Compounds in Aqueous Solution

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1 1 Compounds in Aqueous Solution Many reactions involve ionic compounds, especially reactions in water KMnO 4 in water K + (aq) ) + MnO 4- (aq)

2 2 CCR, page 149

3 3 How do we know ions are present in aqueous solutions? The solutions They are called!"!#$"%#! HCl,, MgCl 2, and NaCl are &. They dissociate completely (or nearly so) into ions.

4 4 & HCl,, MgCl 2, and NaCl are &. They dissociate completely (or nearly so) into ions.

5 5 Acetic acid ionizes only to a small extent, so it is a '( it is a '( CH 3 CO 2 H(aq) ---> CH 3 CO 2- (aq) + H + (aq)

6 6 Acetic acid ionizes only to a small extent, so it is a '( is a '( CH 3 CO 2 H(aq) ---> CH 3 CO 2- (aq) + H + (aq)

7 7 Some compounds dissolve in water but do not conduct electricity. They are called Examples include: sugar ethanol ethylene glycol

8 )* 8 Screen 5.4 & Figure 5.1

9 9 )* Common minerals are often formed with anions that lead to insolubility: sulfide carbonate fluoride oxide Azurite, a copper carbonate Iron pyrite, a sulfide Orpiment, arsenic sulfide

10 10 An acid > > H + in water Some & acids are HCl H 2 SO 4 HClO 4 HNO 3 hydrochloric sulfuric perchloric nitric HNO 3

11 11 An acid > > H + in water HCl(aq) ---> H + (aq) ) + Cl - (aq)

12 12 #+ * HCl Cl - H 2 O hydronium ion H 3 O +

13 ( 13 WEAK ACIDS = weak electrolytes CH 3 CO 2 H acetic acid H 2 CO 3carbonic carbonic acid H 3 PO 4phosphoric phosphoric acid HF hydrofluoric acid Acetic acid

14 14 Nonmetal oxides can be acids CO 2 (aq) + H 2 O(liq) ---> > H 2 CO 3 (aq) SO 3 (aq) + H 2 O(liq) ---> > H 2 SO 4 (aq) and can come from burning coal and oil.

15 ,! see Screen 5.9 and Table Base ---> > OH -- in water NaOH(aq) ---> Na + (aq) ) + OH - (aq) NaOH is a strong base

16 Ammonia, NH Ammonia, NH 3, 16

17 17,! Metal oxides are bases CaO(s) ) + H 2 O(liq) --> > Ca(OH) 2 (aq) CaO in water. Indicator shows solution is basic.

18 18 Know the strong acids & bases!

19 19! Mg(s) ) + 2 HCl(aq) --> > H 2 (g) + MgCl 2 (aq) We really should write Mg(s) ) + 2 H + (aq) ) + 2 Cl - (aq) ---> H 2 (g) + Mg 2+ (aq) + 2 Cl - (aq) ions are SPECTATOR IONS The two Cl - ions are they do not participate. Could have used NO - 3.

20 ! 20 Mg(s) ) + 2 HCl(aq) --> > H 2 (g) + MgCl 2 (aq) Mg(s) ) + 2 H + (aq) ) + 2 Cl - (aq) ---> > H 2 (g) + Mg 2+ (aq) + 2 Cl - (aq) We leave the spectator ions out Mg(s) + 2 H + (aq) ---> > H 2 (g) + Mg 2+ (aq) to give the NET IONIC EQUATION

+ $ Sections 5.2 & 5.4-5.6 5.6 CD-ROM Ch. 5 21 We will look at!-. /! $!# Pb(NO 3 ) 2 (aq) + 2

21 + $ Sections 5.2 & CD-ROM Ch We will look at!-. /! $!# Pb(NO 3 ) 2 (aq) + 2 KI(aq) ----> > PbI 2 (s) + 2 KNO 3 (aq) AX + BY AY + BX The anions exchange places between cations.

22 $ 22 The driving force is the formation of an insoluble compound a precipitate. Pb(NO 3 ) 2 (aq) + 2 KI(aq) -----> 2 KNO 3 (aq) + PbI 2 (s) Net ionic equation Pb 2+ (aq) + 2 I - (aq) ---> > PbI 2 (s)

23 0,$ 23 The driving force is the formation of water. NaOH(aq) ) + HCl(aq) ---> NaCl(aq) ) + H 2 O(liq) Net ionic equation OH - (aq)) + H + (aq) ---> > H 2 O(liq) This applies to ALL reactions of STRONG acids and bases.

24 Acid-Base Reactions CCR, page

25 0,$ 25 A-B B reactions are sometimes called!#$"1# because the solution is neither acidic nor basic at the end. The other product of the A-B A B reaction is a "#, MX. HX + MOH ---> MX + H 2 O M n + comes from base & X n - comes from acid This is one way to make compounds!

26 /02 &$ 26 This is primarily the chemistry of metal carbonates. CO 2 and water ---> > H 2 CO 3 H 2 CO 3 (aq) + Ca > 2 H + (aq) ) + CaCO 3 (s) (limestone) Adding acid reverses this reaction. MCO 3 + acid ---> > CO 2 + salt

27 27 /02 & $ CaCO 3 (s) + H 2 SO 4 (aq) ---> 2 CaSO 4 (s) + H 2 CO 3 (aq) Carbonic acid is unstable and forms CO 2 & H 2 O H 2 CO 3 (aq) ---> > CO 2 + water (Antacid tablet has citric acid + NaHCO 3 )

28 28

29 * $ Sections

30 # & 30 In solution we need to define the - "5!# the component whose physical state is preserved when solution forms "#! the other solution component

31 31 * The amount of solute in a solution is given by its concentration. is given by its concentration Molarity (M) = moles solute liters of solution Concentration (M) = [ ]

32 L of water was used to make 1.0 L of solution. Notice the water left over. CCR, page 177

$,"!6 4788&* 9:. &+' (7 8 " * Step 1: Calculate moles of NiCl 2 6H 2 O 5.00 g 1 mol 237.7 g = 0.

33 $,"!6 4788&* 9:. &+' (7 8 " * Step 1: Calculate moles of NiCl 2 6H 2 O 5.00 g 1 mol g = mol Step 2: Calculate molarity mol L = M [NiCl 2 6 H 2 O ] = M

34 # +* 34 CuCl 2 (aq) --> Cu 2+ (aq) + 2 Cl - (aq) If [CuCl 2 ] = 0.30 M, then [Cu 2+ ] = 0.30 M [Cl - ] = 2 x 0.30 M

35 /"$#% 35 What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. ml of a M solution? Because Conc (M) = moles/volume = mol/v this means that ;95

36 /"$#% 36 What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. ml of a M solution? ;95 Step 1: Calculate moles of acid required. ( mol/l)(0.250 L) = mol Step 2: Calculate mass of acid required. ( mol )(90.00 g/mol) = 1.13 g

37 37 & Weigh out a solid solute and dissolve in a given quantity of solvent. Dilute a concentrated solution to give one that is less concentrated.

38 $,"!6%+4788 " *<8. ' = 38 Add water to the 3.0 M solution to lower its concentration to 0.50 M +

39 $,"!6%+4788 " *<8. ' = 39 H2 O But how much water do we add? 3.0 M NaOH 0.50 M NaOH Concentrated Dilute

40 $,"!6%+4788 " *<8. ' ? 40 How much water is added? The important point is that --->!"# $!"

41 $,"!6%+4788 " *<8.. ' = 41 Amount of NaOH in original solution = M V = (3.0 mol/l)(0.050 L) = 0.15 mol NaOH Amount of NaOH in final solution must also = 0.15 mol NaOH Volume of final solution = (0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L or 300 ml

42 $,"!6%+4788 " *<8. ' = 42 H2 O 3.0 M NaOH 0.50 M NaOH Concentrated Dilute Conclusion: add 250 ml of water to 50.0 ml of 3.0 M NaOH to make 300 ml of 0.50 M NaOH.

43 &) 43 A shortcut C V initial initial = C V final final

44 . ph: a way to express acidity -- the concentration of H + in solution. 44 Low ph: high [H + ] High ph: low [H + ] Acidic solution ph < 7 Neutral ph = 7 Basic solution ph > 7

45 #+. 45 ph = log (1/ [H + ]) = - log [H + ] In a neutral solution, [H + ] = [OH - ] = 1.00 x 10-7 M at 25 o C ph = - log [H + ] = -log (1.00 x 10-7 ) = - (-7) = 7 See CD Screen 5.17 for a tutorial

46 46 If the [H + ] of soda is 1.6 x 10-3 M, the ph is? Because ph = - log [H + ] then ph= - log (1.6 x 10-3 ) ph = - (-2.80) ph = 2.80

47 . 47 If the ph of Coke is 3.12, it is. Because ph = - log [H + ] then log [H + ] = - ph Take antilog and get [H + ] = 10 - ph [H + ] = = 7.6 x 10-4 M

"##.!#$% Section 5.10 48 Zinc reacts with acids to produce H 2 gas. Have 10.

48 "##.!#$% Section Zinc reacts with acids to produce H 2 gas. Have 10.0 g of Zn What volume of 2.50 M HCl is needed to convert the Zn completely?

49 /!!$""2$#.!#$% ""# 49 Mass zinc Mass HCl Moles zinc Stoichiometric factor Moles HCl Volume HCl

50 Zinc reacts with acids to to produce H 2 gas. If If you have 10.0 g of of Zn, what volume of of 2.50 M HCl is is needed to to convert the Zn Zn completely? 50 Step 1: Write the balanced equation Zn(s) ) + 2 HCl(aq) --> > ZnCl 2 (aq) + H 2 (g) Step 2: Calculate amount of Zn 10.0 g Zn 1.00 mol Zn g Zn = mol Zn Step 3: Use the stoichiometric factor

51 Zinc reacts with acids to to produce H 2 gas. If If you have 10.0 g of of Zn, what volume of of 2.50 M HCl is is needed to to convert the Zn Zn completely? 51 Step 3: Use the stoichiometric factor mol Zn 2 mol HCl 1 mol Zn = mol HCl Step 4: Calculate volume of HCl req d mol HCl 1.00 L 2.50 mol = L HCl

52 0,!$!# # 52 H 2 C 2 O 4 (aq) + 2 NaOH(aq) ---> acid base Na 2 C 2 O 4 (aq) + 2 H 2 O(liq) Carry out this reaction using a TITRATION. Oxalic acid, H 2 C 2 O 4

53 *& &' ' +) 53 CCR, page 186

54 # 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3. Indicator shows when exact stoichiometric reaction has occurred. 4. Net ionic equation H + + OH - --> > H 2 O 5. At equivalence point moles H + = moles OH - 54

55 LAB PROBLEM #1: Standardize a solution of NaOH i.e., accurately determine its concentration g of H 2 C 2 O 4 (oxalic acid) requires ml of NaOH for titration to an equivalence point. What is the concentra-tion tion of the NaOH?

56 1.065 g of of H 2 C 2 O 4 (oxalic acid) requires ml of of NaOH for titration to to an an equivalence point. What is is the concentration of of the NaOH? 56 Step 1: Calculate amount of H 2 C 2 O g 1 mol g = mol Step 2: Calculate amount of NaOH req d mol acid 2 mol NaOH 1 mol acid = mol NaOH

57 1.065 g of of H 2 C 2 O 4 (oxalic acid) requires ml of of NaOH for titration to to an an equivalence point. What is is the concentration of of the NaOH? 57 Step 1: Calculate amount of H 2 C 2 O 4 = mol acid Step 2: Calculate amount of NaOH req d = mol NaOH Step 3: Calculate concentration of NaOH mol NaOH L = M [NaOH]] = M

58 LAB PROBLEM #2: Use standardized NaOH to determine the amount of an acid in an unknown. 58 Apples contain malic acid, C 4 H 6 O 5. C 4 H 6 O 5 (aq) + 2 NaOH(aq) ---> Na 2 C 4 H 4 O 5 (aq) + 2 H 2 O(liq) g of apple requires ml of M NaOH for titration. What is weight % of malic acid?

59 76.80 g of apple requires ml of M NaOH for titration. What is is weight % of malic acid? 59 Step 1: Calculate amount of NaOH used. C V = (0.663 M)( L) = mol NaOH Step 2: Calculate amount of acid titrated mol NaOH = mol acid 1 mol acid 2 mol NaOH

60 76.80 g of apple requires ml of M NaOH for titration. What is is weight % of malic acid? 60 Step 1: Calculate amount of NaOH used. = mol NaOH Step 2: Calculate amount of acid titrated = mol acid Step 3: Calculate mass of acid titrated mol acid 134 g mol = 1.54 g

61 76.80 g of apple requires ml of M NaOH for titration. What is is weight % of malic acid? 61 Step 1: Calculate amount of NaOH used. = mol NaOH Step 2: Calculate amount of acid titrated = mol acid Step 3: Calculate mass of acid titrated. = 1.54 g acid Step 4: Calculate % malic acid g g 100% = 2.01%

62 62 A0$$ Thermite reaction Fe 2 O 3 (s) + 2 Al(s) ----> 2 Fe(s) ) + Al 2 O 3 (s) Section 5.7

63 EXCHANGE: Precipitation Reactions 63 EXCHANGE Gas-Forming Reactions REACTIONS EXCHANGE Acid-Base Reactions REDOX REACTIONS

64 64 $! -$!# Oxidation 2 H 2 (g) + O 2 (g) ---> > 2 H 2 O(liq) Mg(s) ) + 2 HCl(aq) --> > MgCl 2 (aq) + H 2 (g) All corrosion reactions are oxidations. 2 Al(s) ) + 3 Cu 2+ (aq) ---> > 2 Al 3+ (aq) + 3 Cu(s) Reduction Fe 2 O 3 (s) + 2 Al(s) --> > 2 Fe(s) ) + Al 2 O 3 (s)

65 65 $! - $!# Cu(s) ) + 2 Ag + (aq) ---> > Cu 2+ (aq) + 2 Ag(s) In all reactions if something has been oxidized then something has also been reduced

66 Why Study Redox Reactions 66 Batteries Corrosion Manufacturing metals Fuels

67 67 $! -$!# Redox reactions are characterized by ELECTRON TRANSFER between an electron donor and electron acceptor. Transfer leads to 1. increase in oxidation number of some element = OXIDATION 2. decrease in oxidation number of some element = REDUCTION

68 - #,!$ 68 The electric charge an element APPEARS to have when electrons are counted by some arbitrary rules: 1. Each atom in free element has ox. no. = 0. Zn O 2 I 2 S 8 2. In simple ions, ox. no. = charge on ion. -11 for Cl - +2 for Mg 2+

69 - #,!$ O has ox. no. = -2 (except in peroxides: in H 2 O 2, O = -1) 4. Ox. no. of H = +1 (except when H is associated with a metal as in NaH where it is -1) 5. Algebraic sum of oxidation numbers = 0 for a compound = overall charge for an ion

70 - #,!$ 70 NH 3 N = ClO - Cl = H 3 PO 4 P = MnO - 4 Mn = Cr 2 O 2-7 Cr = C 3 H 8 C = Oxidation number of F in HF?

71 $&B&$A $A $ 71 Corrosion of aluminum 2 Al(s) ) + 3 Cu 2+ (aq) --> > 2 Al 3+ (aq) + 3 Cu(s) Al(s) --> > Al 3+ (aq) + 3 e - Ox. no. of Al increases as e - are donated by the metal. Therefore, Al is OXIDIZED Al is the REDUCING AGENT in this balanced half- reaction.

72 $&B&$A $A $ 72 Corrosion of aluminum 2 Al(s) ) + 3 Cu 2+ (aq) --> > 2 Al 3+ (aq) + 3 Cu(s) Cu 2+ (aq) + 2 e - --> Cu(s) Ox. no. of Cu decreases as e - are accepted by the ion. Therefore, Cu is REDUCED Cu is the OXIDIZING AGENT in this balanced half- reaction.

73 $&B&$A $A $ Notice that the 2 half-reactions add up to give the overall reaction if we use 2 mol of Al and 3 mol of Cu 2+ 2 Al(s) --> > 2 Al 3+ (aq) + 6 e - 3 Cu 2+ (aq) + 6 e - --> > 3 Cu(s) Al(s) ) + 3 Cu 2+ (aq) ---> > 2 Al 3+ (aq) + 3 Cu(s) 2+. Final eqn.. is balanced for mass and charge. 73

74 AB& $&& # # )7 )7 C C 74 Metals (Cu) are reducing agents HNO 3 is an oxidizing agent Cu + HNO 3 --> Cu 2+ + NO 2 Metals (Na, K, Mg, Fe) are reducing agents 2 K + 2 H 2 O --> 2 KOH + H 2

75 $&B&$A $A $ # )7 C C 75 Reaction Type In terms of oxygen In terms of halogen In terms of electrons Oxidation gain gain loss Reduction loss loss gain

76 !A *$A $ 76 Metal + halogen 2 Al + 3 Br 2 ---> > Al 2 Br 6

77 !A *$A $ 77 Nonmetal (P) + Oxygen Metal (Mg) + Oxygen

78 !A *$A $ 78 Metal + acid Mg + HCl Mg = reducing agent H + = oxidizing agent Metal + acid Cu + HNO 3 Cu = reducing agent HNO 3 = oxidizing agent

Page III-4b-1 / Chapter Four Part II Lecture Notes. Chemistry 221 Professor Michael Russell MAR. Ba(NO3)2(aq)? soluble. BaCl2(aq)?

Page III-4b-1 / Chapter Four Part II Lecture Notes Solution Stoichiometry and Chemical Reactions Chapter 3 & Chapter 4, or Chapter 4 Part II Chemistry 221 Professor Michael Russell Terminology In solution