Part One: Ions in Aqueous Solution

Size: px

Start display at page:

Download "Part One: Ions in Aqueous Solution"

Scott Ralph Francis
6 years ago
Views:

1 A. Electrolytes and Non-electrolytes. CHAPTER FOUR: CHEMICAL REACTIONS Part One: Ions in Aqueous Solution 1. Pure water does not conduct electric current appreciably. It is the ions dissolved in the water that conduct charge. 2. Electrolyte = substance that dissolves in water forming ions Chapter 4 Page 1

2 3. Classification of water-soluble substances. Water-soluble Substances Electrolytes -dissociate to some extent in water to form ions -solution conducts electricity Nonelectrolytes -do not dissociate into ions in water -solution is non-conductive Strong Electrolytes -dissociate to large extent -conduct well Weak Electrolytes -dissociate to small extent -conduct slightly HAc (acetic acid) strong strong most soluble acids bases salt HCl, HNO 3, NaOH NaCl, CaCl 2 H 2 SO 4 a. Dissociation = process in which a solid ionic compound separates into ions in solution. H NaCl 2 O Na + + Cl b. Ionization = process in which a molecular compound separates to form ions. H HCl 2 O H + + Cl c. When one of the ions is H +, the substance is an acid. H HCl 2 O H + + Cl Chapter 4 Page 2

3 d. When OH - are produced, substance is a base. NaOH H 2 O Na + + OH e. When neither ion is H + or OH -, substance is a salt. H KBr 2 O K + + Br B. Strong and weak acids. 1. Strong acids ionize completely in dilute aqueous solution. Example: HCl is a gas when pure, dissolves to large extent in water, and fully ionizes. 2. Know these: H HCl(g) 2 O HCl(aq) H + (aq) + Cl (aq) H 2 O 3. H + ions exist in aqueous solution attached to H 2 O (i.e., hydrated). H + (aq) really means H 3 O + H + H O hydronium ion H Chapter 4 Page 3

4 4. Weak acids ionize only slightly. Example: acetic acid CH 3 COOH(aq) H + (aq) + CH 3 COO - (aq) dominant form 99.5% 0.5% Abbreviation: HAc(aq) H + (aq) + Ac - (aq) 5. Know these: 6. Note that we have introduced the idea of reversible reactions. ( ) Reaction is continually happening to appreciable extent in both directions to produce a balance in which both Reactants and Products are present. C. Bases = most commonly metal hydroxides. (M x (OH) y ) 1. Strong soluble bases: NaOH(s) Na + (aq) + OH (aq) H 2 O dissociate completely Chapter 4 Page 4

5 a. Insoluble bases: metal hydroxides which have low solubility in H 2 O, produce low amount of OH - (aq). Cu(OH) 2, Zn(OH) 2, Fe(OH) 2, Fe(OH) 3 b. Weak bases: soluble in H 2 O but ionize only slightly. Example: NH 3 (aq) + H 2 O(l) molecular compound NH 4+ (aq) + OH - (aq) D. Solubility rules for ionic compounds in aqueous solution. 1. Compounds of Group IA and NH 4 + are soluble. 2. Nitrates, NO 3- ; acetates, CH 3 COO - ; chlorates, ClO 3- ; and perchlorates, ClO 4-, are soluble. 3. Chlorides, Cl -, are soluble EXCEPT AgCl, Hg 2 Cl 2, and PbCl 2. Bromides, Br -, and iodides, I -, similar to chlorides, with some exceptions. As these halide ions (Cl -, Br -, I - ) increase in size, solubilities decrease. 4. Sulfates, SO 4 2-, are soluble EXCEPT PbSO 4, BaSO 4, HgSO 4, CaSO 4, SrSO 4 and Ag 2 SO 4 5. Carbonates, CO 3 2-, are insoluble EXCEPT those of the Group IA metals and NH 4 +. MgCO 3 is moderately soluble. 6. Phosphates, PO 4 3-, and arsenates, AsO 4 3-, are insoluble EXCEPT those of the Group IA metals and NH Sulfides, S 2-, are insoluble EXCEPT Group IA and IIA metals and NH Metal hydroxides, OH -, are insoluble EXCEPT Group IA metals and Group IIA metals below Mg. Chapter 4 Page 5

6 E. Writing Reactions in Aqueous Solution. (Section 4.2) 1. Three ways of writing equations. (illustrated by neutralization of Mg(OH) 2 with HCl in aqueous solution. a. Molecular equations = show complete formulas for all compounds. (best for stoichiometric calculations) 2HCl(aq) + Mg(OH) 2 (s) MgCl 2 (aq) + 2H 2 O (insoluble) (soluble) b. Complete ionic equations = show predominant form of compounds as ions if they strongly dissociate. (the bridge to the net ionic equation) 2[H + (aq) + Cl - (aq)] + Mg(OH) 2 (s) [Mg 2+ (aq) + 2Cl - (aq)] + 2H 2 O c. Net ionic equation = complete ionic equation after spectator ions have been canceled out. 2H + (aq) + Mg(OH) 2 (s) Mg + (aq) + 2H 2 O [This is the essence of the reaction.] 2. To get the net ionic equation: a. Write down the full molecular equation first. b. For each compound, ask: Is it a strong electrolyte? If not, keep full formula. For example, AgCl(s). c. Cancel out ions that appear on both sides. These are spectator ions. 3. Problem: Write down all 3 types of equations for the reaction of sodium sulfide with silver nitrate to produce silver sulfide and sodium nitrate. a. Molecular equation: Na 2 S(aq) + 2AgNO 3 (aq) Ag 2 S(s) + 2NaNO 3 (aq) all Na salts all nitrates insoluble soluble soluble sulfide Chapter 4 Page 6

7 b. Complete ionic equation: c. Net ionic equation: [2Na + (aq) + S 2- (aq)] + 2[Ag + (aq) + NO 3 - (aq)] Ag 2 S(s) + 2[Na + (aq) + NO 3 - (aq)] S 2- (aq) + 2Ag + (aq) Ag 2 S(s) (Simple formation of insoluble silver sulfide.) Part Two: Types of Chemical Reactions There are millions. We will look at three important types: a) precipitation; b) acid-base; and c) oxidation-reduction. Three driving forces for ionic reaction to happen, all in which there is a removal of ions from solution: a. Formation of insoluble solid precipitate. (e.g. Ag 2 S) b. Formation of a very stable nonionized molecule. (e.g. H 2 O) c. Formation of a gas that escapes. (e.g., H 2 ) A. Precipitation Reactions. (Metathesis Reaction I) (Section 4.3) 1. An insoluble solid (precipitate) forms and then settles out of solution. 2. Driving force is strong attraction between cations and anions, resulting in removal of ions from solution. 3. Example: addition of sodium phosphate solution to a calcium chloride solution. What happens? Pictorially: Chapter 4 Page 7

8 Na + Ca 2+ Na + Ca 2+ + PO 4 3- Cl - PO 4 3- Cl - Na + Cl - solid Ca 3 (PO 4 ) 2 2Na 3 PO 4 (aq) + 3CaCl 2 (aq) Ca 3 (PO 4 ) 2 (s) + 6NaCl(aq) Ca 3 (PO 4 ) 2 ppts! Net ionic equation: 2PO 4 3- (aq) + 3Ca 2+ (aq) Ca 3 (PO 4 ) 2 (s) 4. Additional example: Potassium chromate solution added to lead nitrate solution. What happens? K + Pb 2+ soluble Net ionic equation: insoluble CrO 4 2- NO 3 - K 2 CrO 4 (aq) + Pb(NO 3 ) 2 (aq) PbCrO 4 (s) + 2KNO 3 (aq) CrO 4 2- (aq) + Pb 2+ (aq) PbCrO 4 (s) yellow solid B. Acid/Base Reactions. (Metathesis Reaction II) (Section 4.4) 1. General form of neutralization between acid and base: acid + base salt + H 2 O Chapter 4 Page 8

9 2. Driving force here is formation of H 2 O from H + and OH Example - strong acid + strong base: Total ionic reaction: HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) [H + (aq) + Cl - (aq)] + [Na + (aq) + OH - (aq)] [Na + (aq) + Cl - (aq)] + H 2 O Net ionic reaction for strong acid + strong base: H + (aq) + OH - (aq) H 2 O(l) 4. Example - weak acid + strong base: Total ionic reaction: HAc(aq) + NaOH(aq) NaAc(aq) + H 2 O(l) HAc(aq) + [Na + (aq) + OH - (aq)] [Na + (aq) + Ac - (aq)] + H 2 O(l) Net ionic reaction for weak acid + strong base: HAc(aq) + OH - (aq) Ac - (aq) + H 2 O(l) 5. Example of Acid-Base reaction with Formation of a Gas (a Driving Force for Reaction) reaction of insoluble carbonate salt with strong acid: 2HCl(aq) + CaCO 3 (s) H 2 CO 3 (aq) + CaCl 2 (aq) limestone carbonic acid seashells (weak acid) chalk Chapter 4 Page 9

10 Net ionic reaction: 2H + (aq) + CaCO 3 (s) H 2 CO 3 (aq) + Ca 2+ (aq) i.e. calcium carbonate dissolves in acid. driving force: further decomposition of H 2 CO 3 to CO 2 (g) H 2 CO 3 (aq) CO 2 (g) + H 2 O(l) escapes (origin of bubbles) C. Oxidation-Reduction (Redox) Reactions. (introduction) (Section 4.5) 1. Many reactions involve transfer of electrons between species, called oxidationreduction reactions, abbreviated redox reactions. 2. To recognize these, we first need to introduce a concept called the oxidation number or oxidation state. 3. Oxidation originally referred to combining with Oxygen, resulting in increase in oxidation number. D. Oxidation Numbers. (keeping track of electrons) 1. Oxidation number (ox#), or oxidation state, of an element: a. In an ionic compound, oxidation number of an element = number of e - gained or lost by an atom of that element. Example: NaCl compound consists of Na + and Cl - ion pair. ox# of Na = +1 Na + said to be in +1 oxidation state ox# of Cl = -1 CaCO 3 ox# of Ca = +2 Chapter 4 Page 10

11 b. In a molecular compound, ox# is assigned according to rules that aid in balancing redox equations. Meaning is different. Example: CO 2 ox# of O is -2 ox# of C is +4 (+4)(-2) CO 2 Example: CO ox# of O is -2 ox# of C is +2 (+2)(-2) CO C is said to be less fully oxidized to CO than in CO Rules for assigning ox# s: a. Ox# s always assigned on a per atom basis. b. Ox# s of any free uncombined element is zero. Examples: O 2, S 8, Cu(s) Chapter 4 Page 11

12 c. Ox# s of an element in a simple monatomic ion equals the charge of the ion. Na +, Na is +1 d. Sum of ox# s in a compound = 0. e. Sum of ox# s in polyatomic ion = charge of ion. CO 3-2 C is +4; O is -2 sum = (-2) = -2 f. H is always +1 except when combined with a metal in a metal hydride: NaH, CaH 2. (+1)(-1) NaH g. Group IA are always +1. h. Group IIA are always +2. i. Group IIIA are always +3, except in exotic compounds. j. Group VA are -3 in binary compound with a metal or with hydrogen. (+1)(-3) Na 3 N (-3)(+1) NH 3 -sodium nitride -ammonium Otherwise, most often +3 or +5. k. Oxygen is -2 except in OF 2 (+2) and in peroxides. (-1). (+1)(-1) H 2 O 2 (+1)(-1) Na 2 O 2 -example of exception -sodium peroxide l. Halogens are -1 in binary compounds with metals or H. Chapter 4 Page 12

13 3. Examples of assigning ox# in some compounds. (Hint: 1st do elements having hard and fast rules, then others by elimination.) a. Sodium hypochlorite (Clorox bleach): NaOCl (+1) -do Group IA first always; Na (-2) -do O next; O -Cl has to be what to sum to zero? +1 (+1) (-2)(+1) Na O Cl After bleach does its thing, Cl ends up as NaCl. Cl is -1 there. b. Potassium chromate: K 2 CrO 4 (+1) -do Group IA first always; K (-2) -do O; O -Cr has to be what to sum to zero? +6 (+1) (+6)(-2) K 2 Cr O 4 c. Sulfuric acid: H 2 SO 4 (+1)(+6)(-2) H 2 S O 4 H is +1, except in hydrides. O is -2, except in peroxides and OF 2 S is what to make sum = 0? +6 d. Sulfurous acid: H 2 SO 3 (+1)(+4)(-2) H 2 S O 3 Note: S is (+4) here and (+6) in sulfuric acid. Chapter 4 Page 13

14 4. Oxidation originally referred to combining with Oxygen, resulting in increase in ox#. Example - Combustion of methane is an oxidation of the Carbon: (-4)(+1) (0) (+4)(-2) CH 4 + 2O 2 CO 2 + 2H 2 O (-4) (+4) {C} {C} oxidation of C 5. Oxidation = algebraic increase in ox# with corresp. loss of e Reduction = algebraic decrease in ox# with corresp. gain of e Oxidation and reduction always occur simultaneously. Oxygen (0) (-2) Hydrogen (0) (+1) (0) (0) (+1)(-2) 2H 2 + O 2 2 H 2 O reduced oxidized 8. Oxidizing agent: a. Oxidizes other species. b. Is itself reduced. 9. Example: Immerse copper wire in silver nitrate solution. Copper is dissolved as the Cupric ion and silver is deposited as metal. AgNO 3 (aq) + Cu(s) Cu(NO 3 ) 2 (aq) + Ag(s) Net ionic equation: Ag + (aq) + Cu(s) Cu 2+ (aq) + Ag(s) What species is oxidized? Cu (0) (+2) What species is reduced? Ag (+1) (0) What is the oxidizing agent? AgNO 3 or Ag + (aq) Chapter 4 Page 14

15 E. Many Oxidation-Reduction Reactions also fit under other categories, such as: 1. Combination reactions. 2. Decomposition reactions. 3. Displacement reactions. 4. Combustion reactions. F. Combination Reactions. 1. Combination reaction = reaction in which 2 or more substances combine to form a third substance. 2. Element + Element Compound a. 2 Mg(s) + O 2 2 MgO (metal + nonmetal ionic cmpd) b. P 4 (s) + 6 Cl 2 (g) 4 PCl 3 (s) (nonmetal+nonmetal molec cmpd) 3. Compound + Element Compound PCl 3 (s) + Cl 2 (g) PCl 5 (s) 4. Compound + Compound Compound CaO(s) + CO 2 (g) CaCO 3 (s) (this one is NOT redox) G. Decomposition Reactions. 1. Example: 2 KClO 3 2 KCl + 3O 2 Chapter 4 Page 15

16 H. Displacement Reactions. 1. Reaction in which one element displaces another from a compound. M + AX MX + A 2. The more readily a metal M displaces another element and forms a positive ion, the more active that metal is. 3. The metal doing the displacing is being oxidized. 4. Table 4.6 is Activity Series: Chapter 4 Page 16

17 5. [More Active Metal + [Less Active Metal + Salt of Less Active Metal] Salt of More Active Metal] Example: Zinc metal is placed in copper(ii) sulfate solution. Copper metal falls to bottom. Resulting solution contains zinc sulfate, ZnSO 4. Formula unit equation: CuSO 4 (aq) + Zn(s) Cu(s) + ZnSO 4 (aq) Net ionic equation: Cu 2+ (aq) + Zn(s) Cu(s) + Zn 2+ (aq) Displacement reaction: the more active metal, zinc, displaces ions of less active metal, copper, from aqueous solution. 6. [Active Metal + Nonoxidizing Acid] [Hydrogen + Salt of Acid] a. Example: Zn(s) + H 2 SO 4 (aq) ZnSO 4 (aq) + H 2 (g) strong acid soluble salt Net ionic equation: Zn(s) + H + (aq) + HSO 4- (aq) Zn 2+ (aq) + SO 4 2- (aq) + H 2 (g) b. HNO 3 is the common oxidizing acid. It reacts with active metals to produce oxides of nitrogen, but not hydrogen, H 2. c. Very active metals can even displace hydrogen from water. Example: 2K(s) + 2H 2 O(l) 2[K + (aq) + OH - (aq)] + H 2 (g) 7. [Active Nonmetal + [Less Active Nonmetal + Salt of Less Active Nonmetal] Salt of More Active Nonmetal] a. Example: Cl 2 (g) + 2[Na + (aq) + Br - (aq)] 2[Na + (aq) + Cl - (aq)] + Br 2 (l) chlorine sodium bromide sodium chloride bromine Chapter 4 Page 17

18 b. Each halogen will displace less active (heavier) halogens from their binary salts; i.e., the order of increasing activities is: I. Combustion Reactions. (least active) I 2 < Br 2 < Cl 2 < F 2 (most active) 1. Combustion reaction = a reaction in which a substance reacts with oxygen, sometimes with rapid release of heat to product a flame. 2. Oxygen goes from 0 to -2 oxidation state. 3. Of the combusted material, all Carbon ends up as CO or CO 2, all Hydrogen ends up in H 2 O. 4. Example: Combustion of propane: C 3 H 8 (g) + O 2 (g) J. Balancing Simple Redox Reactions. 1. The following redox reaction looks balanced, but what is wrong? Zn(s) + Ag + (aq) Zn 2+ (aq) + Ag(s) The charge is not balanced. 2. Balance using the Half-Reaction method: (separate into two hypothetical half-reactions, one an oxidation, the other a reduction, by specifically putting electrons in the equations. Zn(s) Zn 2+ (aq) + 2 e- oxidation 2[Ag + (aq) + 1 e- Ag(s) ] reduction Zn(s) + 2Ag + (aq) Zn 2+ (aq) + 2Ag(s) 3. More complicated cases are handled in Chapter 19, in connection with electrochemical cells. Chapter 4 Page 18

19 Part Three: Working with Solutions A. Molar Concentrations of Solution. (Section 4.7) 1. Many reactions carried out in solution for convenience. 2. Solution = homogeneous mixture of 2 or more substances. 3. Solvent = the dispersing medium. (substance usually present in greatest abundance) 4. Solute = substance dissolved in the solvent. (usually present in least amount) 5. Aqueous solutions most common. (water is solvent) 6. Concentrations expressed in various ways: a. Mass Percent of solute: % solute = mass of solute mass of solution 100% b. Molarity (M) or molar concentration: molarity = number of moles of solute number of liters of solution B. Mass Percent of Solute. 1. Problem involving percent solute: A saline solution is 5.00% NaCl. Calculate the mass of NaCl present in 400 g of solution. 5.00% means 5.00 g NaCl 100 g solution 5.00 g NaCl 400g solution = 20.0g NaCl present 100 g solution (% solute) mass solution = mass solute 100 Chapter 4 Page 19

20 2. Variation involving density: A saline solution is 5.00% NaCl. Calculate the mass of NaCl present in 400 ml of solution. The density of the solution is 1.10 g/ml. 400 ml solution x 1.10 g/ml = 440 g solution Then work like previous example: 5.00 g NaCl 440g solution = 22.0g NaCl present 100 g solution 3. Another variation: What volume of a 30% glucose solution contains 50 grams of glucose? The density of solution is 1.15 g/ml. 50 g glucose ---> vol of solution required 100g soln 30g glu 1mL soln 1.15g soln 100g soln 1mL soln 50g glu =145mL soln 30g glu 1.15g soln is the % solute by mass is the density of solution C. Molarity, or Molar Concentration (M). 1. Definition: molarity = moles of solute Liters of solution 2. Preparation of solutions of desired concentration: Chapter 4 Page 20

What mass of NaOH would you weigh out to prepare 2.00 L of 0.100 M NaOH? 0.100 M reads 0.

200 moles NaOH 1 Liter of solution 0.200 moles NaOH 40.0 g NaOH mol NaOH = 8.0 g NaOH Figure 4.18 3.

present) = (conc in molarity) x (Volume in Liters) 4. Problem: What mass of HCl is present in a 2.

21 What mass of NaOH would you weigh out to prepare 2.00 L of M NaOH? M reads molar meaning moles NaOH per Liter of solution moles NaOH 2.00 L solution = moles NaOH 1 Liter of solution moles NaOH 40.0 g NaOH mol NaOH = 8.0 g NaOH Figure Important relations: M= n V molarity = moles solute Volume of solution so: n = MV or: (moles of solute present) = (conc in molarity) x (Volume in Liters) 4. Problem: What mass of HCl is present in a 2.5 L jug of concentrated HCl solution (11.6 M)? a. First compute moles HCl present: moles Chapter 4 = conc x volume = 11.6 mol/l x 2.5 L = 29.0 mol Page 21

22 b. Now convert to grams HCl: 29.0 mol x g HCl/mol = 1,057 g HCl 5. Problem: Calculate the molarity of a solution in which 10.0 g NaOH are dissolved in L solution g moles NaOH molarity = Vol solution (L) = 40.0 g/mol L = mol/l = M 6. Commercial phosphoric acid is 85% H 3 PO 4 by mass, and its specific gravity is Calculate its molarity. molarity = moles H 3 PO 4 Vol solution (L) 85% H 3 PO 4 = 85 g H 3 PO 4 100g soln =14.7 mol/l D. Dilution of Solutions. (Section 4.8) 1.69 g soln 1 ml soln 103 ml 1 mol H PO 3 4 L 97.9 g H 3 PO 4 1. Dilution is simply the addition of solvent to an existing solution. Moles of solute remains fixed. 2. Therefore: moles solute before = moles solute after n 1 = n 2 M 1 V 1 = M 2 V 2 molarity x volume = molarity x volume before before after after Chapter 4 Page 22

23 3. Since V is on both sides, can use V in any units as long as consistent. 4. Dilution problem: If 10.0 ml of 3.0 M NaCl is diluted to ml, what is the new molarity of NaCl solution? V 1 M 1 = V 2 M 2 (10.0 ml) x (3.0 M) = (250.0 ml) x M 2 M 2 = 10.0 ml 3.0 M 250 ml = 0.12 M E. Volumetric Analysis (Section 4.10) (Solution Stoichiometry) 1. Carrying out reactions in solution: Problem: What volume of M HCl(aq) would exactly neutralize 400 ml of M NaOH(aq)? moles HCl = moles NaOH (at neutralization) n HCl = n NaOH M HCl V HCl = M NaOH V NaOH 0.100M V HCl = 0.200M L V HCl = M L M = L = 800 ml Chapter 4 Page 23

24 F. Method of Titration: 1. Titration = process in which a solution of one reactant (the titrant) is carefully added to a solution of another reactant. Volume of titrant required for complete reaction is measured. 2. Complete reaction is determined by a color change in the indicator. a. Example: the indicator phenolphthalein is colorless in acidic solution and pink in basic solution. b. The indicator is placed in acidic reactant solution. The base is then added until the first drop produces a pink color that persists. c. When stoichiometric amounts have reacted, we call this the end point or equivalence point of the titration. d. Example: reacting H 2 SO 4 with NaOH, the equivalence point is when: moles of NaOH added = 2 x (moles of H 2 SO 4 present) e. Example: reacting HCl with NaOH, equivalence point is when: moles of HCl = moles of NaOH Chapter 4 Page 24

25 3. Problem: 20.0 ml of HCl solution are required to neutralize 10.0 g of NaOH. What is the molarity of the HCl solution? 4. In the neutralization of phosphoric acid with sodium hydroxide, what volume of M NaOH is required to neutralize 400 ml of 0.50 M H 3 PO 4? 1 mole H 3 PO 4 reacts with 3 mol NaOH L H 3 PO 4 x 0.50 mol/l H 3 PO 4 = mol H 3 PO 4 Requires 3 x mol = mol NaOH mol = MV = M x V V = mol M = 3.00 L Chapter 4 Page 25

26 5. Problem: 50.0 ml of H 2 SO 4 of unknown concentration is titrated with a M NaOH titrant solution. The end point is reached after 40.0 ml of titrant are added. Calculate molarity of H 2 SO 4 solution. H 2 SO NaOH Na 2 SO 4 + H 2 O end point when: (moles of NaOH) = 2 x (moles of H 2 SO 4 ) V b M b = 2 x (V a M a ) 40.0 ml x M = 2 x (50.0 ml x M a ) M a = M H 2 SO 4 6. In the neutralization of 2.00 g of MgO with HCl, 25.0 ml of an HCl solution are required. What is the molarity of the HCl solution? 2 HCl + MgO MgCl 2 + H 2 O Chapter 4 Page 26

Solubility Rules See also Table 4.1 in text and Appendix G in Lab Manual

Solubility Rules See also Table 4.1 in text and Appendix G in Lab Manual Ch 4 Chemical Reactions Ionic Theory of Solutions - Ionic substances produce freely moving ions when dissolved in water, and the ions carry electric current. (S. Arrhenius, 1884) - An electrolyte is a