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승민 함
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1 강의개요 Chapter 4 Reactions in Aqueous Solution ( 수용액중의반응 ) 농도의종류와계산 수용액중의반응 Pb(NO 3 (aq) + 2KI (aq) HCl (aq) + NaOH (aq) Alka-Selter/Water Cu wire/ag(no 3 ) (aq) 침전반응산 - 염기중화반응기체생성반응산화환원반응 용액 (Solution) 4.1 Solute Concentrations; Molarity ( 용질의농도 ; 몰농도 ) ex) What are the molarities of Al 3+ and SO 2-4 in 1.00 M Al 2 (SO 4 ) 3? Al 2 (SO 4 ) 3 (s) 2Al 3+ (aq) + 3SO 4 2- (aq) KMnO 4 (s) K + (aq) + MnO 4- (aq)

2 Preparation of Solution ( 용액의제조 ) 1. 무게를잰다 2. 부피플라스크에넣고소량의증류수로녹인다 3. 물을표선조금밑까지가한다. 4. 흔들어섞는다 5. 증류수를표선까지채운다. 4.2 Precipitation Reactions ( 침전반응 ) Precipitation diagram : enable to determine whether or not a precipitate will form when dilute solutions of two ionic solutes are mixed Stoichiometry : relationships between amounts of reactants and products Precipitation Diagram ( 침전다이아그램 ) Precipitation Diagram 침전반응의결과를예측하는데이용 eg 1) Ba(NO 3 와 Na 2 CO 3 용액을혼합하면어떤일이? 존재하는이온종 : Ba 2+, NO 3-, Na +, CO 3 2- eg 2) BaCl 2, NaOH 용액의혼합존재하는이온종 : Ba 2+, Cl -, Na +, OH - 가능한침전 : Ba(OH, NaCl 불용성침전 : 없음 아무런반응도일어나지않음 ( no reaction) 가능한침전 : BaCO 3, NaNO 3 불용성침전 : BaCO 3 알짜이온방정식 (Net Ionic Equation) : Ba 2+ (aq) + CO 3 2- (aq) BaCO 3 (s)

3 Stoichiometry ( 화학양론 ) 예제 2) Cu 2+ (aq) + 2OH - (aq) Cu(OH (s) ex 1) 0.100M NaOH 50.0 ml 와반응하는데필요한 0.200M CuSO 4 Cu 2+ (aq) + 2OH - (aq) Cu(OH (s) ex 2) 6.52 g Cu(OH (s) 을얻는데필요한 M Cu(NO 3 용액의부피? 1.00 L 1 mol Cu mol OH - /L X 50 ml X X X 1000 ml 2 mol OH - 1 mol CuSO 4 1 mol Cu 2+ 용액의부피? 1mol Cu(OH) 1 mol Cu g Cu(OH X X X g Cu(OH 1 mol Cu(OH 1 mol Cu(NO 3 1 mol Cu 2+ n NaOH n Cu(OH)2 X 1.00L mol CuSO 4 n CuSO4 = L = 12.5 ml X 1.00L = L = 33.4 ml n Cu(NO3) mol Cu(NO Acid-Base Reactions ( 산 - 염기반응 ) Acid : produces H + ions in water solution Base : produces OH - ions in water solution Acid-Base Reactions Acid-Base Titrations : procedure used to determine concentration of an acid or base Acids ( 산 ) 수용액에서 H + 이온형성센산 (strong acids) 은물에서완전해리 : HCl, HBr, HI, HNO 3, HClO 4, H 2 SO 4 HCl(aq) H + (aq) + Cl - (aq) or HCl(aq) + H 2 O H 3 O + (aq) + Cl - (aq) 반응이완결되고 HCl 분자는전혀남지않음약한산 (weak acids) 은물에서부분적해리 HF(aq) H + (aq) + F - (aq) HF수용액은 HF 분자와 H + 와 F - 이온의혼합물

Bases ( 염기 ) 수용액에서 OH - 이온을형성센염기 (strong bases) 는물에서완전이온화 Group I 과무거운 Group II 의수산화물 (hydroxides) Ca(OH (s) Ca 2+ (aq) + 2OH (aq) 약한염기 (weak bases) 는부분적으로이온화되어 OH - 이온형성 NH 3 (aq) + H 2 O NH 4+ (aq)

4 Bases ( 염기 ) 수용액에서 OH - 이온을형성센염기 (strong bases) 는물에서완전이온화 Group I 과무거운 Group II 의수산화물 (hydroxides) Ca(OH (s) Ca 2+ (aq) + 2OH (aq) 약한염기 (weak bases) 는부분적으로이온화되어 OH - 이온형성 NH 3 (aq) + H 2 O NH 4+ (aq) + OH (aq) Equations for Acid-Base Reactions ( 산 - 염기반응식 ) 강산 + 강염기 : HCl + Ca(OH H + (aq) + OH - (aq) H 2 O 약산 + 강염기 : HF + Ca (OH HF(aq) + OH (aq) F (aq) + H 2 O 강산 + 약염기 : HCl + NH 3 H + (aq) + NH 3 (aq) NH 4+ (aq) 당량점과지시약 Acid-Base Titrations ( 산 - 염기적정 ) ex) 10.0 ml HCl 과반응하는데 M Ca(OH ml가사용되었다면 HCl 용액의몰농도는? OH - (aq) + H + (aq) H 2 O n OH- = = = n H+ = n HCl [HCl] = mol L = M

5 4.4 Oxidation-Reduction Reactions ( 산화 - 환원반응 ) Oxidation Number : a number that is assigned to an element in a molecule or ion to reflect, qualitatively, its state of oxidation Oxidation & Reduction : increase and decrease in oxidation number Oxidation Number ( 산화수 ) 다음과같은임의적규칙에의해 가전하 (Pseudocharge) 를정함 : i) 순수한원소로이루어진분자중의원자의산화수는 0 이다. ex) F in F 2, O in O 2 ii) 단원자이온중의원소의산화수는그이온의전하와같다. ex) 철의산화수 : Fe 2+ 에서는 +2, Fe 3+ 에서는 +3 Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) 산화 : Zn(s) Zn 2+ (aq) + 2e - 환원 : 2H + (aq) + 2e - H 2 (g) 계속 iii) 화합물중에서 Group 1 원소 : +1, Group 2 : +2, F: 1. K (in KBr, KMnO4) 의산화수 +1 O 는일반적으로 2, 단과산화물 (H 2 O 2 ) 에서는 1, OF 2 에서는 +2 H 는일반적으로 +1, 단금속의수소화물 (NaH) 에서는 -1 iv) ( 중성 ) 분자에있는원자들의산화수의합은 0, 다원자이온에서는그이온의전하와같다. H 2 SO 4 : (+2) + (ox. no. of S) + 4(-2) = 0; ox. no. of S = +6 2 Cr 2 O 7 : 2(ox. no. of Cr) + 7(-2) = -2; ox no. of Cr = +6 Balancing Half-Equations (Oxidation or Reduction) 산화 (oxidation) = 산화수의증가환원 (reduction) = 산화수의감소 HCl(g) + HNO 3 (l) NO 2 (g) Cl 2 (g) + H 2 O(l) HCl 는산화됨 ; Cl 의산화수 -1 0 HNO 3 는환원됨 ; N 의산화수 +5 +4

6 계속 ClO 3- (aq) + I - (aq) Cl - (aq) + I 2 (s) Balancing Redox Equations ( 산화환원반응의완결 ) ClO 3 - (aq) + I - (aq) Cl - (aq) + I 2 (s) first half-equation: 2I - (aq) I 2 (aq) 2I - (aq) I 2 (aq) + 2e - (1) second half-equation: ClO 3- (aq) + 6e - Cl - (aq) 두개의반쪽반응 (half-equations:) 으로나눔산화 : I - (aq) I 2 (s) 환원 : ClO 3- (aq) Cl - (aq) 전략 : 1 산화되거나환원되는원소의균형을맞춘다. 2 전자를더해산화수의균형을맞춘다. 3 H + (acidic) 과 OH - (basic) 을더해전하의균형을맞춘다. 4 분자를더해산소의수를맞춘다. eg) H 2 O acidic: ClO 3- (aq) + 6H + (aq) + 6e - Cl - (aq) (2) basic: ClO 3- (aq) + 6e - Cl - (aq) + 6OH - (aq) (3) acidic: ClO 3- (aq) + 6H + (aq) + 6e - Cl - (aq) + 3H 2 O (4) basic: ClO 3- (aq) + 3H 2 O+ 6e - Cl - (aq) + 6OH - (aq)(5) Combine half-equations so that electrons cancel acidic: 3[2I - (aq) I 2 (s) + 2e - ] (1)x3 ClO 3- (aq) + 6H + (aq) + 6e - Cl - (aq) + 3H 2 O (4) 6I - (aq) + ClO 3- (aq) + 6H + (aq) 3I 2 (s) + Cl - (aq) + 3H 2 O basic: (1)x3 +(5) 6I - (aq) + ClO 3- (aq) + 3H 2 O 3I 2 (s) + Cl - (aq) + 6OH - (aq) ex 1) Cu(s) + Ag + (aq) Cu 2+ (aq) + Ag(s) ex 2) Fe 2+ (aq) + MnO 4 - (aq) Fe 3+ (aq) + Mn 2+ (aq) 산화 Cu(s) Cu 2+ (aq) + 2 e - 환원 Ag + (aq) + e - Ag(s) 2Ag + (aq) + 2e - 2Ag(s) Cu(s) + 2Ag + (aq) Cu 2+ (aq) + 2Ag(s) Fe 2+ (aq) Fe 3+ (aq) + e - (1) MnO 4- (aq) + 5e - Mn 2+ (aq) (2) 전하균형 MnO 4- (aq) + 5e - + 8H + (aq) Mn 2+ (aq) 원자균형 MnO 4- (aq) + 5e - + 8H + (aq) Mn 2+ (aq) + 4 H 2 O(l) 전자균형 5Fe 2+ (aq) 5Fe 3+ (aq) + 5e - (1) X 5 5Fe 2+ (aq) + MnO 4- (aq) + 8H + (aq) Fe 3+ (aq) + Mn 2+ (aq) + 4 H 2 O(l) (1) X 5 + (2)

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untitled Chapter 4 Reactions in Aqueous Solution 3 4.1 Pb(NO 3 ) 2 (aq) + 2KI (aq) HCl (aq) + NaOH (aq) Alka-Selter/Water Cu wire/ag(no 3 ) (aq) 3 4.2 (Solution) KMnO 4 (s) K + (aq)+ MnO 4- (aq) 3 4.3 (Molarity)