1. Homework Problems Chapter 4 Homework Problems: 2, 8, 11, 14, 20, 22, 25, 28, 29, 30, 32, 42, 44, 48, 52, 54, 64, 68, 82, 84, 94, 124, 132, 150

2. CHAPTER 4 Reactions in Aqueous Solution

3. Solutions A solution is a homogeneous mixture of two or more pure chemical compounds. Solutions may be solids, liquids, or gases. Solvent - The major component of a solution (if there is one) Solute(s) - The minor component(s) of a solution Example: A solution is prepared by adding a small amount of sodium chloride (NaCl) to water. solvent = water (major component) solute = sodium chloride Aqueous solution - Solution where water is the solvent.

4. Electrolytes and Nonelectrolytes Many chemical substances dissolve in water. A substance that forms ions when dissolved in water is called an electrolyte, while a substance that does not form ions when dissolved in water is called a nonelectrolyte. Electrolytes can be divided into two categories: Strong electrolyte - Completely dissociates (falls apart) when added to solution, to form ions. Weak electrolyte - Partially dissociates (falls apart) when added to solution, to form ions.

5. Classification of Compounds as Electrolytes (in Water) We previously divided compounds into two general categories: Ionic compounds - Collections of cations (positive ions) and anions (negative ions) in a crystal structure. Usually form from a metal (or cation group) and a nonmetal (or anion group). Molecular compounds - Exist as individual molecules. Usually form from two or more nonmetals. We may use this as a starting point in classifying compounds as strong electrolytes, weak electrolytes, and nonelectrolytes.

6. Ionic Compounds Ionic compounds - All ionic compounds are strong electrolytes. Examples: NaCl(s)  Na + (aq) + Cl - (aq) KOH(s)  K + (aq) + OH - (aq) Ca(NO 3 ) 2 (s)  Ca 2+ (aq) + 2 NO 3 - (aq) Note that an insoluble ionic compound is considered a strong electrolyte. This is because even though very little of the compound dissolves in water, whatever does dissolve completely ionizes.

7. Molecular Compounds Strong acids - Molecular compounds that are strong acids are also strong electrolytes. Examples: HCl(aq)  H + (aq) + Cl - (aq) HNO 3 (aq)  H + (aq) + NO 3 -(aq) Weak acids and bases - Molecular compounds that are weak acids or weak bases are also weak electrolytes. Note that an acid forms H + ion when added to water, and a base forms OH - ion. Examples: HF(aq)  H + (aq) + F - (aq) NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq) The double arrow “  ” indicates a reaction that proceeds in both the for- ward and reverse direction, establishing equilibrium.

8. Molecular Compounds (Continued) Other molecular compounds - Molecular compounds that are neither acids nor bases are nonelectrolytes. Examples: C 6 H 12 O 6 (s)  C 6 H 12 O 6 (aq) CH 3 OH(l)  CH 3 OH(aq) Note that water (H 2 O) is classified as a nonelectrolyte, even though a small number of H + and OH - ions are formed in pure water. H 2 O(l)  H + (aq) + OH - (aq) For pure water at T = 25 ° there are only ~ 2 x 10 -7 M of ions present. This is small enough that water is classified as a nonelectrolyte.

9. Flow Chart For Classifying Compounds

10. Precipitation Reactons A precipitation reactions is a reaction that forms a solid (a precipitate) out of soluble reactants. Examples: Pb(NO 3 ) 2 (aq) + 2 KCl(aq)  PbCl 2 (s) + 2 KNO 3 (aq) 3 NaOH(aq) + FeBr 3 (aq)  Fe(OH) 3 (s) + 3 NaBr(aq)

11. Solubility (Water) Solid compounds are divided into two general categories. Soluble compound - a compound that will dissolve in water Insoluble compound - a compound that will not dissolve in water The above classification scheme is an oversimplification. Most “insoluble” compounds will dissolve to a slight extent. For “soluble” compounds there will be a limit as to the amount of compound that will dissolve in a given amount of water. We can also discuss solubility in other solvents.

12. Why Ionic Compounds Dissolve in Water The forces holding an ionic compound together are the strong electrical attraction that exists between cations and anions. It is therefore somewhat surprising that ionic compounds will dissolve in water. The reason some ionic compounds will dissolve in water is because the water molecules have a partial negative charge on the oxygen atom (  -) and partial positive charges on the hydrogen atoms (  +), where “  ” indicates a small positive or negative charge. The reason these partial charges exist will be discussed later in the semester. Because cations and anions can strongly interact with the negative and positive partial charges on several water molecules at the same time, it is possible for the ions in an ionic compound to be pulled out of the solid and go into solution.

13. Hydration and Solvation The process of surrounding a cation or anion with water molecules to stabilize them in aqueous solution is called hydration. Often these ions will be surrounded by a definite number of water molecules, called a hydration shell. The more general process of surrounding ions or molecules with solvent molecules in a solution is called solvation. Water is unusual because the interaction of ions with water molecules is much stronger than the interactions that occur in most other solvents. hydration of anion hydration of cation

14. Solubility Rules a) All compounds with group IA cations, or NH 4 + (ammonium ion), are soluble. b) All compounds containing NO 3 - (nitrate ion), ClO 4 - (perchlorate ion), ClO 3 - (chlorate ion), or C 2 H 3 O 2 - (acetate ion) are soluble. c) Most compounds containing Cl -, Br -, and I - are soluble. (EXCEPTIONS: Cations Ag +, Pb 2+, Hg 2 2+ ). d) Most compounds containing SO 4 2- (sulfate ion) are soluble. (EXCEPTIONS: Cations Ag +, Pb 2+, Hg 2 2+, Ca 2+, Sr 2+, Ba 2+ ). e) Most compounds containing OH - (hydroxide ion) are insoluble. (EXCEPTIONS: Cations from group 1A, Ca 2+, Sr 2+, Ba 2+ ). f) Most other compounds are insoluble. (EXCEPTIONS: Compounds covered by rules a-e).

15. Use of Solubility Rules KClO 4 (NH 4 ) 2 SO 4 CuCl 2 PbI 2 NiSO 4 Mg(OH) 2 Zn(IO 3 ) 2

16. Use of Solubility Rules KClO 4 solubleRules a and b (NH 4 ) 2 SO 4 solubleRule a CuCl 2 solubleRule c PbI 2 insolubleRule c NiSO 4 solubleRule d Mg(OH) 2 insolubleRule e Zn(IO 3 ) 2 insolubleRule f

17. Determining Whether a Precipitation Reaction Will Occur When solutions of two soluble substances mix we will not always form a precipitate. We may use the solubility rules to decide if a precipitate will form. 1) Write the ions formed from the starting compounds. 2) Look at all possible combinations of cations and anions. a) If there is a combination of ions that forms an insoluble compound, a precipitate will form and a precipitation reaction occurs. b) If there is not a combination of ions that forms an insoluble compound, then there will be no precipitation reaction.

18. Examples: For each of the following mixtures of two solutions will a precipitate form? Note that all of the solutes are themselves soluble. Solution Asolution Bprecipitate(?) 1)NaCl(aq)AgNO 3 (aq) 2)NaCl(aq)CuBr 2 (aq)

19. Solution Asolution Bprecipitate(?) 1)NaCl(aq)AgNO 3 (aq)yes, AgCl(s) Na + Cl - Ag + NO 3 - NaCl(aq) + AgNO 3 (aq)  NaNO 3 (aq) + AgCl(s) 2)NaCl(aq)CuBr 2 (aq)no Na + Cl - Cu 2+ Br - No reaction.

20. Net Ionic Equations To this point we have written chemical equations in terms of neutral species (molecules, or formula units). However, in aqueous solution it is often more appropriate to write equations in terms of ions. There are three ways we normally write chemical equations: 1) Molecular equation. All reactants and products are written in terms of their chemical formula. 2) Total ionic equation. Strong electrolytes (strong acids, strong soluble bases, and soluble ionic compounds) are written as ions. 3) Net ionic equation. The total ionic equation after removing spectator ions. A spectator ion is an ion that appears on both the reactant and product side of the equation, and which does not participate directly in the chemical reaction.

21. Finding the Total and New Ionic Equations We find the total and net ionic equations as follows. 1) Begin with a balanced molecular equation. 2) Break apart strong electrolytes into ions to obtain the total ionic equation. 3) Cancel ions that appear on both sides of the equation to obtain the net ionic equation. Example: molecularPb(NO 3 ) 2 (aq) + 2 KI(aq)  PbI 2 (s) + 2 KNO 3 (aq)

22. total ionicPb 2+ (aq) + 2 NO 3 - (aq) + 2 K + (aq) + 2 I - (aq)  PbI 2 (s) + 2 K + (aq) + 2 NO 3 - (aq) net ionicPb 2+ (aq) + 2 I - (aq)  PbI 2 (s)

23. By writing the net ionic equation we are able to focus on the important parts of a chemical reaction and observe similarities in reactions. mol.AgNO 3 (aq) + NaCl(aq)  AgCl(s) + NaNO 3 (aq) totalAg + (aq) + NO 3 - (aq) + Na + (aq) + Cl - (aq)  AgCl(s) + Na + (aq) + NO 3 - (aq) netAg + (aq) + Cl - (aq)  AgCl(s) mol.AgNO 3 (aq) + KCl(aq)  AgCl(s) + KNO 3 (aq) totalAg + (aq) + NO 3 - (aq) + K + (aq) + Cl - (aq)  AgCl(s) + K + (aq) + NO 3 - (aq) netAg + (aq) + Cl - (aq)  AgCl(s)

24. Net Ionic Equation When No Precipitate Forms Writing the net ionic equation makes it clear when precipitation reactions do or do not occur. mol.2 KBr(aq) + Zn(NO 3 ) 2 (aq)  2 KNO 3 (aq) + ZnBr 2 (aq) K + Br - Zn 2+ NO 3 - total2 K + (aq) + 2 Br - (aq) + Zn 2+ (aq) + 2 NO 3 - (aq)  2 K + (aq) + 2 NO 3 - (aq) + Zn 2+ (aq) + 2 Br - (aq) All the ions are spectator ions, and so no reaction occurs.

25. Acids and Bases (Arrhenius) Arrhenius definition acid - A substance which, when added to water, forms H + ion base - A substance which, when added to water, forms OH - ion Examples: HCl(aq)  H + (aq) + Cl - (aq) HF(aq)  H + (aq) + F - (aq) NaOH(s)  Na + (aq) + OH - (aq) NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq) The advantage of the Arrhenius definition is that it is simple and easy to implement. The disadvantage is that it is tied in to a particular solvent (water) and is not a general definition.

26. Strong and Weak Acids An acid is a substance that forms H + ion when added to water (Arrhenius definition). A strong acid forms one H + ion for each acid molecule. A weak acid forms much less than one H + ion for each acid molecule. Examples:HClO 4 (aq)  H + (aq) + ClO 4 - (aq)strong HNO 2 (aq)  H + (aq) + NO 2 - (aq)weak There are 7 common strong acids binary ternary HCl HIHClO 3 HNO 3 HBrHClO 4 H 2 SO 4 (1st proton)* All other acids can be assumed to be weak acids. * H 2 SO 4 (aq)  H + (aq) + HSO 4 - (aq)

27. Strong and Weak Bases A base is a substance that forms OH - ion when added to water (Arrhenius definition). A strong soluble base is a soluble hydroxide compound that completely dissociates when added to water. An insoluble base is an insoluble hydroxide compound. There are also a few substances that act as weak bases in solution. These substances form much less than one OH - ion per base molecule. Examples:KOH(s)  K + (aq) + OH - (aq)strong soluble base Fe(OH) 3 (s)  no reactioninsoluble base NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq)weak base There are 8 common strong soluble bases Group 1A - LiOH, NaOH, KOH, RbOH, CsOH Group 2A - Ca(OH) 2, Sr(OH) 2, Ba(OH) 2 All other hydroxide compounds are insoluble bases.

28. Acids and Bases (Bronsted) Bronsted definition acid - a proton (H + ) donor base - a proton (H + ) acceptor HF(aq) + H 2 O(l)  H 3 O + (aq) + F - (aq) acid base In the Bronsted theory, in an acid-base reaction an acid donates a proton, while a base accepts a proton. In addition, in Bronsted theory acids form hydronium ion (H 3 O + ion) instead of hydrogen ion (H + ion) when added to water. The Bronsted definition of an acid and a base is more general than the Arrhenius definition.

29. Examples of Acids (Bronsted) Strong acid HCl(aq) + H 2 O(l)  H 3 O + (aq) + Cl - (aq) acid:HCl base:H 2 O Weak acid HC 2 H 3 O 2 (aq) + H 2 O(l)  H 3 O + (aq) + C 2 H 3 O 2 - (aq) acid:HC 2 H 3 O 2 base:H 2 O The second reaction goes in both directions.

30. Examples of Bases (Bronsted) Strong base NaOH(s)  Na + (aq) + OH - (aq) Considered an ionization reaction in Bronsted theory, not an acid- base reaction. Weak base NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq) base:NH 3 acid:H 2 O The second reaction goes in both directions.

31. Example Chlorous acid (HClO 2 ) is a weak acid. Pyridine (C 5 H 5 N) is a weak base. Indicate the behavior of these two substances when added to water, according to Bronsted theory.

32. Example Chlorous acid (HClO 2 ) is a weak acid. Pyridine (C 5 H 5 N) is a weak base. Indicate the behavior of these two substances when added to water, according to Bronsted theory. HClO 2 (aq) + H 2 O(l)  H 3 O + (aq) + ClO 2 - (aq) C 5 H 5 N(aq) + H 2 O(l)  C 5 H 5 NH + (aq) + OH - (aq) Notice that, depending on the reaction, water can act as either a Bronsted base or a Bronsted acid. A substance that can act as either a Bronsted acid or a Bronsted base is called amphoteric.

33. Polyprotic Acid A polyprotic acid has two or more ionizable protons that can be donated in an acid-base reaction. Monoprotic (one ionizable proton)HCl, HNO 2, CH 3 COOH Diprotic (two ionizable protons)H 2 CO 3, H 2 SO 4 Triprotic (three ionizable protons)H 3 PO 3 Example: H 2 CO 3 (carbonic acid) 1 st protonH 2 CO 3 (aq) + H 2 O(l)  H 3 O + (aq) + HCO 3 - (aq) 2nd protonHCO 3 - (aq) + H 2 O(l)  H 3 O + (aq) + CO 3 2- (aq)

34. Dibasic Base A dibasic base is a strong base that releases two OH - ions per formula unit of base. Example: Ba(OH) 2 (barium hydroxide) Ba(OH) 2 (s)  Ba 2+ (aq) +2 OH - (aq)

35. Acid-Base (Neutralization) Reactions An acid-base reactions (or neutralization reaction) is the reaction of an acid with a base. These reactions usually form a salt and water. salt - An ionic compound that can be formed by the reaction of an acid with a base. Examples: HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) 2 HClO 4 (aq) + Ca(OH) 2 (aq)  Ca(ClO 4 ) 2 (aq) + 2 H 2 O(l) Note that ionic compounds other than hydroxides (OH - ) or oxides (O 2- ) are classified as salts, as they can be formed from acid-base reactions.

36. Strong Acid + Strong Base All strong acid + strong base reactions have the same net ionic equation. mol.2 HClO 4 (aq) + Sr(OH) 2 (aq)  Sr(ClO 4 ) 2 (aq) + 2 H 2 O( ) total2 H + (aq) + 2 ClO 4 - (aq) + Sr 2+ (aq) + 2 OH - (aq)  Sr 2+ (aq) + 2 ClO 4 - (aq) + 2 H 2 O( ) netH + (aq) + OH - (aq)  H 2 O( ) mol.HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O( ) totalH + (aq) + Cl - (aq) + Na + (aq) + OH - (aq)  Na + (aq) + Cl - (aq) + H 2 O( ) netH + (aq) + OH - (aq)  H 2 O( )

37. Net Ionic Equation for Weak Acid or Base Reaction When the acid or base in an acid base reaction is weak, the net ionic equation will be slightly more complicated. mol.HNO 2 (aq) + KOH(aq)  KNO 2 (aq) + H 2 O( ) mol.HBr(aq) + NH 3 (aq)  NH 4 Br(aq)

38. mol.HNO 2 (aq) + KOH(aq)  KNO 2 (aq) + H 2 O( ) totalHNO 2 (aq) + K + (aq) + OH - (aq)  K + (aq) + NO 2 - (aq) + H 2 O( ) netHNO 2 (aq) + OH - (aq)  NO 2 - (aq) + H 2 O( ) mol.HBr(aq) + NH 3 (aq)  NH 4 Br(aq) totalH + (aq) + Br - (aq) + NH 3 (aq)  NH 4 + (aq) + Br - (aq) netH + (aq) + NH 3 (aq)  NH 4 + (aq) Notice that we do not break apart the weak acid or weak base, as they are weak electrolytes, and therefore exist mainly as molecules rather than ions.

39. Oxidation Number An oxidation number is a number assigned to an atom in a molecule or ion indicating whether that atom is electron rich (negative oxidation number), electron poor (positive oxidation number), or neutral (oxidation number of zero). Oxidation numbers do not indicate charges of atoms, but changes in oxidation number indicate whether atoms have gained or lost electron density. A reaction where some of the oxidation numbers change in going from reactants to products is called an oxidation-reduction (redox) reaction.

40. Rules For Assigning Oxidation Numbers 1) The sum of the oxidation numbers of a species is equal to the charge of the species. 2) Atoms in elemental forms have an oxidation number of 0. 3) Hydrogen has an oxidation number of +1 when bonded with nonmetals (molecular compounds) and -1 when combined with metals (ionic compounds). 4) Oxygen usually has an oxidation number of -2. Exceptions: Elemental forms, peroxides (H 2 O 2 ), OF 2. H 2 O 2 H +1, O -1OF 2 F -1, O +2 5) Halogens. F in compounds is always -1. Other halogens are often -1, but in compounds with oxygen or other halogens the oxidation number may take on different values. 6) For assigning oxidation numbers in ionic compounds, it is useful to break the compounds up into ions.

41. Examples: Assign oxidation numbers for each atom in the following substances. H 2 O Fe(NO 3 ) 2 SO 4 2- O 3 HClO 2 MnO 4 -

42. H 2 OH +1; O -2 Fe(NO 3 ) 2 = Fe 2+ and NO 3 - so Fe +2, O -2, and so N +5 SO 4 2- O -2, and so S +6 O 3 elemental form, so O = 0 HClO2 = H + and ClO 2 - (even though molecular) so H +1, O -2, and so Cl +3 MnO 4 - O -2, and so Mn +7

43. Common Oxidation Numbers

44. Oxidation-Reduction Reactions In an oxidation-reduction (redox) reaction the oxidation numbers of some of the atoms involved in the reaction change in going from reactants to products. 2 Fe 2 O 3 (s) + 3 C(s)  4 Fe(s) + 3 CO 2 (g) +3 -2 0 0 +4 -2 We use the following terms in reference to redox reactions. Oxidation - An increase in the value for the oxidation number. Reduction - A decrease in the value for the oxidation number. There will always be one oxidation and one reduction process per reaction. C0 to +4, and so is oxidized. Fe+3 to 0, and so is reduced.

45. Oxidizing and Reducing Agents We call the species that is oxidized in a redox reaction the reducing agent, and the species that is reduced in a redox reaction an oxidizing agent. This seems confusing, but is based on the idea that there must be both an oxidation and a reduction taking place in a redox reaction. C(s) + O 2 (g)  CO 2 (g) 0 0 +4 -2 In this reaction carbon is oxidized (from 0 to +4) and so C is a reducing agent, and oxygen is reduced (from 0 to -2) and so O 2 is an oxidizing agent. SO 2 (g) + 2 Fe 3+ (aq) + 2 H 2 O(l)  2 Fe 2+ (aq) + SO 4 2- (aq) + 4 H + (aq) +4 -2 +3 +1 -2 +2 +6 -2 +1 In this reaction S is oxidized (from +4 to +6) and so SO 2 is a reducing agent, and Fe is reduced (from +3 to +2) and so Fe 3+ is an oxidizing agent.

46. Activity Series The higher the element in the activity series, the easier it is to oxidize. The lower the anion in the activity series, the easier it is to reduce.

47. Example A solution contains cobalt II ions (Co 2+ ). Which of the following metals would be expected to react if placed in the solution? Cu, Fe, Ni, Zn

48. A solution contains cobalt II ions (Co 2+ ). Which of the following metals would be expected to react if placed in the solution? Cu, Fe, Ni, Zn Metals above cobalt in the activity series have a stronger tendency to undergo oxidation and form ions. Those below have a weaker tendency to form ions than cobalt. So - Will react - Fe, Zn Co 2+ (aq) + Fe(s)  Co(s) + Fe 2+ (aq) Co 2+ (aq) + Zn(s)  Co(s) + Zn 2+ (aq) Will not react - Cu, Ni

49. Hydrogen Ion and Active Metals A solution containing hydrogen ion (H + ) will, according to the activity series, oxidize any metal above it (so Pb, Sn, Ni, …) Sn(s) + 2 H + (aq)  Sn 2+ (aq) + H 2 (g) Some metals in the activity series (those at or near the top of the series) are so reactive that they are never found in the metallic state in nature. Such metals are called active metals. The alkali metals and alkaline earth metals are active metals. Metals at or near the bottom of the activity series are not very reactive, and so are often found in the metallic state in nature. Such metals are called noble metals. Examples are gold (Au), platinum (Pt), and mercury (Hg).

50. Balancing Simple Oxidation-Reduction Reactions The following three step procedure can be used to balance simple oxidation-reduction reactions. A variation of this method will be used later to balance more complicated oxidation-reduction reactions. 1) Find the oxidation numbers for all reactants and products 2) Write a half-reaction for oxidation, and a half-reaction for reduction. Use electrons for charge balance. a) Oxidation - will produce electrons b) Reduction - will consume electrons 3) Combine the two half reactions to get the net ionic equation. This may mean multiplying one or both of the half reactions to get the electrons to cancel.

51. Example Balance the following oxidation-reduction reactions. a) Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) b) Cr(s) + Pb 2+ (aq)  Cr 3+ (aq) + Pb(s)

52. a) Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) 0 +2 +2 0 oxidationZn(s)  Zn 2+ (aq) + 2 e - reduction Cu 2+ (aq) + 2 e -  Cu(s) NETZn(s) + Cu 2+ (aq)  Zn 2+ (aq) +Cu(s) b) Cr(s) + Pb 2+ (aq)  Cr 3+ (aq) + Pb(s) 0 +2 +3 0 oxidation(Cr(s)  Cr 3+ (aq) + 3 e - ) x 2 reduction (Pb 2+ (aq) + 2 e -  Pb(s) ) x 3 NET2 Cr(s) + 3 Pb 2+ (aq)  2 Cr 3+ (aq) +3 Pb(s)

53. Molarity The most common unit of concentration for laboratory work is molarity. By definition, the molarity of a solution (M) is M = moles solute liters of solution Unfortunately, we use the same symbol, M, for molarity and for molecular mass. The meaning of M must therefore be determined from the context. To avoid confusion, we often use square brackets “[ ]” to indicate molarity (for example, [KNO 3 ] for the concentration of a potassium nitrate solution, in moles/L. Example: Find the molarity of NaCl in a solution prepared by dissolving 5.000 g NaCl (M = 58.44 g/mol) in water to form a solution with final volume V = 100.0 mL.

54. moles NaCl = 5.000 g 1 mol = 0.08556 mol 58.44 g liters of solution = 100.0 mL 1 L = 0.1000 L 1000.0 mL So [NaCl] = 0.08556 mol NaCl = 0.8556 mol/L = 0.8556 M NaCl 0.1000 L soln

55. Solution Preparation Solutions of known concentration can easily be prepared using an analytical balance and standard laboratory glassware.

56. Stock Solutions and Dilutions A stock solution is a solution of high concentration used to prepare more dilute solution by combining known amounts of the stock solution with pure solvent. If the concentration of the stock solution is known and the method used to carry out the preparation of the dilute solution is also known, the concentration of the dilute solution may be found. Example: A solution is prepared by mixing 20.00 mL of the 0.8556 M NaCl solution discussed above with water to form a dilute solution with a final volume equal to 500.0 mL. What is the concen- tration of this dilute solution?

57. A solution is prepared by mixing 20.00 mL of the 0.8556 M NaCl solution discussed above with water to form a dilute solution with a final volume equal to 500.0 mL. What is the concentration of this dilute solution? moles NaCl = 0.02000 L 0.8556 mol = 0.01711 mol NaCl L stock soln M dilute soln = 0.01711 mol NaCl = 0.03422 mol/L 0.5000L

58. General Relationship For Dilution of Stock Solutions Based on the method used in the above calculation a general relationship useful in preparing dilutions of stock solutions may be found. Let M c = molarity of initial (concentrated) solution L c = volume, in liters, of initial (concentrated) solution M d = molarity of final (dilute) solution L d = volume, in liters, of final (dilute) solution ThenM c L c = M d L d This relationship is based on conservation of mass, in this case, the mass (or number of moles) of solute in the solutions. It will work for any volume unit, as long as the same unit is used for L c and L d.

59. We can check this relationship using our previous example. A solution is prepared by mixing 20.00 mL of the 0.8556 M NaCl solution discussed above with water to form a dilute solution with a final volume equal to 500.0 mL. What is the concentration of this dilute solution? We previously found M = 0.03422 mol/L. Since M c L c = M d L d, then if we divide both sides by L d, we get M d = M c L c = (0.8556 M) (20.00 mL) = 0.03422 mol/L L d (500.0 mL) the same as before.

60. One of the main uses of this relationship is to use stock solutions of known concentration and volumetric glassware (pipettes, flasks) to prepare more dilute solutions with particular concentrations. Example: A student has a 0.4846 M aqueous stock solution of glucose, and wishes to prepare 100.0 mL of a 0.1000 M glucose solution. How can she do this?

61. A student has a 0.4846 M aqueous stock solution of glucose, and wishes to prepare 100.0 mL of a 0.1000 M glucose solution. How can she do this? We could use the method used earlier, but is is faster and more simple to use the relationship M c L c = M d L d M c = 0.4846 MM d = 0.1000 M L c = ?L d = 100.0 mL So(0.4846 M) L c = (0.1000 M) (100.0 mL) L c = (0.1000 M) (100.0 mL) = 20.6 mL (0.4846 M) So add 20.6 mL of the stock solution to a 100.0 mL volumetric flask, and then add water to obtain the final volume of solution.

62. Serial Dilution It is difficult to prepare very dilute solutions (concentration < 10 -3 M) directly, as it would involve precisely weighing out a small mass of solute. Such dilute solutions are prepared by the process of successive, or serial, dilution of a stock solution. Example: Starting with a 0.1000 M stock solution of KNO 3, prepare a dilute solution with a concentration of 1.00 x 10 -5 M.

63. Stock solution0.1000 M Take 5.00 mL of stock solution and dilute to a final volume of 500.0 mL. M d = M c L c = (0.1000 M) (5.00 mL) = 0.001000 M L d (500.0 mL) Dilute this new solution in the same way as in the previous step M d = M c L c = (0.00100 M) (5.00 mL) = 1.00 x 10 -5 M L d (500.0 mL)

64. Solution Stoichiometry If we know the concentration of a solution of a strong electrolyte, we can use the stoichiometry of the dissolution reaction to find the concentration of individual ions in solution. Example. A student has a 0.0200 M solution of copper II nitrate (Cu(NO 3 ) 2 ). What are the concentrations of Cu 2+ ion and NO 3- ion in the solution? Dissolution equation is Cu(NO 3 ) 2 (s)  Cu 2+ (aq) + 2 NO 3 - (aq) [Cu 2+ ] = 0.0200 M Cu(NO 3 ) 2 1 mol Cu 2+ = 0.0200 M Cu 2+ ion 1 mol Cu(NO 3 ) 2 [NO 3 - ] = 0.0200 M Cu(NO 3 ) 2 2 mol NO 3 - = 0.0400 M NO 3 - ion 1 mol Cu(NO 3 ) 2

65. Gravimetric Analysis In gravimetric analysis, we use mass measurements to determine composition or concentrations for unknowns. One common gravimetric method is to add a compound to a solution to form a precipitate, which can then be filtered out to determine the mass of solid formed. Example: An aqueous solution contains an unknown concentra- tion of silver nitrate (AgNO 3 ). Excess solid sodium chloride (NaCl) is added to a 200.00 mL sample of the silver nitrate solution, and 0.184 g of precipitate forms. What is the concentration of silver nitrate in the original solution?

66. AgNO 3 (aq)  Ag + (aq) + NO 3 - (aq) AgNO 3 (aq) + NaCl(aq)  AgCl(s) + NaNO 3 (aq) M(AgCl) = 143.32 g/mol mol AgNO 3 = 0.184 g AgCl 1 mol AgCl 1 mol AgNO 3 143.32 g 1 mol AgCl = 0.00128 mol AgNO 3 [AgNO 3 ] = 0.00128 = 0.00642 M 0.2000L

67. Titration The term titration refers to a reaction between two substances where one reactant is slowly added to a fixed amount of a second reactant until complete reaction occurs. Titrations are often used to find the concentrations of solutions. In an acid-base titration, for example, base is slowly added to a fixed amount of acid until complete reaction occurs (the titration can also be carried out by the slow addition of acid to a fixed amount of base). An indicator (a molecule that has a different color in acid and base solution) is used to visually estimate the equivalence point of the titration (the point where the reaction is complete). This should occur as close as possible to the end point of the titration, the point where the indicator color change occurs.

68. Calculations Involving Titrations Since titrations are chemical reactions, we may carry out calculations associated with titrations in the same way as other calculations involving chemical reactions. We require a balanced chemical equation and information about the amounts of the acid and base solutions used in the titration and the concentration of one of the solutions (or the number of moles of one of the reactants). Example: A 20.00 mL sample of an HCl solution of unknown concentration is titrated with a 0.3145 M NaOH solution. The equivalence point for the titration occurs after the addition of 28.33 mL of the NaOH solution. What is the concentration of the HCl solution (in units of mol/L)? HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l)

69. A 20.00 mL sample of an HCl solution of unknown concentration is titrated with a 0.3145 M NaOH solution. The equivalence point for the titration occurs after the addition of 28.33 mL of the NaOH solution. What is the concentration of the HCl solution (in units of mol/L)? HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) Strategy: 1) Find the number of moles of NaOH (the solution where you know both the volume and concentration). 2) Use the balanced chemical equation to find the number of moles of HCl. 3) Find the concentration of the HCl solution, using the number of moles (found in step 2) and the volume of HCl solution (given in the problem).

70. A 20.00 mL sample of an HCl solution of unknown concentration is titrated with a 0.3145 M NaOH solution. The equivalence point for the titration occurs after the addition of 28.33 mL of the NaOH solution. What is the concentration of the HCl solution (in units of mol/L)? HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O( ) n(NaOH) = 0.02833 L 0.3145 mol = 8.910 x 10 -3 mol NaOH L soln n(HCl) = 8.910 x 10 -3 mol NaOH 1 mol HCl = 8.910 x 10 -3 mol HCl 1 mol NaOH M(HCl) = 8.910 x 10 -3 mol HCl = 0.4455 mol/L = 0.4455 M 0.02000 L soln

71. Oxidation-Reduction (Redox) Titration An oxidation-reduction titration (redox titration) is a titration based on an oxidation-reduction reaction. Calculations are done in the same way as in acid-base titrations. Example: A 20.00 mL sample of a solution containing Fe 2+ ion is titrated with a 0.02418 M solution of BrO 3 - ion. The equivalence point of the titration occurs after the addition of 38.67 mL of the BrO 3 - solution. What is the concentration of the Fe 2+ solution? 6 Fe 2+ (aq) + BrO 3 - (aq) + 6 H + (aq)  6 Fe 3+ (aq) + Br - (aq) + 3 H 2 O( )

72. Example: A 20.00 mL sample of a solution containing Fe 2+ ion is titrated with a 0.02418 M solution of BrO 3 - ion. The equivalence point of the titration occurs after the addition of 38.67 mL of the BrO 3 - solution. What is the concentration of the Fe 2+ solution? 6 Fe 2+ (aq) + BrO 3 - (aq) + 6 H + (aq)  6 Fe 3+ (aq) + Br - (aq) + 3 H 2 O( ) Strategy: 1) Find the number of moles of BrO 3 - (the solution where you know both the volume and concentration). 2) Use the balanced chemical equation to find the number of moles of Fe 2+. 3) Find the concentration of the Fe 2+ solution, using the number of moles (found in step 2) and the volume of Fe 2+ solution (given in the problem).

73. 6 Fe 2+ (aq) + BrO 3 - (aq) + 6 H + (aq)  6 Fe 3+ (aq) + Br - (aq) + 3 H 2 O( ) 20.00 mL Fe 2+ soln38.67 mL BrO 3 - soln 0.02418 M BrO 3 - moles BrO 3 - = 0.03867 L soln 0.02418 mol = 9.350 x 10 -4 mol BrO 3 - L soln moles Fe 2+ = 9.350 x 10 -4 mol BrO 3 - 6 mol Fe 2+ = 5.610 x 10 -3 mol Fe 2+ 1 mol BrO 3 - molarity of Fe 2+ soln = 5.610 x 10 -3 mol Fe 2+ = 0.2805 M 0.02000L soln

74. End of Chapter 4 A physicist, a biologist and a chemist went to the ocean for the first time. The physicist saw the ocean and was fascinated by the waves. He said he wanted to do some research on the fluid dynamics of the waves and walked into the ocean. He was drowned and never returned. The biologist saw the ocean and said that he wanted to do research on the marine flora and fauna he saw there. He walked into the ocean and also never returned. The chemist waited for a long time and then wrote down the following observation. “Physicists and biologists are soluble in ocean water”. - anonymous Two questions to distinguish chemists from non-chemists: (1) How do you pronounce “unionized”? (2) What is a mole?