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Ch19.1 – Acids and Bases Acids - corrosive, taste sour, put electrolytes in soln, react with metals Ex1) Single displacement reaction: H 2 SO 4(aq) + Zn.

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Presentation on theme: "Ch19.1 – Acids and Bases Acids - corrosive, taste sour, put electrolytes in soln, react with metals Ex1) Single displacement reaction: H 2 SO 4(aq) + Zn."— Presentation transcript:

1 Ch19.1 – Acids and Bases Acids - corrosive, taste sour, put electrolytes in soln, react with metals Ex1) Single displacement reaction: H 2 SO 4(aq) + Zn (s) →

2 Ch19.1 – Acids and Bases Acids - corrosive, taste sour, put electrolytes in soln, react with metals Ex1) Single displacement reaction: H 2 SO 4(aq) + Zn (s) → ZnSO 4(aq) + H 2(g) - produce hydrogen ions when dissolved in water. Ex2) Decomposition: HCl (s) →

3 Ch19.1 – Acids and Bases Acids - corrosive, taste sour, put electrolytes in soln, react with metals Ex1) Single displacement reaction: H 2 SO 4(aq) + Zn (s) → ZnSO 4(aq) + H 2(g) - produce hydrogen ions when dissolved in water. Ex2) Decomposition: HCl (s) → H + (aq) + Cl - (aq) What actually happens is the H + bonds with water: HCl (s) + H(OH) →

4 Acids - corrosive, taste sour, put electrolytes in soln, react with metals Ex1) Single displacement reaction: H 2 SO 4(aq) + Zn (s) → ZnSO 4(aq) + H 2(g) - produce hydrogen ions when dissolved in water. Ex2) Decomposition: HCl (s) → H + (aq) + Cl - (aq) What actually happens is the H + bonds with water: HCl (s) + H(OH) → H 3 O + (aq) + Cl - (aq) - will cause some substances (called indicators) to change colors in presence of acid

5 Naming acids 1. Salt that ends in –ide, name acid by putting hydro in front, use the root of the anion, and end it with –ic Ex1) HCl (s) :

6 Naming acids 1. Salt that ends in ide, name acid by putting hydro in front, use the root of the anion, and end it with ic Ex1) HCl (s) : hydrogen chloride  hydrochloric acid 2. Polyatomic ions in salts, if ion ends in ate, then give it the ending ic Ex2) H 2 SO 4(s) :

7 Naming acids 1. Salt that ends in ide, name acid by putting hydro in front, use the root of the anion, and end it with ic Ex1) HCl (s) : hydrogen chloride  hydrochloric acid 2. Polyatomic ions in salts, if ion ends in ate, then give it the ending ic Ex2) H 2 SO 4(s) : hydrogen sulfate  sulfuric acid 3. Polyatomic ions in salts, if ion ends in ite, then give it the ending ous Ex3) H 2 SO 3(s) :

8 Naming acids 1. Salt that ends in ide, name acid by putting hydro in front, use the root of the anion, and end it with ic Ex1) HCl (s) : hydrogen chloride  hydrochloric acid 2. Polyatomic ions in salts, if ion ends in ate, then give it the ending ic Ex2) H 2 SO 4(s) : hydrogen sulfate  sulfuric acid 3. Polyatomic ions in salts, if ion ends in ite, then give it the ending ous Ex3) H 2 SO 3(s) : hydrogen sulfite  sulfurous acid 4. Polyatomic one above most common, add per in front Ex4) H 2 SO 5(s) :

9 Naming acids 1. Salt that ends in ide, name acid by putting hydro in front, use the root of the anion, and end it with ic Ex1) HCl (s) : hydrogen chloride  hydrochloric acid 2. Polyatomic ions in salts, if ion ends in ate, then give it the ending ic Ex2) H 2 SO 4(s) : hydrogen sulfate  sulfuric acid 3. Polyatomic ions in salts, if ion ends in ite, then give it the ending ous Ex3) H 2 SO 3(s) : hydrogen sulfite  sulfurous acid 4. Polyatomic one above most common, add per in front Ex4) H 2 SO 5(s) : hydrogen persulfate  persulfuric acid 5. Polyatomic two below most common, add hypo in front Ex5) H 2 SO 2(s) :

10 Naming acids 1. Salt that ends in ide, name acid by putting hydro in front, use the root of the anion, and end it with ic Ex1) HCl (s) : hydrogen chloride  hydrochloric acid 2. Polyatomic ions in salts, if ion ends in ate, then give it the ending ic Ex2) H 2 SO 4(s) : hydrogen sulfate  sulfuric acid 3. Polyatomic ions in salts, if ion ends in ite, then give it the ending ous Ex3) H 2 SO 3(s) : hydrogen sulfite  sulfurous acid 4. Polyatomic one above most common, add per in front Ex4) H 2 SO 5(s) : hydrogen persulfate  persulfuric acid 5. Polyatomic two below most common, add hypo in front Ex5) H 2 SO 2(s) : hydrogen hyposulfite  hyposulfurous acid

11 Exs:HF HNO 3 HNO 2 H 2 CO 3 H 2 S HCLO 3 HCLO 4 HCLO 2 HCLO

12 Bases - corrosive, taste bitter (like aspirin), produce electrolytes, - put hydroxide ions in solution Ex: Decomposition: NaOH (S) 

13 Bases - corrosive, taste bitter (like aspirin), produce electrolytes, - put hydroxide ions in solution Ex: Decomposition: NaOH (S)  Na + (aq) + OH - (aq) - react with the acids to form water and a salt Ex: Double displacement (neutralization): HCl (aq) + NaOH (aq)  - cause indicators to change color

14 Bases - corrosive, taste bitter (like aspirin), produce electrolytes, - put hydroxide ions in solution Ex: Decomposition: NaOH (S)  Na + (aq) + OH - (aq) - react with the acids to form water and a salt Ex: Double displacement (neutralization): HCl (aq) + NaOH (aq)  NaCl (aq) + H(OH) (l) - cause indicators to change color Naming Bases Ex: Ca(OH) 2 :

15 Bases - corrosive, taste bitter (like aspirin), produce electrolytes, - put hydroxide ions in solution Ex: Decomposition: NaOH (S)  Na + (aq) + OH - (aq) - react with the acids to form water and a salt Ex: Double displacement (neutralization): HCl (aq) + NaOH (aq)  NaCl (aq) + H(OH) (l) - cause indicators to change color Naming Bases Ex: Ca(OH) 2 : calcium hydroxide Ch19 HW#1 1 – 3

16 Lab19.1 – Characteristics of Acids - due in 2 days - Ch19 HW#1 due at beginning of period

17 Ch19 HW# ID Acid, Base, both a) bitter tasteb) electrolytec) indicator color change d) sour tastee) reacts with metal from H 2(g) 2. Name a) HF b) H 2 PO 4 c) KOH d) H 2 SO 4 e) HNO 2 f) Mg(OH) 2

18 Ch19 HW# ID Acid, Base, both a) bitter tasteb) electrolytec) indicator color change base both both d) sour tastee) reacts with metal from H 2(g) acid 2. Name a) HF (Must get 3b) H 2 PO 4 out of 6 forc) KOH candy)d) H 2 SO 4 e) HNO 2 f) Mg(OH) 2

19 Ch19 HW# ID Acid, Base, both a) bitter tasteb) electrolytec) indicator color change base both both d) sour tastee) reacts with metal from H 2(g) acid 2. Name a) HF (hydrogen fluoride)  b) H 2 PO 4 (hydrogen phosphate)  c) KOH  d) H 2 SO 4 (hydrogen sulfate)  e) HNO 2 (hydrogen nitrite)  f) Mg(OH) 2 

20 Ch19 HW# ID Acid, Base, both a) bitter tasteb) electrolytec) indicator color change base both both d) sour tastee) reacts with metal from H 2(g) acid 2. Name a) HF (hydrogen fluoride)  hydrofluoric acid b) H 2 PO 4 (hydrogen phosphate)  c) KOH  d) H 2 SO 4 (hydrogen sulfate)  e) HNO 2 (hydrogen nitrite)  f) Mg(OH) 2 

21 Ch19 HW# ID Acid, Base, both a) bitter tasteb) electrolytec) indicator color change base both both d) sour tastee) reacts with metal from H 2(g) acid 2. Name a) HF (hydrogen fluoride)  hydrofluoric acid b) H 2 PO 4 (hydrogen phosphate)  phosphoric acid c) KOH  d) H 2 SO 4 (hydrogen sulfate)  e) HNO 2 (hydrogen nitrite)  f) Mg(OH) 2 

22 Ch19 HW# ID Acid, Base, both a) bitter tasteb) electrolytec) indicator color change base both both d) sour tastee) reacts with metal from H 2(g) acid 2. Name a) HF (hydrogen fluoride)  hydrofluoric acid b) H 2 PO 4 (hydrogen phosphate)  phosphoric acid c) KOH  potassium hydroxide d) H 2 SO 4 (hydrogen sulfate)  e) HNO 2 (hydrogen nitrite)  f) Mg(OH) 2 

23 Ch19 HW# ID Acid, Base, both a) bitter tasteb) electrolytec) indicator color change base both both d) sour tastee) reacts with metal from H 2(g) acid 2. Name a) HF (hydrogen fluoride)  hydrofluoric acid b) H 2 PO 4 (hydrogen phosphate)  phosphoric acid c) KOH  potassium hydroxide d) H 2 SO 4 (hydrogen sulfate)  sulfuric acid e) HNO 2 (hydrogen nitrite)  f) Mg(OH) 2 

24 Ch19 HW# ID Acid, Base, both a) bitter tasteb) electrolytec) indicator color change base both both d) sour tastee) reacts with metal from H 2(g) acid 2. Name a) HF (hydrogen fluoride)  hydrofluoric acid b) H 2 PO 4 (hydrogen phosphate)  phosphoric acid c) KOH  potassium hydroxide d) H 2 SO 4 (hydrogen sulfate)  sulfuric acid e) HNO 2 (hydrogen nitrite)  nitrous acid f) Mg(OH) 2 

25 Ch19 HW# ID Acid, Base, both a) bitter tasteb) electrolytec) indicator color change base both both d) sour tastee) reacts with metal from H 2(g) acid 2. Name a) HF (hydrogen fluoride)  hydrofluoric acid b) H 2 PO 4 (hydrogen phosphate)  phosphoric acid c) KOH  potassium hydroxide d) H 2 SO 4 (hydrogen sulfate)  sulfuric acid e) HNO 2 (hydrogen nitrite)  nitrous acid f) Mg(OH) 2  magnesium hydroxide

26 3.Formulas a) chromic acid b) iron ( II ) hydroxide c) hydroiodic acid d) lithium hydroxide e) chlorous acid f) perchloric acid

27 3.Formulas a) chromic acid chromate: CrO 4 -2 H 2 CrO 4 b) iron ( II ) hydroxide Fe +2 (OH) -1  Fe(OH) 2 c) hydroiodic acid H + I -1  HI d) lithium hydroxide e) chlorous acid f) perchloric acid

28 3.Formulas a) chromic acid chromate: CrO 4 -2 H 2 CrO 4 b) iron ( II ) hydroxide Fe +2 (OH) -1  Fe(OH) 2 c) hydroiodic acid H + I -1  HI d) lithium hydroxide Li + OH -  LiOH e) chlorous acid chlorite: ClO 2 -1 HClO 2 f) perchloric acid perchlorate: ClO 4 -1 HClO 4

29 Ch19.2 – K w (The Equilibriium Constant for Water) - Water molecules collide, sometimes hydrogens are knocked off:

30 Ch19.2 – K w (The Equilibriium Constant for Water) - Water molecules collide, sometimes hydrogens are knocked off: This is the self-ionization of water, doesn’t happen very much. For 25˚ with a neutral pH: [H 3 O] + = 1x10 -7 M[OH] - = 1x10 -7 M

31 Ch19.2 – K w (The Equilibriium Constant for Water) - Water molecules collide, sometimes hydrogens are knocked off: This is the self-ionization of water, doesn’t happen very much. For 25˚ with a neutral pH: [H 3 O] + = 1x10 -7 M[OH] - = 1x10 -7 M Their combined concentration, when multiplied = 1x M [H] +. [OH] - = 1x M This stays constant! Which means if one gets bigger, the other gets smaller to keep constant. Another name for [H 3 O] + is just [H + ], which is what we recognize as an acid.

32 Their combined concentration, when multiplied = 1x M [H] +. [OH] - = 1x M This stays constant! Which means if one gets bigger, the other gets smaller to keep constant. Neutral water has both acid & base in it, but they’re in small amounts and cancel each other. When [H + ] > 1 x then [OH - ] < 1 x Acidic When [H + ] 1 x Basic

33 Ex1) A solution has [H + ] = 2.7x10 -3 M. Is it acidic, basic, or neutral? What is the [OH - ] concentration?

34 Ex1) A solution has [H + ] = 2.7x10 -3 M. Is it acidic, basic, or neutral? What is the [OH - ] concentration? Acidic [H] +. [OH] - = 1x M [2.7x10 -3 ]. [OH] - = 1x M [OH] - = 3.7x M Ex2) If [H + ] = 1.1x M, is the solution acid, basic, neutral? Find [OH - ]

35 Ex1) A solution has [H + ] = 2.7x10 -3 M. Is it acidic, basic, or neutral? What is the [OH - ] concentration? Acidic [H] +. [OH] - = 1x M [2.7x10 -3 ]. [OH] - = 1x M [OH] - = 3.7x M Ex2) If [H + ] = 1.1x M, is the solution acid, basic, neutral? Find [OH - ] Basic [H + ]. [OH - ] = 1x [1.1x ] [OH - ] = 1x [OH - ] = 9.1x10 -5 M Ex3) If [OH - ] = 4.4x10 -8 M, acidic, basic, neutral? find [H + ]

36 Ex1) A solution has [H + ] = 2.7x10 -3 M. Is it acidic, basic, or neutral? What is the [OH - ] concentration? Acidic [H] +. [OH] - = 1x M [2.7x10 -3 ]. [OH] - = 1x M [OH] - = 3.7x M Ex2) If [H + ] = 1.1x M, is the solution acid, basic, neutral? Find [OH - ] Basic [H + ]. [OH - ] = 1x [1.1x ] [OH - ] = 1x [OH - ] = 9.1x10 -5 M Ex3) If [OH - ] = 4.4x10 -8 M, acidic, basic, neutral? find [H + ] [H + ]. [OH - ] = 1x [H + ][4.4x10 -8 ] = 1x [H + ] = 2.3x10 -7 (slightly acidic) Ch19 HW#2 4 – 7

37 4) Concentration of [H + ] & [OH - ] in: a) A basic solution: b) An acidic soln: c) A neutral soln: 5) Classify: a) [H + ] = 6x M b) [OH - ] = 3x10 -2 M c) [H + ] = 2x10 -7 M d) [OH - ] = 1x10 -7 M

38 Ch19 HW#2 4 – 7 4) Concentration of [H + ] & [OH - ] in: a) A basic solution: [OH - ] > 1x10 -7 M, [H + ] < 1x10 -7 M b) An acidic soln: [H + ] > 1x10 -7 M, [OH - ] < 1x10 -7 M c) A neutral soln: [H + ] = 1x10 -7 M, [OH - ] = 1x10 -7 M 5) Classify: a) [H + ] = 6x M b) [OH - ] = 3x10 -2 M c) [H + ] = 2x10 -7 M d) [OH - ] = 1x10 -7 M

39 Ch19 HW#2 4 – 7 4) Concentration of [H + ] & [OH - ] in: a) A basic solution: [OH - ] > 1x10 -7 M, [H + ] < 1x10 -7 M b) An acidic soln: [H + ] > 1x10 -7 M, [OH - ] < 1x10 -7 M c) A neutral soln: [H + ] = 1x10 -7 M, [OH - ] = 1x10 -7 M 5) Classify: a) [H + ] = 6x MBasic b) [OH - ] = 3x10 -2 MBasic c) [H + ] = 2x10 -7 MSlightly acidic d) [OH - ] = 1x10 -7 MNeutral

40 6) [OH - ] = 1x10 -3 M. Find [H + ] 7) [H + ] = 3x10 -6 M. Find [OH - ]

41 6) [OH - ] = 1x10 -3 M. Find [H + ] [H + ]. [OH - ] = 1x [H + ][1x10 -3 ] = 1x [H + ] = 1x (Basic) 7) [H + ] = 3x10 -6 M. Find [OH - ]

42 6) [OH - ] = 1x10 -3 M. Find [H + ] [H + ]. [OH - ] = 1x [H + ][1x10 -3 ] = 1x [H + ] = 1x (Basic) 7) [H + ] = 3x10 -6 M. Find [OH - ] [H] +. [OH] - = 1x M [3x10 -6 ]. [OH] - = 1x M [OH] - = 3x10 -9 M (Acidic)

43 Ch19.3 – It Time for Log! (pH Calculations) # /10 (.1) 1/100 (.01) 1/1000 (.001) 10 ? Log Value

44 Ch19.3 – It Time for Log! (pH Calculations) # /10 (.1) 1/100 (.01) 1/1000 (.001) 10 ? Log Value Exs: Log Antilog

45 Ch19.3 – It Time for Log! (pH Calculations) # /10 (.1) 1/100 (.01) 1/1000 (.001) 10 ? Log Value pH Info pH = -log[H 3 O + ]pH < 7 (acidic) pOH = -log[OH - ]pH = 7 (neutral) pH + pOH = 14pH > 7 (basic) Exs: Log Antilog = = =0.05

46 pH Info Ex 1) The hydrogen-ion concentration of a solution is 1x mol/L. What is the pH of the solution Ex 2) Given [H + ] = 9.27x M, find pH and pOH. Ex3) The pH of a solution is 6.0. Find [H + ] for this solution. pH = -log[H 3 O + ]pH < 7 (acidic) pOH = -log[OH - ]pH = 7 (neutral) pH + pOH = 14pH > 7 (basic)

47 pH Info Ex 1) The hydrogen-ion concentration of a solution is 1x mol/L. What is the pH of the solution pH = -log[H 3 O + ] pH = -log[1x M] pH = 10 (Basic) Ex 2) Given [H + ] = 9.27x M, find pH and pOH. pH = -log[H 3 O + ]pH + pOH = 14 pH = -log[9.27x M] pOH = 14 pH = (Basic)pOH = This is power This is sig digs of tenfor logs Ex3) The pH of a solution is 6.0. Find [H + ] for this solution. pH = -log[H 3 O + ]pH < 7 (acidic) pOH = -log[OH - ]pH = 7 (neutral) pH + pOH = 14pH > 7 (basic)

48 Ex3) The pH of a solution is 6.0. Find [H + ] for this solution. Ex4) Given [H 3 O + ] = 4.28x10 -7 M, find pH, pOH, and [OH - ].

49 Ex3) The pH of a solution is 6.0. Find [H + ] for this solution. pH = -log[H 3 O + ] 6.0 = -log[H 3 O + ] [H 3 O + ] = 1x10 -6 M Ex4) Given [H 3 O + ] = 4.28x10 -7 M, find pH, pOH, and [OH - ]. pH = -log[H + ] pH = -log[4.28 x10 -7 ] pH = 6.37 pH + pOH = pOH = 14 pOH = 7.63 [H + ][OH - ] = 1x [4.28 x10 -7 ][OH - ] = 1x [OH - ] = 2.34 x10 -8 M Ch19 HW#3 8 – 11 Common SubstancespH 1M HCl M HCl1.0 Stomach2.0 Lemon Juice3.0 Coffee, soda4-5 Milk, Urine5-7 Pure water7 Sea water, Blood8 Ammonia NaOH13 1.0M NaOH14

50 Ch19 HW#3 8 – Determine pH: a) [H + ] = 1x10 -6 Mb) [H + ] = m pH = -log(1x10 -6 ) pH = -log(.0001) pH = pH = c) [OH - ] = 1x10 -2 md) [OH - ] = 1x pOH = -log(1x10 -2 ) pOH = -log(1x ) pOH = pOH = pH = pH = pH = -log[H + ]pOH = -log[OH - ] [H + ] = antilog(-pH)[OH - ] = antilog(-pOH) pH + pOH = 14[H + ]. [OH - ] = 1x10 -14

51 Ch19 HW#3 8 – Determine pH: a) [H + ] = 1x10 -6 Mb) [H + ] = m pH = -log(1x10 -6 ) pH = -log(.0001) pH = 6 pH = 4 c) [OH - ] = 1x10 -2 md) [OH - ] = 1x pOH = -log(1x10 -2 ) pOH = -log(1x ) pH = -log[H + ]pOH = -log[OH - ] [H + ] = antilog(-pH)[OH - ] = antilog(-pOH) pH + pOH = 14[H + ]. [OH - ] = 1x10 -14

52 Ch19 HW#3 8 – Determine pH: a) [H + ] = 1x10 -6 Mb) [H + ] = m pH = -log(1x10 -6 ) pH = -log(.0001) pH = 6 pH = 4 c) [OH - ] = 1x10 -2 md) [OH - ] = 1x pOH = -log(1x10 -2 ) pOH = -log(1x ) pOH = 2 pOH = 11 pH = 4 pH = 3 pH = -log[H + ]pOH = -log[OH - ] [H + ] = antilog(-pH)[OH - ] = antilog(-pOH) pH + pOH = 14[H + ]. [OH - ] = 1x10 -14

53 9. [H + ] concentrations given pH: a) pH = 4.0b) pH = 11.0 [H + ] = antilog (-4.0) [H + ] = antilog (-11.0) c) pH = 8.0 [H + ] = antilog (-8.0) pH = -log[H + ]pOH = -log[OH - ] [H + ] = antilog(-pH)[OH - ] = antilog(-pOH) pH + pOH = 14[H + ]. [OH - ] = 1x10 -14

54 9. [H + ] concentrations given pH: a) pH = 4.0b) pH = 11.0 [H + ] = antilog (-4.0) [H + ] = antilog (-11.0) [H + ] = 1x10 -4 M [H + ] = 1x M c) pH = 8.0 [H + ] = antilog (-8.0) [H + ] = 1x10 -8 M pH = -log[H + ]pOH = -log[OH - ] [H + ] = antilog(-pH)[OH - ] = antilog(-pOH) pH + pOH = 14[H + ]. [OH - ] = 1x10 -14

55 10. [OH - ] concentrations given pH: a) pH = 6.0b) pH = 9.0 pOH = 8.0 pOH = 5.0 [OH - ] = antilog (-8.0) [OH - ] = antilog (-5.0) c) pH = 12.0 pOH = 2.0 [OH - ] = antilog (-2.0)

56 10. [OH - ] concentrations given pH: a) pH = 6.0b) pH = 9.0 pOH = 8.0 pOH = 5.0 [OH - ] = antilog (-8.0) [OH - ] = antilog (-5.0) [OH - ] = 1x10 -8 M [OH - ] = 1x10 -5 M c) pH = 12.0 pOH = 2.0 [OH - ] = antilog (-2.0) [OH - ] = 1x10 -2 M

57 11. Find pH : a) [H + ] = 1x10 -4 mb) [H + ] = pH = -log (1x10 -4 ) pH = -log (0.001) c) [OH - ] = 0.01d) [OH - ] = 1x10 -6 m pOH = -log (0.01) pOH = -log (1x10 -6 )

58 11. Find pH : a) [H + ] = 1x10 -4 mb) [H + ] = pH = -log (1x10 -4 ) pH = -log (0.001) pH = 4 pH = 3 c) [OH - ] = 0.01d) [OH - ] = 1x10 -6 m pOH = -log (0.01) pOH = -log (1x10 -6 ) pOH = 2 pOH = 6 pH = 12 pH = 8

59 Ch19.4 – Acid Theories Arrhenius Theory of Acids and Bases (1887) Acid-produces H + in a water solution (begins with an “H”) Base- produces OH - in a water solution (ends with an “OH”) Monoprotic Acid - contains 1 ionizable hydrogen Diprotic Acid - contains 2 ionizable hydrogens Triprotic Acid - contains 3 ionizable hydrogens Ex) ID each. If acid, which type? a) HNO 3 b) NaOH c) HC 2 H 3 O 2

60 Ch19.4 – Acid Theories Arrhenius Theory of Acids and Bases (1887) Acid-produces H + in a water solution (begins with an “H”) Base- produces OH - in a water solution (ends with an “OH”) Monoprotic Acid - contains 1 ionizable hydrogen Diprotic Acid - contains 2 ionizable hydrogens Triprotic Acid - contains 3 ionizable hydrogens Ex) ID each. If acid, which type? a) HNO 3 Nitric Acid1 ionizable H Monoprotic b) NaOHSodium hydroxide (Base) c) HC 2 H 3 O 2 Acetic Acid1 H! Monoprotic

61 Metals react with water to produce basic solutions. (Kick out a hydrogen) Ex) Na + HOH → Ex) Al + HOH → Nonmetals react with water to produce acidic solutions (slam together) Ex) SO 3 + HOH → Ex) CO 2 + HOH → Ex) SO 2 + HOH → Ch19 HW# 4 12 – 15

62 12. Calculate pH a) [H + ] = 5.0x10 -6 b) [H + ] = 8.3x pH = -log(5.0x10 -6 ) pH = -log(8.3x ) c) [OH - ] = 2x10 -5 d) [OH - ] = 4.5x pOH = -log(2x10 -5 ) pOH = -log(4.5x )

63 Ch19 HW#4 12 – Calculate pH a) [H + ] = 5.0x10 -6 b) [H + ] = 8.3x pH = -log(5.0x10 -6 ) pH = -log(8.3x ) pH = 5.3 pH = 9.1 c) [OH - ] = 2x10 -5 d) [OH - ] = 4.5x pOH = -log(2x10 -5 ) pOH = -log(4.5x )

64 Ch19 HW#4 12 – Calculate pH a) [H + ] = 5.0x10 -6 b) [H + ] = 8.3x pH = -log(5.0x10 -6 ) pH = -log(8.3x ) pH = 5.3 pH = 9.1 c) [OH - ] = 2x10 -5 d) [OH - ] = 4.5x pOH = -log(2x10 -5 ) pOH = -log(4.5x ) pOH = 4.7 pOH = 10.3 pH = 9.3 pH = 3.7

65 13. Find [H + ] a) pH = 5.0b) pH = = -log[H + ] 5.8 = -log[H + ] [H + ] = antilog (-5) [H + ] = antilog (-5.8) c) pH = 12.20d) pH = = -log[H + ] 2.64 = -log[H + ]

66 13. Find [H + ] a) pH = 5.0b) pH = = -log[H + ] 5.8 = -log[H + ] [H + ] = antilog (-5) [H + ] = antilog (-5.8) [H + ] = 1x10 -5 M [H + ] = 1.58x10 -6 M c) pH = 12.20d) pH = = -log[H + ] 2.64 = -log[H + ]

67 13. Find [H + ] a) pH = 5.0b) pH = = -log[H + ] 5.8 = -log[H + ] [H + ] = antilog (-5) [H + ] = antilog (-5.8) [H + ] = 1x10 -5 m [H + ] = 1.58x10 -6 m c) pH = 12.20d) pH = = -log[H + ] 2.64 = -log[H + ] [H + ] = 6.31x M [H + ] = M 14) ID Acid a) H 2 CO 3 b) H 3 PO 4 c) HCl

68 13. Find [H + ] a) pH = 5.0b) pH = = -log[H + ] 5.8 = -log[H + ] [H + ] = antilog (-5) [H + ] = antilog (-5.8) [H + ] = 1x10 -5 m [H + ] = 1.58x10 -6 m c) pH = 12.20d) pH = = -log[H + ] 2.64 = -log[H + ] [H + ] = 6.31x M [H + ] = M 14) ID Acid a) H 2 CO 3 b) H 3 PO 4 c) HCl Carbonic acidPhosphoric AcidHydrochloric Acid Diprotic Triprotic Monoprotic

69 15. Write equation a) Potassium metal reacts w/ water… K (s) + H(OH)  b) Calcium metal reacts with water… Ca (s) + H(OH)  c) Nitrogen dioxide dissolve in water… NO 2(g) + H(OH)  d) Nitrogen trioxide reacts with water… NO 3(g) + H(OH) 

70 15. Write equation a) Potassium metal reacts w/ water… ___ K (s) + ___H(OH)  ___K(OH) + ___H 2(g) b) Calcium metal reacts with water… Ca (s) + H(OH)  c) Nitrogen dioxide dissolve in water… NO 2(g) + H(OH)  d) Nitrogen trioxide reacts with water… NO 3(g) + H(OH) 

71 15. Write equation a) Potassium metal reacts w/ water… ___ K (s) + ___H(OH)  ___K(OH) + ___H 2(g) b) Calcium metal reacts with water… ___Ca (s) + ___H(OH)  ___ Ca(OH) 2 + ___H 2(g) c) Nitrogen dioxide dissolve in water…(slam together) NO 2(g) + H(OH)  d) Nitrogen trioxide reacts with water… NO 3(g) + H(OH) 

72 15. Write equation a) Potassium metal reacts w/ water… ___ K (s) + ___H(OH)  ___K(OH) + ___H 2(g) b) Calcium metal reacts with water… ___Ca (s) + ___H(OH)  ___ Ca(OH) 2 + ___H 2(g) c) Nitrogen dioxide dissolve in water…(slam together) ___NO 2(g) + ___H(OH)  ___ H 2 NO 3 (nitric acid) d) Nitrogen trioxide reacts with water… NO 3(g) + H(OH) 

73 15. Write equation a) Potassium metal reacts w/ water… ___ K (s) + ___H(OH)  ___K(OH) + ___H 2(g) b) Calcium metal reacts with water… ___Ca (s) + ___H(OH)  ___ Ca(OH) 2 + ___H 2(g) c) Nitrogen dioxide dissolve in water…(slam together) ___NO 2(g) + ___H(OH)  ___ H 2 NO 3 (nitric acid) (acid rain) d) Nitrogen trioxide reacts with water… ___NO 3(g) + ___H(OH)  ___ H 2 NO 4 (pernitric acid)

74 Ch19.4 Bronsted-Lowry Theory of Acids and Bases Acid - hydrogen ion donor (proton donor) Ex: HClH x Cl in water → H + + [ x Cl ] – Base - hydrogen ion acceptor (proton acceptor) Ex: OH – + H + → HOH NH 3 + H + → NH 4 +

75 Conjugates - in reversible reactions, there are acid-base pairs on both sides, based on B-L theory. Ex: HBr + KOH → HOH KBr Ex: NH 3 + HOH → NH OH - Ex: HCl + HOH → H 3 O + + Cl - Water is amphoteric - a substance that can act as an acid or base depending on who its with. ← ← ←

76 Measuring pH - use indicators They are made of weak acids or bases that dissociate at known pH ranges Ex: Methyl Red pH range: If placed in an acid with pH < 4.8 it has red color If placed in a weak acid with pH between , it has an orange color. If placed in really weak acid (pH between 6 and 7) or in neutral solution pH=7 or in a basic solution pH>7 it is yellow. Ex: phenolphthalein pH range pH above 10.0 (strong base) → pink pH between (weak base) → pale pH below 8.2 (acid, neutral, or really weak base) → clear

77 HW#18) Indigo - Carmine indicator is a weak acid. It basically dissociates like this: HIn (aq) H + (aq) + In – (aq) (Yellow)(Blue) Use LeChatlier’s Principle to predict color change when adding: a) hydrochloric acid HIn (aq) H + (aq) + In – (aq) b) sodium hydoxide HIn (aq) H + (aq) + In – (aq) Ch19 HW#5 16 – 19

78 Lab19.2 – Determination of pH - due in 2 days - Ch19 HW#5 beginning of period

79 Ch19 HW#5 16 – 19 16) B-L Theory Acid - Base - Advantage over Arrhenius – 17) Ionize HNO 3 & Na 2 CO 3 in water & label a) HNO 3 + H(OH) H 3 O + + NO 3 - b) Na 2 CO 3 + H(OH) H 3 O + + HCO ) A pH meter is more precise than indicator solns. Why?

80 Ch19 HW#5 16 – 19 16) B-L Theory Acid - proton donor Base - proton acceptor Advantage over Arrhenius - more comprehensive Ex: NH 3 is a base 17) Ionize HNO 3 & Na 2 CO 3 in water & label a) HNO 3 + H(OH) H 3 O + + NO 3 - b) Na 2 CO 3 + H(OH) H 3 O + + HCO ) A pH meter is more precise than indicator solns. Why?

81 Ch19 HW#5 16 – 19 16) B-L Theory Acid - proton donor Base - proton acceptor Advantage over Arrhenius - more comprehensive Ex: NH 3 is a base 17) Ionize HNO 3 & Na 2 CO 3 in water & label a) HNO 3 + H(OH) H 3 O + + NO 3 - Acid Base conj acid conj base b) Na 2 CO 3 + H(OH) H 3 O + + HCO ) A pH meter is more precise than indicator solns. Why?

82 Ch19 HW#5 16 – 19 16) B-L Theory Acid - proton donor Base - proton acceptor Advantage over Arrhenius - more comprehensive Ex: NH 3 is a base 17) Ionize HNO 3 & Na 2 CO 3 in water & label a) HNO 3 + H(OH) H 3 O + + NO 3 - Acid Base conj acid conj base b) Na 2 CO 3 + H(OH) H 3 O + + HCO 3 - Base Acid conj conj base acid 19) A pH meter is more precise than indicator solns. Why?

83 Ch19 HW#5 16 – 19 16) B-L Theory Acid - proton donor Base - proton acceptor Advantage over Arrhenius - more comprehensive Ex: NH 3 is a base 17) Ionize HNO 3 & Na 2 CO 3 in water & label a) HNO 3 + H(OH) H 3 O + + NO 3 - Acid Base conj acid conj base b) Na 2 CO 3 + H(OH) H 3 O + + HCO 3 - Base Acid conj conj base acid 19) A pH meter is more precise than indicator solns. Why? Tell exact pHOnly work over a certain range & then guess based on color.

84 Ch20.1 – Neutralizations HCl (aq) + NaOH (aq)

85 Ch20.1 – Neutralizations HCl (aq) + NaOH (aq) H(OH) + NaCl Acids & Bases neutralizes each other. Produce a salt & water All are double replacement reactions. Common salts listed in Ch20 (on test.) Titrations – measuring precise amounts to neutralize acids & bases.

86 Ex1) How many moles of sulfuric acid would be required to neutralize 0.50 mol of sodium hydroxide? H 2 SO 4 + NaOH 

87 Titrations – measuring precise amounts to neutralize acids & bases. Ex1) How many moles of sulfuric acid would be required to neutralize 0.50 mol of sodium hydroxide? H 2 SO NaOH  Na 2 SO 4 + 2H(OH)

88 Ex2) A 25 mL solution of H 2 SO 4 is neutralized by 18 mL of 1.0 M NaOH using phenolphthalein as an indicator. What is the concentration of the H 2 SO 4 solution? H 2 SO NaOH  Na 2 SO 4 + 2H(OH)

89 Ex2) A 25 mL solution of H 2 SO 4 is neutralized by 18 mL of 1.0 M NaOH using phenolphthalein as an indicator. What is the concentration of the H 2 SO 4 solution? H 2 SO NaOH  Na 2 SO 4 + 2H(OH)

90 HW#5) How many mLs of 0.45 M hydrochloric acid must be added to 25.0 mL of 1.00 M potassium hydroxide to make a neutral solution? Lab20.1 – Titration Volume of NaOH:_X__ Molarity of NaOH:_0.2_ Volume of HCl:_Y__ Molarity of HCl:_?__ Ch20 HW#1 1 – 6

91 1) Word eqn for neutralization rxn: 2) ID products and balance: a) HNO 3 + KOH b) HCl + Ca(OH) 2 → c) H 2 SO 4 + NaOH → 3) How many moles in sodium hydroxide to neutralize 0.2 mol phosphoric acid? NaOH + H 3 PO 4 → Na 3 PO 4 + H(OH)

92 Ch20 HW#1 1 – 6 1) Word eqn for neutralization rxn: Acid + Base Salt + Water 2) ID products and balance: a) HNO 3 + KOH b) HCl + Ca(OH) 2 → c) H 2 SO 4 + NaOH → 3) How many moles in sodium hydroxide to neutralize 0.2 mol phosphoric acid? NaOH + H 3 PO 4 → Na 3 PO 4 + H(OH)

93 Ch20 HW#1 1 – 6 1) Word eqn for neutralization rxn: Acid + Base Salt + Water 2) ID products and balance: a) HNO 3 + KOH H(OH) + KNO 3 b) 2HCl +1 Ca(OH) 2 → 2H(OH) + 1CaCl 2 c) 1H 2 SO 4 + NaOH → 2H(OH) + 1Na 2 SO 4 3) How many moles in sodium hydroxide to neutralize 0.2 mol phosphoric acid? NaOH + H 3 PO 4 → Na 3 PO 4 + H(OH)

94 Ch20 HW#1 1 – 6 1) Word eqn for neutralization rxn: Acid + Base Salt + Water 2) ID products and balance: a) HNO 3 + KOH H(OH) + KNO 3 b) 2HCl +1 Ca(OH) 2 → 2H(OH) + 1CaCl 2 c) 1H 2 SO 4 + NaOH → 2H(OH) + 1Na 2 SO 4 3) How many moles in sodium hydroxide to neutralize 0.2 mol phosphoric acid? 3 NaOH + 1 H 3 PO 4 → 1 Na 3 PO H(OH)

95 4) How many grams of potassium hydroxide to neutralize 1.56mol nitric acid? KOH + HNO 3 KNO 3 + HOH 5) Molarity of phosphoric acid if 15.0 mL is neutralized by 38.5 mL of.15Mm NaOH? NaOH + H 3 PO 4 → Na 3 PO 4 + HOH

96 4) How many grams of potassium hydroxide to neutralize 1.56mol nitric acid? KOH + HNO 3 KNO 3 + HOH 5) Molarity of phosphoric acid if 15.0 mL is neutralized by 38.5 mL of.15M NaOH? NaOH + H 3 PO 4 → Na 3 PO 4 + HOH

97 4) How many grams of potassium hydroxide to neutralize 1.56mol nitric acid? KOH + HNO 3 KNO 3 + HOH 5) Molarity of phosphoric acid if 15.0 mL is neutralized by 38.5 mL of.15M NaOH? 3 NaOH + H 3 PO 4 → Na 3 PO HOH

98 Ch20.1 Neutralizations cont Ex1) What is the molarity of a sodium hydroxide solution if 25mLs of it neutralizes 35 mLs of a.50 M H 2 S?

99 Ch20.1 Neutralizations cont Ex1) What is the molarity of a sodium hydroxide solution if 25mLs of it neutralizes 35 mLs of a.50 M H 2 S? 1 H 2 S + 2NaOH → 1Na 2 S + 2H(OH)

100 HW#7) 55.2 mLs of a 2.0M HCl solution neutralizes 83.2mLs of a Ca(OH) 2 solution. Find its concentration. HCl + Ca(OH) 2 → CaCl 2 + H(OH)

101 HW#7) 55.2 mLs of a 2.0M HCl solution neutralizes 83.2mLs of a Ca(OH) 2 solution. Find its concentration. 2 HCl + Ca(OH) 2 → CaCl H(OH)

102 HW#8) How many mLs of a 3M NaOH solution are needed to neutralize 100 mLs of a 6M H 2 SO 4 solution?

103 2 NaOH + H 2 SO 4  Na 2 SO HOH

104 HW#9) How many mLs of a 2.5M H 3 PO 4 solution are needed to neutralize 50 mLs of a 1M KOH solution?

105 H 3 PO KOH → K 3 PO H(OH) Ch20 HW#2 7 – 9

106 Ch19,20 Mid Chapter Review pH Review On white board for test: pH= -log[H 3 O + ]pOH= -log[OH - ] [H 3 O + ]= antilog(–pH) [OH - ]= antilog(–pOH) pH + pOH =14 [H 3 O + ] [OH - ] = 1x Ex1) Given [OH - ] = 1.72 x10 -3 M, find pOH, pH, [H 3 O + ]

107 Ch19,20 Mid Ch Review Assignment 1. Name each acid or base when given the formula and vise versa. a. H 2 SO 4 b. HCL c. NaOH d. HNO 3 e. Magnesium hydroxide f. Phosphoric acid g. Sulfurous acid h. hydrobrombic acid i. HF j. HClO 3 k. H 2 CO 3 l. Nitrous acid m. aluminum hydroxide n. hydroselenic acid o. strontium hydroxide

108 2. Classify a solution as neutral, acidic, or basic: a. [H + ] = 1x10 -5 M b. [OH - ] = 1x10 -5 M c. [H + ] = 1x Calc the pH of a solution given hydroxide-ion concentration: [OH - ] is 1x10 -4 M 4. Skip (Repeat of #3) 5. A solution has a pH of 3. Calc the [H + ] concentration.

109 6. Identify the conjugate acid-base pairs in the following reaction: a. NH 3 + H 2 O  NH OH - HW:b. HNO 3 + H 2 O  H 3 O + + NO 3 - c. CH 3 COOH + H 2 O  H 3 O + + CH 3 COOH - d. Skip (Repeat of b) HW:e. H 2 O + CH 3 COOH -  CH 3 COOH + OH -

110 7. Match each solution with its correct description a. Dilute, weak acid(1) 18M H 2 SO 4(aq) b. Dilute, stong base (2) 0.5M NaOH (aq) c. concentrated, strong acid(3) 15M NH 3(aq) d. dilute, strong acid(4) 0.1M HC 2 H 3 O 2(aq) e. concentrated, weak base(5) 0.1M HCl (aq) HW: 8. What is the pH of a solution with [H + ] = 1x10 -3 ? HW: 9. A solution has a pOH of What is the pH of the solution? HW: 10. A solution has a hydroxide ion concentration, [OH - ] = 9.1x10 -9 M. What is the pH?

111 11. Write the equation for the neutralization reaction of potassium hydroxide by sulfuric acid. 12. Use the last equation to find the volume of 0.80M KOH to neutralize 15.0 mL of 0.65M H 2 SO 4.

112 13. What is the molarity of H 2 SO 4 if 25.0 mL neutralizes 15.0 mL of 0.100M NaOH? HW: 14. What volume of 0.12M Ba(OH) 2 is needed to neutralize 12.2 mL of 0.25M HCl?

113 Demo Lab20.1 Neutralization Lab20.1 – Titration Volume of NaOH:___ Molarity of NaOH:_0.2_ Volume of HCl:___ Molarity of HCl:_?__

114 Lab20.1 Neutralization - due in 2 days

115 Ch21.1 – Oxidation-Reduction Reactions (Redox) Oxidation - originally meant oxygen combining with an element to form an oxide. Ex1)Iron + Oxygen Iron(III)Oxide(a form of rust) Carbon (s) + Oxygen (g) Carbon Dioxide (g) Methane (CH 4(g) ) + Oxygen Carbon Dioxide + Water vapor

116 Ch21.1 – Oxidation-Reduction Reactions (Redox) Oxidation - originally meant oxygen combining with an element to form an oxide. Ex1)Iron + Oxygen Iron(III)Oxide(a form of rust) lost 3e - 4 Fe + 3O 2 2 Fe 2 O 3 gained 2e - Carbon (s) + Oxygen (g) Carbon Dioxide (g) Methane (CH 4(g) ) + Oxygen Carbon Dioxide + Water vapor

117 Ch21.1 – Oxidation-Reduction Reactions (Redox) Oxidation - originally meant oxygen combining with an element to form an oxide. Ex1)Iron + Oxygen Iron(III)Oxide(a form of rust) lost 3e - 4 Fe + 3O 2 2 Fe 2 O 3 gained 2e - Carbon (s) + Oxygen (g) Carbon Dioxide (g) lost 4e - C + O 2 CO 2 gained 2e - Methane (CH 4(g) ) + Oxygen Carbon Dioxide + Water vapor

118 Ch21.1 – Oxidation-Reduction Reactions (Redox) Oxidation - originally meant oxygen combining with an element to form an oxide. Ex1)Iron + Oxygen Iron(III)Oxide(a form of rust) lost 3e - 4 Fe + 3O 2 2 Fe 2 O 3 gained 2e - Carbon (s) + Oxygen (g) Carbon Dioxide (g) lost 4e - C + O 2 CO 2 gained 2e - Methane (CH 4(g) ) + Oxygen Carbon Dioxide + Water vapor lost 8e - CH 4 + O 2 CO 2 + H 2 O gained 2e -

119 Reduction - originally meant the loss of oxygen from a compound. Iron(III)oxide + Carbon Iron + Carbon dioxide

120 Reduction - originally meant the loss of oxygen from a compound. Iron(III)oxide + Carbon Iron + Carbon dioxide gained 3e - Fe 2 O 3 + C Fe + CO 2 loses 4e - Redox doesn’t just have to be with oxygen. Other elements act similarly. So use gain/loss of electrons to study redox. OIL RIG OIL – Oxidation Is Losing electrons RIG – Reduction is Gaining Electrons (reduced in charge) Ex2) Mg + S → Mg +2 + S -2

121 Reduction - originally meant the loss of oxygen from a compound. Iron(III)oxide + Carbon Iron + Carbon dioxide Fe 2 O 3 + C Fe + CO 2 Redox doesn’t just have to be with oxygen. Other elements act similarly. So use gain/loss of electrons to study redox. OIL RIG OIL – Oxidation Is Losing electrons RIG – Reduction is Gaining Electrons (reduced in charge) Ex) Mg + S → Mg +2 + S -2 Oxidized: Mg Reduced: S

122 Reduction - originally meant the loss of oxygen from a compound. Iron(III)oxide + Carbon Iron + Carbon dioxide gained 3e - Fe 2 O 3 + C Fe + CO 2 loses 4e - Redox doesn’t just have to be with oxygen. Other elements act similarly. So use gain/loss of electrons to study redox. OIL RIG OIL – Oxidation Is Losing electrons RIG – Reduction is Gaining Electrons (reduced in charge) gained 2e - Ex2) Mg + S → Mg +2 + S -2 loses 2e - Oxidized: Mg Reduced: S

123 Ex3) What is oxidized/reduced: 2 Ag(NO 3 ) (aq) + Cu (s) Cu(NO 3 ) 2(aq) + 2Ag (s) Oxidized: ___ Reduced: ___

124 Ex3) What is oxidized/reduced: gained 1e – 2 Ag +1 (NO 3 ) -1 (aq) + Cu (s) Cu +2 (NO 3 ) -1 2 (aq) + 2 Ag (s) lost 2e – Oxidized: Cu That makes Cu the reducing agent. Reduced: Ag That makes AgNO 3 the oxidizing agent.

125 Ex4) 3 CuCl Al 2 AlCl Cu 3 Cu +2 Cl Al 2 Al +3 Cl Cu Oxidized: ___ Reduced: ___ HW#3) (refer to electronegativities table) a) H 2(g) + Cl 2(g) 2HCl (g) Oxidized: Reduced:

126 Ex4) 3 CuCl Al 2 AlCl Cu 3 Cu +2 Cl Al 2 Al +3 Cl Cu Oxidized: Al Al is the reducing agent. Reduced: Cu CuCl 2 is the oxidizing agent. HW#3) (refer to electronegativities table) a) H 2(g) + Cl 2(g) 2HCl (g) Oxidized: Reduced: Ch21 HW#1 1-3

127 Ch21 HW#1 1,2 1) What chemical process accompanies a reduction reaction? 2) a) H 2 + Cl 2 → 2HCl b) 2Li + F 2 → 2 LiF c) 2Ba + O 2 → 2BaO d)N 2 + 2O 2 → 2NO 2 e) H 2 + S→ H 2 S f) CaO + Al → Al 2 O 3 + Ca g) Zn+Cu(SO 4 ) -2 →Cu + (Zn)(SO 4 )- 2

128 Ch21 HW#1 1,2 1) What chemical process accompanies a reduction reaction? OILRIG 2) a) H 2 + Cl 2 → 2HCl b) 2Li + F 2 → 2 LiF c) 2Ba + O 2 → 2BaO d)N 2 + 2O 2 → 2NO 2 e) H 2 + S→ H 2 S f) CaO + Al → Al 2 O 3 + Ca g) Zn+Cu(SO 4 ) -2 →Cu + (Zn)(SO 4 )- 2

129 Ch21 HW#1 1,2 1) What chemical process accompanies a reduction reaction? oxidation OILRIG 2) a) H 2 + Cl 2 → 2HCl H Cl b) 2Li + F 2 → 2 LiF c) 2Ba + O 2 → 2BaO d)N 2 + 2O 2 → 2NO 2 e) H 2 + S → H 2 S f) CaO + Al → Al 2 O 3 + Ca g) Zn + Cu(SO 4 ) → Cu + (Zn)(SO 4 )

130 Ch21 HW#1 1,2 1) What chemical process accompanies a reduction reaction? oxidation OILRIG 2) a) H 2 + Cl 2 → 2HCl H Cl b) 2Li + F 2 → 2 LiF Li F c) 2Ba + O 2 → 2BaOBa O d)N 2 + 2O 2 → 2NO 2 e) H 2 + S → H 2 S f) CaO + Al → Al 2 O 3 + Ca g) Zn + Cu(SO 4 ) → Cu + (Zn)(SO 4 )

131 Ch21 HW#1 1,2 1) What chemical process accompanies a reduction reaction? oxidation OILRIG 2) a) H 2 + Cl 2 → 2HCl H Cl b) 2Li + F 2 → 2 LiF Li F c) 2Ba + O 2 → 2BaOBa O d)N 2 + 2O 2 → 2NO 2 N O e) H 2 + S → H 2 SHS f) CaO + Al → Al 2 O 3 + Ca g) Zn + Cu(SO 4 ) → Cu + (Zn)(SO 4 )

132 Ch21 HW#1 1,2 1) What chemical process accompanies a reduction reaction? oxidation OILRIG 2) a) H 2 + Cl 2 → 2HCl H Cl b) 2Li + F 2 → 2 LiF Li F c) 2Ba + O 2 → 2BaOBa O d)N 2 + 2O 2 → 2NO 2 N O e) H 2 + S → H 2 SHS f) CaO + Al → Al 2 O 3 + Ca AlCa g) Zn + Cu(SO 4 ) → Cu + (Zn)(SO 4 ) ZnCu

133 Ch21.2 – Half Reactions Ex1) NO 2 + ClO -  NO Cl - OA ____ Red____ Reduction ½ Reaction RA____ Ox____ Oxidation ½ Reaction

134 Ex2) Fe +2 + Cr 2 O 7 -2  Fe +3 + Cr +3 OA____ Red____ Red ½ Reaction RA____ Ox____ Ox ½ Reaction

135 Lab21.1 Redox MnO Fe +2  Fe +3 + Mn +4 OA____ Red____ Reduction ½ Reaction RA____ Ox____ Oxidation ½ Reaction

136 Ch21 HW#2 3(a – d) 3a) Mg + H 2 SO 4  MgSO 4 + H 2 Ox____ Red____ b) 2H 2 + O 2  2H 2 O Ox____ Red____ c) 2 KClO 3  2KCl + 3O 2 Ox____ Red____ d) ZnO + 2HCl  ZnCl 2 + H 2 O Ox____ Red____

137 Lab21.1 Redox - due in 2 days - Ch21 HW#2 due at beginning of period

138 Ch22.1 Electrochemistry Zn (s) + CuSO 4(aq) ZnSO 4(aq) + Cu (s)

139 Ch22.1 Electrochemistry lose 2e- (oxidized) Zn (s) + CuSO 4(aq) ZnSO 4(aq) + Cu (s) gained 2e- (reduced) Oxidized ½ Reaction: Zn (s) Zn (aq) +2 +2e- Reduced ½ Reaction: Cu +2 (aq) + 2e - Cu (s) Activity Series Table given on HW Electrons flow down the chart!

140 Volitaic cell (wet cell battery) - convert chemical potential energy into electrical energy Electrodes - metals in voltaic cells Anode - negative electrode, electrons produced here (oxidation - OIL) Cathode - positive electrode, electrons head here (reduction - RIG) electrons transfer Na + Cl - Na + Cl - Na + Cl - Na + Cl - Na + thru the wire Cl - (salt bridge) Cl - CuNa + (necessary to maintain Na + Zn Cl - an ion charge balance) Cl - CuNa + Na + Zn Cl - Cu Zn CuSO 4 soln ZnSO 4 soln

141 Reduction Half Cell Oxidation Half Cell Cu e – → Cu Zn → Zn e – Zn → Zn e - E˚= +.76V Cu e - → CuE˚= +.34 V E cell =+1.10 V electrons transfer Na + Cl - Na + Cl - Na + Cl - Na + Cl - Na + thru the wire Cl - (salt bridge) Cl - CuNa + (necessary to maintain Na + Zn Cl - an ion charge balance) Cl - CuNa + Na + Zn Cl - Cu Zn Cu +2 SO 4 -2 Zn +2 SO 4 -2 SO 4 -2 Cu +2 SO 4 -2 Zn +2 CuSO 4 soln ZnSO 4 soln E˚ → standard conditions 25˚C, I Molar

142 Ex1) What is the cell reaction and cell potential for a voltaic cell of the following half cells? Fe +3 + e - → Fe +2 E˚= +.77 V Ni e - → Ni E˚= –.25V

143 HW #3) What is the cell reaction and cell potential for a voltaic cell of the following half cells? Cu e - → CuE˚= +.34 V Al e - → Al E˚= –1.66V Ch22 HW#1 1 – 4

144 Ch19,20 Review 1. Name these acids: a. HF b. HClO 3 c. H 2 CO 3 _ 2. Write formulas: a. Nitrous acid b. Aluminum hydroxide c. Hydroselenic acid d. Strontium hydroxide e. Phosphoric acid

145 3. Calculate pH, acid or basic? a. [H + ] = 1 x mol/L b. [OH - ] = 2 x mol/L 4. What are hydroxide ion concentrations? a. pH = 4.0 b. pH = 8.0

146 5. Calc pH (a,b) a. [H + ] = 2.4 x b. [OH - ] = 9.1 x and [H + ] on (c,d) c. pH = 13.2 d. pOH = Identify the acid-base pairs: a. HNO 3 + H 2 O  H 3 O + + NO 3 - b. CH 3 COOH + H 2 O  H 3 O + + CH 3 COO - c. NH 3 + H 2 O  NH OH -

147 7. What is the molarity of H 2 SO 4 if 25.0 mL neutralizes 15.0 mL of 0.100M NaOH? 8. A voltaic cell is constructed using electrodes w/ following half-reactions. Mn  Mn +2 (aq) + 2e - E 0 Mn +2 = V Pb +2 (aq) + 2e -  Pb (s) E 0 Pb +2 = –0.13 V Write a cell reaction. Calculate the standard cell potential.

148 9. A voltaic cell is constructed using electrodes with the following half-reactions. Fe (s)  Fe e - E 0 Fe +2 = V Sn +2 (aq) + 2e -  Sn (s) E 0 Sn +2 = –0.14 V Determine the cell reaction and the standard cell potential.


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