) 2 (PO 4. ) 3 NaOCl LiBr KNO 3

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Frederica Hancock
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1 Salts 1

2 Salts are solid crystalline substances at room temperature that contains the cation of a base and the anion of an acid, e.g.: NaCl Mg 3 (PO 4 ) 2 Al 2 (SO 4 ) 3 NaOCl LiBr KNO 3 2

3 Some common salts are summarized in the table, below: Salt CaSO 4 ½H 2 O MgSO 4 7H 2 O Na 2 B 4 O 7 10H 2 O NaHCO 3 NaNO 2 AgNO 3 Name Plaster of Paris Epsom salts Borax Baking soda Preservative Antiseptic/germicide 3

4 The formation of salts necessarily depend on their solubility -- or the lack thereof -- in water. Solubility rules that actually help make chemical reactions make sense are tabulated, following slide: 4

5 Rule Alkali metal and NH 4+ salts are all soluble. Nitrates and acetates are all soluble. Chlorides, bromides and iodides are all soluble. Exceptions Some cations in analytical group 5 are moderately insoluble AgOAc is moderately insoluble Those salts of Pb 2+, Ag +, Hg 2 2+ ; BiOCl and SbOCl Sulfates are soluble. Those salts of Ca 2+, Sr 2+, Ba 2+, Ag +, Pb 2+, Hg 2 2+ Carbonate and sulfite salts are generally insoluble. Those of the alkali metals and NH 4 + Sulfides are generally insoluble. Those of the alkali metals and NH 4+ ; alkaline earth sulfides and Cr 2 S 3 and Al 2 S 3 are decomposed by water Hydroxides are generally insoluble. All other salts are insoluble. Alkali metals and NH 4+ ; Barium, strontium and calcium hydroxides are moderately soluble. 5

6 As a general rule, solubility is defined as being dissolved in aqueous solution to about 3-5%. Soluble salts are electrolytes, i.e., they will conduct an electrical current. The rules of electrolytes are summarized in the table, next slide, as well: 6

7 Summary of Strong and Weak Electrolytes Rule Most acids are weak electrolytes. Most bases are weak electrolytes. Most salts are strong electrolytes Exception The common strong acids: hydrochloric, hydrobromic, hydroiodic, nitric, sulfuric, chloric and perchloric The strong basic hydroxides: Li, Na, K, Rb, Cs, Ca, Sr, Ba hydroxides The most importantly weakly ionized salt is HgCl 2 ; occasionally, the following are listed without general agreement: Hg(CN) 2, CdCl 2, CdBr 2, CdI 2 and Pb(OAc) 2 7

8 Preparation of Salts 8

9 Mechanism Direct union of their elements. Reactions of acids with metals, metal hydroxides or metal oxides. Reactions of basic anhydrides with acid anhydrides. Reaction of acids with salts. Reaction of salts with other salts. Representative Reactions 2Na + Cl 2 2NaCl Fe + S FeS Zn + H 2 SO 4 ZnSO 4 + H 2 Fe(OH) 3 + 3HCl FeCl 3 + 3H 2 O CuO + H 2 SO 4 CuSO 4 + H 2 O BaO + SO 3 BaSO 4 CaO + CO 2 CaCO 3 BaCO 3 + 2HCl BaCl 2 + H 2 O + CO 2 BaCl 2 + H 2 SO 4 BaSO 4 + 2HCl AgNO 3 + NaCl AgCl + NaNO 3 ZnCl 2 + Na 2 S ZnS + 2NaCl 9

10 MO y X Synthesis. - Synthesis of the Oxyhalides Cl 2 (g) + 2NaOH(aq) + cold NaOCl(aq) + NaCl(aq) + H 2 O(l) The chlorine is bubbled through the lye; NaOCl is used for bleach, e.g., Clorox. 3Cl 2 (g) + 6NaOH(aq) + heat NaClO 3 (aq) + 5NaCl(aq) + 3H 2 O(l) NaClO 3 is used to bleach paper pulp. NaClO 3 +oxygen + electrolytic conditions NaClO 4 NaClO 4 is an oxidizer, e.g., 3NH 4 ClO 4 (s) + 3Al(s; powder) Al 2 O 3 (s) + AlCl 3 (g) + 3NO + 6H 2 O 1.4*10 6 pounds of ammonium perchlorate are used per launch to get the space shuttle off the ground. 10

11 Ionic Reactions and Ionic Equations 11

12 When reactions between ions occur, at least one kind of ion is removed from the "field of action". Simply put, its concentration decreases as the reaction proceeds. There are three ways to remove ions: 1. Formation of an insoluble precipitate 2. Formation of a weakly ionized substance, and 3. Oxidation or reduction of an ion Let's examine each way individually: 12

13 1: Formation of An Insoluble Precipitate Excess chloride ion "drives" this reaction to the right. 13

14 2: Formation of A Weakly Ionized Substance 14

15 3: Oxidation or Reduction of An Ion Note, also, that most of these reactions, one way or another, follow the solubility rules discussed previously in both CHEM 121 and

16 Rules for Writing Ionic Equations 16

17 1: Ionic formulas are written for a strong electrolyte in solution (review Electrolyte Rules, previously in 121 and 122), e.g.: 17

18 2: Molecular formulas are written for: a) Elements, gases, solids and non-electrolytes, e.g.: b) Weak electrolytes in solution, e.g.: c) Solid strong electrolytes or precipitates, e.g.: 18

19 Writing Ionic Equations 19

20 When writing these equations, do so to answer the following three (3) questions: 1. What kind of reaction is it? Double decomposition? Redox? 2. What are the possible products of the reaction? 3. Are any of the possible products or reactants insoluble or weakly ionized? 20

21 Double Decomposition Reactions (Precipitate and Weak Electrolyte Reactions) -- Examples a) KCl + Na(NO 3 ) NR Even the products would be soluble and, hence, no reaction occurs. b) Silver nitrate and hydrochloric acid -- precipitate formation Since the hydrogen ion and the nitrate ion are spectators, the net ionic reaction is the result. c) Hydrochloric acid and calcium acetate -- weak electrolyte formation 21

22 d) Acetic acid and sodium hydroxide -- conversion of one weak electrolyte to another e) Lead chloride and sodium sulfate -- conversion of one precipitate to another f) Hydrogen sulfide and copper (II) nitrate in acidic solution competition between a weak electrolyte and a precipitate 22

23 Double Decomposition Reactions (by Proton Transfer) 23

24 In this type of reaction, there must be a proton donor (D 1 ) and a proton acceptor (A 1 ). Donors dissociate as follows: D1 A 1 + H + Acceptors accept the proton as follows: H + + A 2 D 2 The overall reaction, then, is obtained by adding the two up (similarly to adding half reactions in redox: D 1 + A 2 A 1 + D 2 remember conjugate acid-base pairs from CHEM 121? Keeping with this form of these reactions, in the reactions, following slide, are examples of protolytic reactions -- Donors on the left of both reactants and products and Acceptors on the right of both reactants and products. 24

25 25

26 Double Decomposition Reactions (by Reactions Involving Precipitates and Complex Ions) 26

27 a) Silver chloride and ammonia -- Solution by Complex Ion Formation b) Diamino silver chloride and potassium iodide Dissociation of A Complex Ion to Form a Precipitate c) Tetra-amino cadmium sulfate and potassium cyanide Conversion of One Complex to Another d) Diamino silver chloride and nitric acid Decomposition of A Complex with Precipitate Formation 27

28 Redox Equations/Reactions 28

29 Examples: Fe 3+ + V 2+ V 3+ + Fe 2+ Cl 2 + 2Br - Br 2 + 2Cl - Remember: an increase in the oxidation number is oxidation; a reduction in the oxidation number is a reduction reaction. The graphic, below, helps put this in the proper perspective: The negatively charged end is the reduced end and the positively charged end is the oxidized end of the above graphic. 29

30 The Table, Below, Summarizes Typical Reducing Agents: Typical Reducing Agents Reducing Agents Conditions Oxidized Form Change in Oxidation Number Zn Acidic Zn 2+ 0 to 2+ Al Alkaline Al(OH) 4-0 to 3+ H 2 S Acidic S -2 to 0 Sn 2+ HCl SnCl to +4 S 2 O 4 2- Alkaline SO to +4 (S) HNO 2 Acidic NO to +5 (N) H 2 O 2 Acidic O 2-1 to 0 (O) 30

31 The table, below, summarizes typical oxidizing agents: Typical Oxidizing Agents Oxidizing Agent Conditions Reduced Form Change in Oxidation Number Cu 2+ Cu +2 to 0 Fe 3+ Fe to +2 Br 2 Acidic Br - 0 to -1 Cr 2 O 2-7 Acidic Cr to +3 (Cr) MnO - 4 Acidic Mn to +2 (Mn) MnO - 4 Alkaline MnO 2 +7 to +4 (Mn) NO M HNO 3 NO 2 +5 to +4 (N) NO - 3 Alkaline NH 3 +5 to -3 (N) ClO - 3 HNO 3 ClO 2 +5 to +4 (Cl) H 2 O 2 Acidic H 2 O -1 to -2 (O) H 2 O 2 Alkaline OH - -1 to -2 (O) 31

32 Colligative Properties 32

33 Aka collective properties because they are bound together through their common origin. Each of these properties is proportional to ONLY the number of solute molecules present NOT on the size or molar mass of the molecules. 33

34 For The Time Being, The Following Assumptions are Necessary 1. The solutions are IDEAL, 1. i.e., the vapor pressure of a solution of liquids is equal to the sum of each individual liquid s vapor pressure and 2. the vapor pressure of the defined solute varies with its concentration in a linear manner: P 1 = k X 1 34

35 Ideal Solution -- Graphically 35

36 Non-Ideal Solution Positive deviation from NON-Ideal solution of CS 2 with acetone. 36

37 NON-Ideal Solution Negative Deviation from NON- Ideal solution of CHCl 3 with Acetone 37

38 Assumptions 2. The solutions are dilute concentrations less than 0.2 m 3. We will consider ONLY (at this time) nonelectrolyte solutions 38

39 Summary of Colligative Properties 1. Lower of the Vapor Pressure of the Solvent 2. Boiling Point Elevation 3. Freezing Point Depression 4. Osmotic Pressure 39

40 Vapor Pressure Lowering -- Graphically 40

41 Vapor Pressure -- Arithmetic An IDEAL solution, i.e., one that contains a solvent 1 and a NON-volatile solute 2 (e.g., sucrose in water) follows Raoult s Law: The vapor pressure of a component of a solution is equal to the product of its mole fraction and the vapor pressure of the pure liquid. 41

42 Vapor Pressure -- Example P 1 = X 1 P 1 * P 1* = vapor pressure of the pure liquid 1 X 1 = mol fraction of 1 P 1 = experimentally obtained pressure P 2 = X 2 P 2 * P 2* = vapor pressure of the pure liquid 2 X 2 = mol fraction of 2 P 2 = experimentally obtained pressure 42

43 How to Calculate Mol Fraction Calculate the mole fraction of each component in an ideal solution of glycerine (92.1 g; MW 92.1 g/mol) and ethanol (184 g; MW 46 g/mol) at 40 C. 1mol (92.1g glycerine) = mol glycerine g glycerine 1mol (184 g EtOH ) 4 46 = mol EtOH g EtOH 1mol X glycerine = = 0.2 (1 + 4) mol 4mol X EtOH = = 0.8 (1 + 4) mol 43

44 Example, Cont d Determine the vapor pressure of the EtOH component of the solution if the vapor pressure of pure EtOH at 40 C is torr. P EtOH = X EtOH P * EtOH P EtOH = (0.8)(135.3torr) P EtOH = torr 44

45 Cont d The reduction in vapor pressure is = 27.1 torr and is calculated also as follows: Δ P Δ = P X glycerine P = 27.1torr * EtOH Δ P = (0.2)(135.3torr) 45

46 In other words, the glycerine molecules in this solution lower the frequency of escape of EtOH molecules from the surface of the liquid The greater the number of molecules of SOLUTE, the less the vapor pressure exerted by the SOLUTION 46

47 Boiling Point Elevation Boiling point the temperature at which a pure solvent s or solution s vapor pressure = atmospheric pressure NOTE: with reduction in vapor pressure comes an increase in boiling point (BP) Arithmetically: ΔT = K m b 47

48 ΔT = rise in BP in C K b = boiling point elevation constant = 0.51 C/m -- ALWAYS m = molality of the solution 48

49 Molarity vs Molality M Molarity Dependent upon temperature m molality Independent of temperature mol of solute M = m = L of solution mol of solute kg of solution 49

50 Example 500 ml water are mixed with 5.85 g NaCl. Calculate the M and m of this solution. ρ H2 O at this temperature is g/mol g NaCl (500 ml water) 1mol NaCl 23 g Na 35.5 g Cl + = g 1kg ml 1000 g = = 0.1mol NaCl = 0.493kg water 50

51 Cont d mol 0.1mol M = = = 0.2 M L 0.5 L mol 0.1mol m = = = 0.203m kg 0.493kg 51

52 Example Determine the boiling point of a solution of 15 g ethylene glycol (C 2 H 4 (OH) 2 ) in 1 kg water. (15 1mol g ethylene glycol) 62 g Ethylene glycol 0.24mol m = = 0.24 m 1kg ΔT T BP = K b = 100 m = (0.51)(0.24) = o C o C o = = 0.24mol C o C 52

53 Another Example Determine the boiling point of a solution of 186 g ethylene glycol in 1 kg water. (186 1mol Ethylene glycol g Ethylene glycol) 62 g Ethylene glycol 3mol m = = 3m 1kg ΔT = T BP K b m = (0.51)(3) = 1.53 = = o o C C = 3mol 53

54 Another Example The boiling point of a solution of water and ethylene glycol is 105 C. How much ethylene glycol is present in 1 kg of water? (9.8mol 9.8mol kg ΔT = K 1kg = 9.8mol Ethylene glycol 62 g Ethylene glycol Ethylene glycol) 1mol B m o o 0.51 C 5 C = ( X ) m o 5 C m = 9.8m o 0.51 C = g Ethylene glycol 54

55 Freezing Point Depression ΔT = K f m ΔT= change in temperature the freezing point depression K f = freezing point constant = 1.86 C/m m = molality of the solution 55

56 Example Determine the freezing point of a solution of 50 g NaCl in 0.5 kg water. (50 g NaCl) ΔT m = 100 1mol 23 g Na g Cl o 0.855mol 0.5kg C 3.18 o C = 1.71m = (1.86)(1.71) = 3.18 = 96.8 = 0.855mol o o C C 56

57 What is Origin of BP and FP constants? Phase Diagram Left shift for FP depression Right shift for BP elevation When compared against pure water and nonelectrolyte solution 57

58 Osmosis/Osmotic Pressure Osmosis the movement of water from a region of [higher] to [lower] Osmotic pressure the force causing the movement of water 58

59 M R T Π = osmotic pressure in atm M = molarity of the solution R = gas constant = L-atm/mol-K T = temperature in K (absolute temperature) 59

60 Example What is the osmotic pressure of a 0.4 M solution of albumin at 40 C? T Π = Π = M R T = = 313 K (0.4)( )(313) Π = 10.3atm 60

61 Osmotic Pressure Significance 61

62 Physiological Application Iso- Osmotic 62

63 Physiological Application -- Starvation Blood is hypo-osmotic (hypo-albuminemia) which leads to increased tissue water retention = edema 63

64 Arithmetic Application Step 1 The following data was obtained during an osmotic pressure experiment: 20 g Hb/L; height of column = 77.8 mm; T = 298K. From this data, determine the MW of Hemoglobin. Pr P = essure = Ah ρ g = A Force Area h ρ g A = cross-sectional area of the tube h = height of the column ρ = density of the solution g = acceleration due to gravity 64

65 65 Arithmetic Application Step 2 atm dyne cm cm dyne cm dyne atm cm dyne s cm g cm dyne s cm g ce s cm g pure water cm g s cm g cm h sin (980.7)(1) (7.78) = Π = = = = = Π = = = ρ

66 Arithmetic Application Step 3 MW = Π = M R T c M = MW g c = L MW = molecular weight in g / mol c RT Π = MW c RT MW = Π (20)( )(298) = g / mol ,000 g Hb / mol 66

67 Uncommon Colligative Properties -- 1 Atomic Weight Determination in Solution From Specific Heats: the product of the specific heat and atomic weight of solid elements is very nearly a constant, i.e., about 6.4 This approximately valid for all solid elements with atomic weights > 40 and for most metallic elements It does NOT hold for such elements as C, Si, P, S 67

68 Example Determine the approximate atomic weight of Fe. Its specific heat is cal/g/. atomic mass specific heat = 6.4 atomic mass = 6.4 specific heat = = 56 g / mol Actual molecular weight of Fe is g/mol Gives us an approximation, however. Called the Law of Dulong and Petit 68

69 Uncommon Colligative Properties -- 2 From Equivalent Weights Exact atomic weights come from experimental determination of Exact equivalent weight Approximate atomic weight 69

70 Example Approximate atomic weight of C = g Exact equivalent weight of C = ought for = 4.02 oxidation state of C to be 4 it' s not because is incorrect exact atomic wt : = g mol 70

71 Colligative Properties of Electrolyte Solutions These properties are influenced by the # of ions present in solution, therefore: ΔT f of 0.01 m NaCl ought to be 2X ΔT f of 0.01 m sucrose (NaCl Na + + Cl - ) 71

72 van t Hoff factor -- #1 i = # of actual particles # of particles in solution in solution BEFORE dissociation E. g., NaCl i = 2 1 = 1 from Na + and 1 from Cl 1 from NaCl i = 2 72

73 van t Hoff Factor -- #2 Given N units of an electrolyte and if α = the degree of dissociation, then for MX #M + + #X - There will be N(1-α) undissociated units and N(#M + + #X - ) α ions in solution at equilibrium 73

74 74 van t Hoff Factor Re-Written 1 ) # (# 1 ) # (# 1 ) # (# ) (1 + = + + = + + = X M i and X M i N X M N N i α α α α α

75 i is also calculable from Π For two solutions, 0.01 m CaCl 2 with an osmotic pressure of atm and 0.01 m sucrose with an osmotic pressure of atm, at 298K, calculate the van t Hoff factor and α for CaCl 2. Assume ideal behavior. Remember that Π is directly proportional to the # of particles present. i = 0.605atm 0.224atm = 2.70 # M α = (# M CaCl CaCl # X i 1 + # X Ca Cl = 3 = = = 0.85 ) is only 85% dissociated 75

76 Modifications of Equations for Electrolytes ΔT = K b i m ΔT = K f i m Π = M i R T Where i = van' t Hoff factor! 76

SCHOOL YEAR CH- 13 IONS IN AQUEOUS SOLUTIONS AND COLLIGATIVE PROPERTIES SUBJECT: CHEMISTRY GRADE : 11 TEST A

SCHOOL YEAR CH- 13 IONS IN AQUEOUS SOLUTIONS AND COLLIGATIVE PROPERTIES SUBJECT: CHEMISTRY GRADE : 11 TEST A SCHOOL YEAR 2017-18 NAME: CH- 13 IONS IN AQUEOUS SOLUTIONS AND COLLIGATIVE PROPERTIES SUBJECT: CHEMISTRY GRADE : 11 TEST A Choose the best answer from the options that follow each question. 1. A solute