chapter 4: reactions in aqueous solutions

1. Chapter 4: Reactions in Aqueous Solutions Chapter Outline 4.1 Solute Concentrations, Molarity 4.2 Precipitation Reactions 4.3 Acid-base Reactions 4.4 Oxidation-reduction Reactions

2. Solute Concentrations; Molarity The concentration of a solute in solution can be expressed in terms of its molarity: Molarity (M) = moles of solute/liters of solution What is the molarity of a solution containing 1.20moles of substance A in 2.50L of solution? [A] = 1.20moles/2.50l = 0.480 mol/l = 0.480m

3. Molarity (Cont’d) How many grams of K2CrO4 is needed to make 1 liter of a 0.100M solution. 0.100moles K2CrO4 x 194.2g K2CrO4/1 mole K2CrO4 = 19.4g K2CrO4 The molarity of a solution can be used to calculate: The number of moles of solute in a given volume. The volume of solution containing a given number of moles.

4. Example 1 The bottle labeled “concentrated HCl” in the lab contains 12.0mol HCl per liter of solution. That is, [HCl] = 12.0 M. How many moles of HCl are there in 25.0ml of this solution? What volume (V) of concentrated HCl must be taken to contain 1.00mol of HCl?

5. Example 1 (Cont’d) The required conversion factors are: 12.0mol HCl/1 L or 1 L/12.0mol HCl nHCl = 25.0mL x 1L/1000mL x 12.0mol HCl/1 L = 0.300mol HCl V = 1.00mol HCl x 1 L/12.0mol HCl = 0.0833L (83.3mL)

6. Ionization When an ionic solid dissolves in water, cations and anions separate: NaCl(s) ? Na+(aq) + Cl-(aq) In water, this molecule is completely ionized. Ionic solids are often referred to as strong electrolytes. Their ions are excellent conductors of electricity.

7. Ionization (Cont’d) Consider ionization of MgCl2: MgCl2 ? Mg2+(aq) + 2Cl-(aq) 1 mole of MgCl2 yield 1 mole of Mg2+ and 2 moles of Cl-. It follows then that: Molarity of Mg2+ = molarity of MgCl2 Molarity of Cl- = 2 x molarity of MgCl2

8. Ionic Solids Containing Polyatomic Ions (NH4)3PO4(s) ? 3NH4+(aq) + PO43-(aq) Molarity NH4+ = 3 x molarity of (NH4)3PO4 Molarity PO43- = molarity of (NH4)3PO4

9. Example 2 Give the concentrations in moles per liter of each ion in: (a) 0.080M K2SO4 (b) 0.40M FeCl3 (a) K2SO4(s) ? 2K+(aq) + SO42-(aq) The conversion factors are 2mol K+/1mol K2SO4 and 1mol SO42-/1mol K2SO4.

10. Example 2 (Cont’d) [K+] = 0.080mol K2SO4/1L x 2mol K+/1mol K2SO4 = 0.16M K+ [SO42-] = 0.080mol K2SO4/1L x 1mol SO42-/ 1mol K2SO4 = 0.080mol SO42-

11. Example 2 (Cont’d) (b) FeCl3(s) ? Fe3+(aq) + 3Cl-(aq) [Fe3+] = 0.40mol FeCl3/1L x 1mol Fe3+/ 1mol FeCl3 = 0.40mol Fe3+ [Cl-] = 0.40mol FeCl3/1L x 3mol Cl-/ 1mol FeCl3 = 1.2M Cl-

12. Precipitation Reactions A precipitation reaction occurs when water solutions of two different ionic compounds are mixed and an insoluble solid separates out of solution. The precipitate is itself ionic; the cation comes from one solution and the anion from another. To predict the occurrence of these reactions, we must know which ionic substances are insoluble in water.

13. Predicting Precipitates Figure 4.4: Precipitation Diagram. If a cation in solution 1 mixes with an anion in solution 2 to form an insoluble compound (colored squares), that compound will precipitate. Cation-anion combinations that lead to the formation of a soluble compound (white squares) will not give a precipitate.

14. Examples NiCl2 and NaOH A precipitate of Ni(OH)2, an insoluble compound, will form. NaCl, a soluble compound, will not precipitate.

15. Predicting Precipitation Reactions Consider the following pairs of solutions: (a) CuSO4 and NaNO3 (b) Na2CO3 and CaCl2 (a) Ions present: Cu2+, SO42-; Na+, NO3- Possible precipitates: Cu(NO3)2, Na2SO4 According to table, both of these are soluble, no precipitate forms.

16. Predicting (Cont’d) (b) Ions present: Na+, CO32-; Ca2+, Cl- Possible precipitates: NaCl, CaCO3 Sodium chloride is soluble, but calcium carbonate is not. When these two are mixed, calcium carbonate precipitates.

17. Net Ionic Equations The precipitation reaction that occurs when solutions of Na2CO3 and CaCl2 are mixed can be represented by a simple equation. Ca2+(aq) + CO32-(aq) ? CaCO3(s) Note the equation only includes the ions that participate in the reaction. Na and Cl are “spectator” ions which are present in solution before and after the precipitation of calcium carbonate.

18. Net Ionic Equations A net ionic equation is one in which only the ions involved in the reaction are shown. Like all equations, net ionic equations must show: (i) Atom balance: there must be the same number of atoms of each element on both sides of the equation. (ii) Charge balance: there must be the same total charge on both sides of the equation.

19. Examples of Net Ionic Equation Write a net ionic equation if possible for the following pairs of solutions: (a) NaOH and Cu(NO3)2 Ions present: Na+, OH-; Cu2+, NO3- NaNO3 is soluble, but Cu(OH)2 is not. Equation: Cu2+(aq) + 2OH-(aq) ? Cu(OH)2(s)

20. Examples (Cont’d) (b) BaCl2 and Ag2SO4 Ions present: Ba2+, Cl-; Ag+, SO42- Possible precipitates: AgCl, BaSO4 Both compounds are insoluble, so two reactions occur. Ba2+(aq) + SO42-(aq) ? BaSO4(s) Ag+(aq) + Cl-(aq) ? AgCl (s)

21. Stoichiometry We can apply the approach introduced in chapter 3 to calculate mole-mass relationships in solution reactions represented by net ionic equations. Limiting reagent problems involving net ionic reactions are solved in the same manner.

22. Examples Consider the net ionic equation derived from the reaction that occurs when solutions of NaOH and Cu(NO3)2 are mixed. Cu2+(aq) + 2OH-(aq) ? Cu(OH)2 (s) What volume of 0.106M Cu(NO3)2 solution is required to form 6.52g of solid Cu(OH)2?

23. Example (Cont’d) Strategy: The first 3 steps in the path are mass Cu(OH)2 ? moles Cu(OH)2 ? moles Cu2+ Because 1 mole of Cu(NO3)2 produces one mole of Cu2+, it follows that moles Cu(NO3)2 = moles Cu2+ V(CuNO3)2 = 6.52g Cu(OH)2 x 1mole Cu(OH)2/97.57g Cu(OH)2 x 1mole Cu2+/1mol Cu(OH)2 x 1mole Cu(NO3)2/1mole Cu2+ x 1.00L Cu(NO3)2/0.106moles Cu(NO3)2 = 0.630L Cu(NO3)2

24. Acid-Base Reactions An acid is a species that produces H+ ions in water solution A base is a species that produces OH- ions in water solution Strong Acids (HCl) completely ionize in water while weak acids do not. HCl (aq) ? H+(aq) + Cl-(aq)

25. Strong Acids and Bases There are six common strong acids: Hydrochloric Acid (HCl) Hydrobromic Acid (HBr) Hydroiodic Acid (HI) Nitric Acid (HNO3) Perchloric Acid (HClO4) Sulfuric Acid (H2SO4)

26. Writing Equations for Strong and Weak Acids Strong: HB(aq) ? H+(aq) + B-(aq) Weak: HB(aq) H+(aq) + B-(aq) Strong Bases: Lithium Hydroxide (LiOH) Sodium Hydroxide (NaOH) Potassium Hydroxide (KOH) Calcium Hydroxide (Ca(OH)2)

27. Strong Bases A strong base in water completely ionizes to OH- and a cation. NaOH(s) ? Na+(aq) + OH-(aq) A weak base produces OH- ions by reacting with water. NH3(aq) + H2O ? NH4+(aq) + OH-(aq)

28. Equations for Acid-Base Reactions I. Strong Acid-Strong Base: Consider a solution of HNO3 (strong acid) mixed with a solution of NaOH (strong base). HNO3(aq) ? H+(aq) + NO3-(aq) NaOH(aq) ? Na+(aq) + OH-(aq) H+(aq) + OH-(aq) ? H2O This reaction is referred to as neutralization.

29. Equations for Acid-Base Reactions (Cont’d) II. Weak Acid-Strong Base: Two-step reaction. (1) HB(aq) H+(aq) + B-(aq) (2) H+(aq) + OH-(aq) ? H2O Add the equations together: HB(aq) + OH-(aq) ? B-(aq) + H2O

30. III. Strong Acid-Weak Base: a solution of HCl added to a solution of NH3. (1) NH3(aq) + H2O NH4+(aq) + OH-(aq) (2) H+(aq) + OH-(aq) ? H2O Add the two equations together: H+(aq) + NH3 ? NH4+

31. Acid-Base Titrations Acid-base reactions in water are often used to determine the concentration of a dissolved species (H+, OH-) or its percentage in a solid mixture. Titration is measuring the volume of a standard solution (solution of known concentration) required to react with a measured amount of sample.

32. Example of Titration CH3COOH(aq) + OH-(aq) ? CH3COO- + H2O The objective of the titration is to determine the point at which the reaction is complete (acid solution is neutralized). This point is called the equivalence point.

33. Oxidation-Reduction reactions An oxidation-reaction (red-ox) reaction involves the transfer of electrons between two species in aqueous solution. In a redox reaction, one species loses (donates) electrons, while the other species gains (accepts) electrons.

34. Example of Red-Ox Reaction Zn(s) + 2H+(aq) ? Zn2+ + H2(g) A red-ox reaction can be split into two half-reactions: oxidation and reduction. Oxidation: Zn(s) ? Zn2+ + 2e- Reduction: 2H+(aq) + 2e- ? H2(g)

35. Example (Cont’d) Oxidation and reduction always occur together; you cannot have one without the other. There is no net change in the number of electrons in a red-ox reaction. Those electrons given off in the oxidation are taken on by another species in the reduction.

36. Terms in Red-Ox Reactions The ion or molecule that accepts electrons is called the oxidizing agent. The ion or molecule that donates electrons is called the reducing agent.

37. Oxidation Number The concept of oxidation number is used to simplify the electron bookkeeping in redox reactions. (1) For a monoatomic ion (Na+, S2-), the oxidation number is simply the charge of the ion. (2) In a molecule or polyatomic ion, the oxidation number of an element is a “pseudo-charge,” obtained in a rather arbitrary fashion.

38. Rules for Assigning Oxidation Numbers I. The oxidation number of an element in an elementary substance is 0. For example, the oxidation number of chlorine in Cl2 or of phosphorous in P4 is 0. II.The oxidation number of an element in a monoatomic ion is equal to the charge of that ion. In the ionic compound, NaCl, sodium has an oxidation number of +1 and chlorine an oxidation number of –1.

39. Rules (Cont’d) III. Certain elements have the same oxidation number in all or almost all of their compounds. Group 1 metals are always +1 ions when they form compounds and Group 2 elements are always +2 ions when they form compounds. Oxygen is usually –2 (except peroxide) while hydrogen is ordinarily +1 (can be –1).

40. Rules (Cont’d) IV. The sum of the oxidation numbers in a neutral species is 0; in a polyatomic ion, it is equal to the charge of that ion. What is the oxidation number of S in Na2SO4? First look for elements whos oxidation number is always or almost always the same. In Na2SO4, Na is +1 and O is –2. Sodium sulfate must be neutral, so the sum of the oxidation numbers must be 0. 0 = 2(+1) + x + 4(-2); x = +6

41. Working Definitions of Oxidation and Reduction Oxidation is defined as an increase in the oxidation number. Reduction is defined as a decrease in the oxidation number. Zn(s) + 2H+(aq) ? Zn2+ + H2(g) Zn is oxidized (oxid. #: 0 ? +2) H+ is reduced (oxid. #: +1 ? 0)

42. Example 2Al(s) + 3Cu2+(aq) ? 2Al3+ + 3Cu(s) Al is oxidized (oxid. #: 0 ? +3) Cu2+ is reduced (oxid. #: +2 ? 0)

43. Balancing Half-Reactions Before you can balance an overall redox reaction, you must balance the half reactions. Fe2+(aq) ? Fe3+(aq) (oxid. # Fe: +2 ? +3) Fe2+(aq) ? Fe3+ + e-

44. Rules for Balancing Half-Reactions 1. Balance the atoms of the element being oxidized or reduced. 2. Balance the oxidation number by adding electrons. Add them to the left for a reduction; add them to the right for an oxidation. 3. Balance charge by adding H+ ions in acidic conditions and OH- in basic conditions. 4. Balance hydrogen by adding H2O molecules. 5. Check to make sure that oxygen is balanced.

45. Example Balance the following half-equations: (a) MnO4-(aq) ? Mn2+ (acidic solution) (b) Cr(OH)3(s) ? CrO42- (basic solution) MnO4-(aq) ? Mn2+ i) Equation is atom balanced ii) Because Mn is reduced from +7 to +2, 5 electrons must be added to the left hand side. MnO42-(aq) + 5e- ? Mn2+(aq)

46. Example (Cont’d) iii) There is a charge of –6 on the left and +2 on the right. To balance, add 8 H+ to the left to give a charge of +2 on both sides. MnO42-(aq) + 8H+(aq) + 5e- ? Mn2+(aq) iv) To balance the 8H+ ions on the left, add 4 H2O molecules to the right. MnO42-(aq) + 8H+(aq) + 5e- ? Mn2+(aq) + 4H2O

47. Example (Cont’d) (b) Cr(OH)3(s) ? CrO42- (basic solution) i) There is one Cr atom on each side: Cr(OH)3(s) ? CrO42- (basic solution) ii) Because oxidation # of Cr increases from +3 to +6, add the electrons to the right. Cr(OH)3(s) ? CrO42-(aq) + 3e-

48. Example (Cont’d) Cr(OH)3(s) ? CrO42-(aq) + 3e- iii) There is a charge of 0 on the left and a charge of –5 on the right. To balance charge, add 5 OH- ions to the left. Cr(OH)3(s) + 5OH-(aq) ? CrO42-(aq) + 3e- iv) There are 8 hydrogens on the left and none on the right. Add four waters to balance the hydrogens. Cr(OH)3(s) + 5OH-(aq) ? CrO42-(aq) + 4H2O + 3e-

49. Balancing Redox Equations 4 Step Procedure: 1. Split the equation into two half-reactions, one for reduction, the other for oxidation. 2. Balance one of the half-equations with respect to both atoms and charge as described above. 3. Balance the other half-reaction. 4. Combine the two half-reactions in such a way as to eliminate electrons.

50. Example Balance the following redox equation: Fe2+(aq) + MnO42-(aq) ? Fe3+(aq) + Mn2+(aq) in acidic conditions (a) (1) oxidation: Fe2+(aq) ? Fe3+(aq) reduction: MnO42-(aq) ? Mn2+(aq) (2), (3) The balanced half-reactions, as obtained previously, are: Fe2+(aq) ? Fe3+(aq) + 1e- MnO42-(aq) + 8H+(aq) + 5e- ? Mn2+(aq) + 4H2O

51. Example (Cont’d) (4) To eliminate electrons, multiply the oxidation half-reaction by 5 and add to the reduction half-reaction: 5[Fe2+(aq) ? Fe3+ + 1e-] MnO42-(aq) + 8H+(aq) + 5e- ? Mn2+(aq) + 4H2O 5Fe2+ + MnO42-(aq) + 8H+(aq) ? 5Fe3+(aq) + Mn2+ + 4H2O