base is the common-ion effect. This is an example of Le Chatlier’s Principle.

20. The Common Ion Effect • What affect does the addition of its conjugate base have on the weak acid equilibrium? On the pH? Used in making buffered solutions

21. The Common-Ion Effect • Common Ion: Two dissolved solutes that contain the same ion (cation or anion). • The presence of a common ion suppresses the ionization of a weak acid or a weak base. • Common-Ion Effect: is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.

22. The Common-Ion Effect Q: Calculate the pH of a 0.20 M CH3COOH solution with no salt added. Q: Calculate the pH of a solution containing 0.20 M CH3COOH and 0.30 M CH3COONa. Q: What is the pH of a solution containing 0.30 M HCOOH, before and after adding 0.52 M HCOOK?

23. Buffer Solutions • A Buffer Solution:is a solution of: (1) a weak acid or a weak base and (2) its salt; both components must be present. • A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. • Buffers are very important to biological systems.

24. Buffer Solutions

25. How do buffers work?

26. Adding strong acid or base to a buffer Adding acid: H3O+ + HA or A- → Adding base: OH- + HA or A-→ Calculating pH: • Stoichiometry of added acid or base • Equilibrium problem (H-H equation)

27. Henderson–Hasselbalch equation • To determine the pH, we apply I.C.E. and then the Henderson–Hasselbalch equation. • When the concentration of HA and salt are high (≥0.1 M) we can neglect the ionization of acid and hydrolysis of salt.

28. Henderson–Hasselbalch equation Henderson-Hasselbalch equation

29. Buffer Solutions • Buffer solutions must contain relatively high acid and base component concentrations, the buffer capacity. • Acid and base component concentrations must not react together. • The simplest buffer is prepared from equal concentrations of acid and conjugate base.

30. Buffer Solutions • Buffer Preparation: Use the Henderson–Hasselbalch equation in reverse. • Choose weak acid with pKa close to required pH. • Substitute into Henderson–Hasselbalch equation. • Solve for the ratio of [conjugate base]/[acid]. • This will give the mole ratio of conjugate base to acid. The acid should always be 1.0.

31. Buffer Solutions Q: Describe how you would prepare a “phosphate buffer” with a pH of about 7.40. Q: How would you prepare a liter of “carbonate buffer” at a pH of 10.10? You are provided with carbonic acid (H2CO3), sodium hydrogen carbonate (NaHCO3), and sodium carbonate (Na2CO3).

32. Buffer Solutions Q: Calculate the pH of a buffer system containing 1.0 M CH3COOH and 1.0 M CH3COONa. What is the pH of the system after the addition of 0.10 mole of gaseous HCl to 1.0 L of solution? Q: Calculate the pH of 0.30 M NH3/0.36 NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?

33. Acid-Base Titrations • Titration – a reaction used to determine concentration (acid-base, redox, precipitation) • Titrant – solution in buret; usually a strong base or acid • Analyte – solution being titrated; often the unknown • equivalence point (or stoichiometric point): mol acid = mol base • Found by titration with an indicator • Solution not necessarily neutral • pH dependent upon salt formed • pH titration curve – plot of pH vs. titrant volume

34. Solubility Equilibria • Solubility Product: is the product of the molar concentrations of constituent ions and provides a measure of a compound’s solubility. MX2(s) → M2+(aq) + 2 X–(aq) Ksp = [M2+][X–]2

35. Solubility Equilibria Al(OH)3 1.8 x 10–33 BaCO3 8.1 x 10–9 BaF2 1.7 x 10–6 BaSO4 1.1 x 10–10 Bi2S3 1.6 x 10–72 CdS 8.0 x 10–28 CaCO3 8.7 x 10–9 CaF2 4.0 x 10–11 Ca(OH)2 8.0 x 10–6 Ca3(PO4)2 1.2 x 10–26 Cr(OH)3 3.0 x 10–29 CoS 4.0 x 10–21 CuBr 4.2 x 10–8 CuI 5.1 x 10–12 Cu(OH)2 2.2 x 10–20 CuS 6.0 x 10–37 Fe(OH)2 1.6 x 10–14 Fe(OH)3 1.1 x 10–36 FeS 6.0 x 10–19 PbCO3 3.3 x 10–14 PbCl2 2.4 x 10–4 PbCrO4 2.0 x 10–14 PbF2 4.1 x 10–8 PbI2 1.4 x 10–8 PbS 3.4 x 10–28 MgCO3 4.0 x 10–5 Mg(OH)2 1.2 x 10–11 MnS 3.0 x 10–14 Hg2Cl2 3.5 x 10–18 HgS 4.0 x 10–54 NiS 1.4 x 10–24 AgBr 7.7 x 10–13 Ag2CO3 8.1 x 10–12 AgCl 1.6 x 10–10 Ag2SO4 1.4 x 10–5 Ag2S 6.0 x 10–51 SrCO3 1.6 x 10–9 SrSO4 3.8 x 10–7 SnS 1.0 x 10–26 Zn(OH)2 1.8 x 10–14 ZnS 3.0 x 10–23

36. Solubility Equilibria Q: The solubility of calcium sulfate (CaSO4) is found experimentally to be 0.67 g/L. Calculate the value of Ksp for calcium sulfate. Q: The solubility of lead chromate (PbCrO4) is 4.5 x 10–5 g/L. Calculate the solubility product of this compound. Q: Calculate the solubility of copper(II) hydroxide, Cu(OH)2, in g/L.

37. Solubility Equilibria • Ion Product (Q): solubility equivalent of the reaction quotient. It is used to determine whether a precipitate will form. Q < Ksp UnsaturatedQ = Ksp SaturatedQ > Ksp Supersaturated; precipitate forms.

38. Factors that Affect Solubility • Common-Ion Effect • LeChatelier’s Principle revisited Addition of a product ion causes the solubility of the solid to decrease, but the Ksp remains constant. • pH • LeChatelier’s Principle again! Basic salts are more soluble in acidic solution. Acidic salts are more soluble in basic solution. Environmental example: CaCO3 – limestone Stalactites and stalagmites form due to changing pH in the water and thus solubility of the limestone.

39. Solubility Equilibria Q: Exactly 200 mL of 0.0040 M BaCl2 are added to exactly 600 mL of 0.0080 M K2SO4. Will a precipitate form? Q: If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will precipitation occur?

40. The Common-Ion Effect and Solubility • The solubility product (Ksp) is an equilibrium constant; precipitation will occur when the ion product exceeds the Ksp for a compound. • If AgNO3 is added to saturated AgCl, the increase in [Ag+] will cause AgCl to precipitate. Q = [Ag+]0 [Cl–]0 > Ksp

41. The Common-Ion Effect and Solubility

42. The Common-Ion Effect and Solubility

43. The Common-Ion Effect and Solubility Q: Calculate the solubility of silver chloride (in g/L) in a 6.5 x 10–3 M silver chloride solution. Q: Calculate the solubility of AgBr (in g/L) in:(a) pure water(b) 0.0010 M NaBr

44. Complex Ion Equilibria and Solubility • A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. • Most metal cations are transition metals because they have more than one oxidation state. • The formation constant (Kf) is the equilibrium constant for the complex ion formation.

45. Complex Ion Equilibria and Solubility

46. Complex Ion Equilibria and Solubility

47. Complex Ion Equilibria and Solubility ION Kf Ag(NH3)2+ 1.5 x 107 Ag(CN)2– 1.0 x 1021 Cu(CN)42– 1.0 x 1025 Cu(NH3)42+ 5.0 x 1013 Cd(CN)42– 7.1 x 1016 CdI42– 2.0 x 106 ION Kf HgCl42– 1.7 x 1016 HgI42– 3.0 x 1030 Hg(CN)42– 2.5 x 1041 Co(NH3)63+ 5.0 x 1031 Zn(NH3)42+ 2.9 x 109

48. Complex Ion Equilibria and Solubility Q: A 0.20 mole quantity of CuSO4 is added to a liter of 1.20 M NH3 solution. What is the concentration of Cu2+ ions at equilibrium? Q: If 2.50 g of CuSO4 are dissolved in 9.0 x 102 mL of 0.30 M NH3, what are the concentrations of Cu2+, Cu(NH3)42+, and NH3 at equilibrium?

49. Complex Ion Equilibria and Solubility Q: Calculate the molar solubility of AgCl in a 1.0 M NH3 solution. Q: Calculate the molar solubility of AgBr in a 1.0 M NH3 solution.

50. Complex Ion Equilibria and Solubility