산과염기 (acid and base) · 2017-11-13 · • ph and poh • weak acid and k a • weak base...
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Chapter 13
산과 염기(Acid and Base)
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• Brønsted-Lowry Acid-Base Model
• The Ion Product of Water
• pH and pOH
• Weak Acid and Ka
• Weak Base and Kb
• Acid Base Properties of Salt Solution
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Brønsted-Lowry Acid-Base Model
Acid is a proton donor, base is a proton acceptor
HB(aq)+A-(aq) HA(aq)+B-(aq)
HB, HA 는 산; A-, B- 는 염기.
HB – B- 와 HA – A- 는 서로 짝산 (conjugate acid )-짝염기(conjugate base) 쌍
양쪽성양성자성(amphiprotic) : H2O, HCO3-… .
←→
+− →←++
OHOHOH HH32
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13.2 The Ion Product of Water
모든 수용액 (Aqueous solution)은 H3O+(H+)과 OH– 사이에 평형상태:
H2O(l) + H2O(l) H3O+(aq) + OH–(aq)
KW = [H3O+]×[OH–] = 1.0 ×10-14 at 25°C :이온화곱상수
(ion product const. of water)
←→
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1. 순수한 물 : [H+] = [OH–] = 1.0 ×10–7 M; 중성용액
2. 산성용액 : [H+] > 1.0 ×10–7 M > [OH–]
3. 염기성용액 : [OH–] > 1.0 ×10–7 M > [H+]
4. 바닷물 , [H+] = 5 ×10-9 M; [OH–] = ?[OH–] = (KW) / (5 ×10-9) = 2 ×10-6 M
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13.3 pH and pOH
pH = –1og10[H+], [H+] = 10-pH
중성용액 : pH = 7.0산성용액 : pH < 7.0
염기성용액 : pH > 7.0
pOH = –1og10[OH–] ;
KW = [H3O+]×[OH–] = 1.0 ×10-14 at 25°CpH + pOH =-log [H3O+]-log[OH–] = -log([H3O+]×[OH–] )=14
pH + pOH = 14.00 at 25°C
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pH
[H+] = 2.4 × 10-6 M ; pH의 계산
pH = –1og10(2.4 ×10–6) = 5.62
pH = 8.68 ; [H+] 의 계산
[H+] = 10–8.68 = 2.1 ×10–9 M
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pH 와 [H+]의 관계
See, Τable 13.1
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센산과 센염기의 pHStrong Acids
HCl, HBr, HI, HClO4, HNO3, H2SO4
Strong Bases
LiOH, NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2
HCl 수용액
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분자HF(aq) + H2O H3O+(aq) + F –(aq)
유기산(RCOOH), HNO2
←→
양이온
NH4+(aq) + H2O H3O+(aq) + NH3(aq)
Zn(H2O)42+(aq) + H2O H3O+(aq) + Zn(H2O)3(OH)+(aq)
전이금속이온(Al3+… )←→
←→
13.4 Weak Acid and Their Eqilibrium Consts.
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Acid Equilibrium Constants, Ka
평형상수 : Ka 의 계산
←→
Ka =[H+] ×[B- ]
[HB]
HB(aq) H+(aq) + B–(aq)
pKa = -log10Ka
약산의 세기 ↑ : Κa ↑ : pKa↓ HNO2 pKa = 3.22, HCN pKa=9.24
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Ka 계산의 예
0.100 M HC2H3O2 용액의 pH 는 2.87; Ka=?
0.1-x x x
[H+] =10-2.87 = 1.3 ×10–3 M = [C2H3O2–]
[HC2H3O2] = 0.100 M – 0.0013 M = 0.099 M
Ka = = 1.7 × 10–5(1.3 × 10–3)2
0.099
←→HC2H3O2(aq) H+(aq) + C2H3O2
–(aq)
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용액 중 [H+] 계산의 예
0.200 M HC2H3O2 용액중의 [H+] 계산, Ka=1.8 × 10–5
0.200 – x x x
[H+] = x 라면, [C2H3O2–] = x , [HC2H3O2] = 0.200 – x 임
0.200 - x ≅ 0.200 라고 가정; x = 1.9 × 10–3 M
이 경우, % ionization =
HC2H3O2(aq) == H+(aq) + C2H3O2–(aq)
52
108.1200.0
−×=− x
x
%0.5%0.1100200.0109.1 3
p=×× −
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용액 중 [H+] 계산의 예
0.100 M HF (Ka= 6.9 × 10–4) 용액 중 [H+] 계산
HF(aq) + H2O == H3O+(aq) + F –(aq)
0.100 – x x x
= 6.9 × 10–4
0.100 – x ≅ 0.100 라고 놓고 풀면 : x = 8.3 × 10–3 > 5%
연속근사법: (cf, 이차방정식)
= 6.9 × 10–4; x = 8.0 × 10–3
x2
0.100 – x
x2
0.100 – 0.008
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Polyprotic Weak Acid
H3PO4 H+ + H2PO4- Ka1 = 7.1 ×10-3
H2PO4- H+ + HPO4
2- Ka2 = 6.2 ×10-8
HPO42- H+ + PO4
3- Ka3 = 4.5 ×10-13
Ka1 > Ka2 > Ka3
H2CO3
←→
←→
←→
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13.5 Weak Bases and their Equilibrium Constants
분자
NH3(aq) + H2O OH– (aq) + NH4+(aq)
아민(CH3NH2… .)←→
약산의 음이온
F– (aq) + H2O HF(aq) + OH– (aq)←→
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NH3 : Kb =
base equilibrium constants, Kb
[NH4+] × [OH–][NH3]
NH3(aq) + H2O == OH– (aq) + NH4+(aq)
pKb = -log10Kb
약염기의 세기 ↑ : Κb ↑ :pKb ↓
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Ka 와 Kb 의 관계
Ka × Kb = KW = 1.0 × 10–14
Ka Kb
HF 6.9 × 10–4 F– 1.4 × 10–11
HAc 1.8 × 10–5 Ac– 5.6 × 10–10
NH4+ 5.6 × 10–10 NH3 1.8 × 10–5
염기의 세기는 그 짝산의 세기와 직접적으로 반비례한다.
See, Table 13.4
pKa + pKb = 14
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Kb
F– 1.4 × 10–11
Ac– 5.6 × 10–10
NH3 1.8 × 10–5
산의 세기
Ka
HF 6.9 × 10–4
HAc 1.8 × 10–5
NH4+ 5.6 × 10–10
HF > HAc > NH4+
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약염기 용액 중 [OH–] 의 계산
약염기 용액 중 [OH-]의 계산 : 0.10M NaF 의 pH ?
F– (aq) + H2O == HF(aq) + OH– (aq) : Kb=1.4 × 10–11
0.10-x x x
= 1.4 × 10–11 ; = 1.4 × 10–11
[OH–] = 1.2 × 10–6 M ; pH = 8.08
[HF][OH–][F–]
[OH–]2
0.10
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Acid-Base Properties of Salt Solutions
Ø염 (salt)
Ø 양이온방관자(Spectator): 강 염기로부터 파생된
Li+, Na+, K+; Ca2+, Sr2+, Ba2+
산성(Acidic): 전이금속을 포함한 모든 다른 양이온들 Al3+, NH4+… .
Ø 음이온방관자 : 강산으로부터 파생된
C1–, Br–, I–, NO3–, SO4
2–
염기성(Basic): 약산으로부터 파생된 음이온, F–, NO2– 등
Ø 총체적 결과 : 강한 것이 이긴다..
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총체적 결과
Salt Cation Anion
NaNO3 Na+(Sp.) NO3–(Sp.) 중성
KF K+(Sp.) F–(Ba.) 염기성
FeCl2 Fe2+(Ac.) Cl–(Sp.) 산성
산성 양이온과 염기성 음이온을 가진 화합물이라면, Ka, Kb값을 서로 비교하여 결정
NH4F: Ka = 5.6 × 10–10, Kb = 1.4 × 10–11 ; 산성