Chapter 3 Molecular Shape and Structure
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
CHAPTER 10 ACIDS AND BASES
THE NATURE OF ACIDS AND BASES
10.1 BrØnsted-Lowry Acids and Bases
In aqueous or nonaqueous solution, Acid : Proton donor Base : Proton acceptor
+− HCl(aq) +⎯H23O(l) ⎯→+H O (aq) Cl (aq) Fig. 10.1 Dissolution of HCl in water.
Proton-transfer reaction
⎯⎯→ +− HCN(aq) ++H23O(l)←⎯⎯ H O (aq) CN (aq)
2+ − Ca (aq) ++HCO32(aq) H O(l)
⎯⎯→ + ←⎯⎯ H33O (aq) + CaCO (s)
A strong acid is fully deprotonated. A weak acid is only partially deprotonated in solution.
Fig. 10.2 Stalactites(종유석,鍾乳石) and Stalagmites(석순,石筍)
2– ▷ Oxide, O , is a strong base. ▷ Ammonia, NH3, is a weak base.
2−− + − O(aq)+⎯H2O(l) ⎯→2 OH(aq) NH32(aq) ++H O(l) R NH4(aq) OH (aq)
Fig. 10.3 An oxide ion in water Fig. 10.4 Equilibrium structure of aqueous ammonia solution 1 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
▶ Conjugate acids and bases
+ Acid ⎯⎯donates⎯ H⎯→ Conjutgate base
⎯⎯→ + − HCN(aq) ++HO2 (l)←⎯⎯ H3O (aq) CN(aq) acid 1 base 2 acid 2 base 1
+ Base ⎯accept⎯⎯⎯s H → Conjutgate acid
+ − NH(3 aq)++HO24()laR NH ( q) OH(aq) base 1 acid 2 acid 1 base 2
Ex. 10.1 Writing the formulas of conjugate acids and bases
− 2− (a) The conjugate base of HCO3 is CO3 2− − Æ The conjugate acid of CO3 is HCO3 . (b) The conjugate base of OH− is O2− . Æ The conjugate acid of O2− is OH− .
☺ Brønsted-Lowry definition applies to nonaqueous solvents
−+ CH33COOH(l) ++NH (l) R CH3CO2(am) NH4(am)
☺ Brønsted-Lowry definition applies also to the gas phase.
HCl(gg) +⎯NH34( ) ⎯→ NH Cl(s)
Fig. 10.5 White powder (NH4Cl) formed from NH3(g) and HCl(g)
Most general definition: Lewis acid : Electron pair acceptor Lewis base : Electron pair donor
2− O is a Lewis base: NH3 is a Lewis base
2 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
Fig. 10.6 Actions of acids (left) and bases (right) in Arrhenius, BrØnsted, and Lewis definitions.
Lewis acid (CO2) + Lewis base (H2O) Æ Brønsted acid (H2CO3)
10.3 Acidic, Basic, and Amphoteric Oxides
▶ Acidic oxide (CO2): Molecular compound (nonmetal), Reacts with water to form a Brønsted acid
2 NaOH(aq) + CO22(g) ⎯⎯→+Na CO3(aq) H2O(l)
NaOH(aq) + CO23(g) ⎯⎯→ NaHCO (aq)
▶ Basic oxide (CaO, MgO): Ionic salt (metal)
Reacts with water to form a Brønsted base
Reacts with acid to form a salt and water
CaO(sl) + H22O( ) ⎯⎯→Ca(OH) (aq)
MgO(s) + 2 HCl(l) ⎯⎯→+MgCl22(aq) H O(l)
▶ Amphoteric oxide (Al2O3): Reacts with both acid and base
Al23O (sl) + 6 HCl( ) ⎯⎯→+2 AlCl3(aq) 3 H2O(l)
2 NaOH(aq) + Al O (s) + 3 H O(l) → 2 Na[Al(OH) ](aq) 23 2 4 Fig. 10.7 Amphoteric oxides of metalloid elements.
3 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
Fig. 10.8 The d-block elements forming amphoteric oxides with intermediate oxidation states.
10.4 Proton Exchange Between Water Molecules
▶ Amphiprotic (양쪽성양성자성): acting both as a proton donor and as a proton acceptor
ZZX +− o −1 2 H23O(la)YZZ HO()q+ OH(aq), ∆=H r +56 kJ ⋅mol (A)
Æ Autoprotolysis (자체양성자이전반응) or Autoionization
aa+− HO3 OH K = 2 a ()HO2
Kaw = +a− : Autoprotolysis constant or water product constant HO3 OH
o o1− aJ ≈ [J]/ c, where c =⋅1 mol L
+− −77− −14o Kw ==[H3O ][OH ] ()1.0 ×10 ×(1.0 ×10 ) =1.0×10 at 25 C
+ Fig. 10.9 Autoprotolysis of water Fig. 10.10 The product of the concentrations of hydronium (H3O ) and hydroxide (OH–) ions in water is a constant.
4 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10 Ex. 10.2 Calculating the concentrations of ions in a solution of a metal hydroxide.
o + − A solution of 0.0030 M Ba(OH)2(aq) at 25 C. [H3O ] = ? [OH ] = ?
Decide whether the Ba, alkaline earth metal compound is fully Æ Ba(OH)2 dissociates almost completely dissociated in solution. in water to produce OH– ions
– 2+− Find the mole ratio of [OH ] Ba(OH)2 (sa) ⎯⎯→+Ba ( q) 2OH (aq) – to solute concentration. 1 mol Ba(OH)2 ≈ 2 mol OH Calculate [OH–] from the [OH−−] = 2 ×⋅0.0030 mol L 1 solute concentration. =⋅0.0060 mol L−1
Rearrange −14 + Kw 1.0 ×10 +− [H3O ] ==− Kw = [H3O ][OH ] [OH ] 0.0060 −−12 1 + =×1.7 10 mol ⋅L to find [H3O ].
10.5 The pH Scale
+ pH =−log a + , pH =−log[H3O ] HO3
pH < 7 (acidic) pH =−log[1.0 ×10−7 ] =7.00 (pure water at 25oC) pH > 7 (basic)
Ex. 10.3 Calculating a pH from a concentration.
+ −−81 (a) What is the pH of human blood in which [H3O ] = 4.0×⋅10 mol L ?
+ −8 pH =−log[H3O ] pH =−log[4.0 ×10 ] =7.40
(b) 0.020 M HCl(aq)
+ pH =−log[H3O ] pH =−log0.020 =1.70
(c) 0.040 M KOH(aq) [OH−−] ==[KOH] 0.040 mol ⋅L 1 K 1.0 ×10−14 [H O+−] ==w =2.5×10 13 3 [OH − ] 0.040
+ −13 pH =−log[H3O ] pH =−log(2.5×10 ) =12.60
5 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
Fig. 10.11 A pH-meter is a voltmeter measuring the pH electronically.
(a) Orange juice (b) Lemon juice Fig. 10.12 Typical pH values of common aqueous solutions.
10.6 The pOH of Solutions
pXl=− ogX
pOH =−loga OH−
pOH =−log[OH − ]
−14 pKKww=−log =−log(1.0×10 ) =14.00
+ − log Kw = log[H3O ][OH ]
+− −−log[H3O ] log[OH ] =−log Kw Fig. 10.13 The pH and pOH scales. Always, pH + pOH = 14 pH +=pOH pKw =14
6 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
WEAK ACIDS AND BASES
Fig. 10.14 Magnesium metal dissolved in (a) HCl and (b) HAc with the same concentrations. The rate of hydrogen evolution depends on the concentration of hydronium ions.
10.7 Acidity and Basicity Constants
▶ Molecular composition of a solution of a weak acid (or base)
- acid (or base) molecules or ions
+ – - small concentrations of H3O (or OH ) ions and the conjugate base of the acid
– + - very small concentrations of OH (or H3O ) ions maintaining autoprotolysis equilibrium
⎯⎯→ + − CH32COOH(aq) ++H O(l)←⎯⎯ H3O (aq) CH3CO2(aq)
aa+− K = HO3 CH32CO aa CH32COOH H O
+ − [H33O ][CH CO2] −5 o Ka ==1.8×10 at 25 C [CH3COOH]
⎯⎯→ +− HA()aq ++H23O(l)←⎯⎯ HO()aq A(aq)
+− [H3O ][A ] Ka = , plKKaa=− og [HA] Fig. 10.15 A solution of a weak acid.
⎯⎯→ + − NH32(aq) ++H O(l)←⎯⎯ NH4(aq) OH (aq)
+ − aa+− NH OH [NH ][OH ] K = 4 Æ K =4 =1.8×10−5 at 25oC , p4K = .75 aa b [NH ] b NH32H O 3 7 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10 + − ⎯⎯→ +− [BH ][OH ] B(aq) ++H 2O(l) ←⎯⎯ BH (aq) OH (aq) , Kb = , pKKbb= −log [B]
10.8 The Conjugate Seasaw
▶ Relative strength of an acid (or a base) and its conjugate base (or acid):
Æ the stronger the acid (or base), the weaker its conjugate base (or acid)
+ ⎯⎯→ + NH42()aq ++HO(l)←⎯⎯ H3O()aq NH3(aq),
+ [H33O ][NH ] Ka = + [NH4 ]
[H O+ ][ NH ] [ NH + ][OH− ] KK = 3 3 × 4 = [H O+ ][OH− ] ab + [ NH ] 3 [ NH4 ] 3
KKab×=Kw (11a) log KKab+=log log Kw
pKKab+=p pKw (11b)
+ pKKaw(NH43) =−p pKb(NH ) =14.00 −4.75 =9.25
Fig. 10.17 Conjugate acid-base pairs.
8 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
Ex. 10.5 Deciding which of two species is the stronger acid or base.
(a) HF / HIO3 (b) HNO2 / HCN
Ka(HIO3) > Ka(HF) Ka(HNO2) > Ka(HCN)
pKa(HIO3)=0.77 < pKa(HF)=3.45 pKa(HNO2)=3.37 < pKa(HCN)=9.31
10.9 Molecular Structure and Acid Strength
Binary acid, HA
The greater the electronegativity of A, the stronger the acid HA.
The acid strengths increase (while bond strengths decrease) down the group.
The more polar the H-A bond, the stronger the acid across a row.
9 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
10.10 The Strengths of Oxoacids and Carboxylic Acids
Proton of an –OH group in an oxoacid is acidic due to high polarity of the O–H bond.
▶ Phosphorus acid, H3PO3
Æ (HO)2PHO
Æ Donates H’s from –OH groups
Æ No donation of H from –PH group
Electronegativity difference Æ
between O and P
▶ Oxoacids with the different central halogen atoms but with the same number of O atoms
Æ The greater the electronegativity of the halogen, the stronger the oxoacid.
▶ Oxoacids with the same central atom with the different number of O atoms
Æ The greater the number of O atoms attached to the central atom, the stronger the oxoacid.
Æ The greater the oxidation number of the central atom, the stronger the oxoacid.
10 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
▶ Strength of organic acids
(1) Acid strength: Carboxylic acid >> ethanol
1) Greater number of O atoms attached to the central atom.
2) Electron delocalization in the conjugate base stabilizes it.
– Æ carboxylate ion is much weaker base than the ethoxide ion, CH3CH2O
(2) Acid strength: Trichloroacetic acid (pKa = 0.52) >> Acetic acid (pKa = 4.75)
Æ Electronegativity difference: Cl > H Æ Electron withdrawing power: –CCl3 > –CH3
11 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
THE pH OF SOLUTIONS OF WEAK ACIDS AND BASES
10.11 Solutions of Weak Acids
concentration of A− percentage deprotonated =×100 % initial concentration of HA
[H O+ ] percentage deprotonated =×3 100 % [HA]initial
TOOLBOX 10.1 HOW TO CALCULATE THE pH OF A SOLUTION OF A WEAK ACID
▶ Set up an equilibrium table.
12 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
+ – Acid, HA H3O Conjugate base, A
Step 1 Initial concentration [HA]initial 0 0
Step 2 Change in concentration –x +x +x
Step 3 Equilibrium concentration [HA]initial – x x x
+ ▶ Use the value of Ka, calculate x (= [H3O ]) .
Ex. 10.7 Calculating the pH and percentage deprotonation of a weak acid.
Calculate the pH and percentage deprotonation of 0.1 M CH3COOH(aq).
[H O+ ][CH CO − ] ⎯⎯→ + − 332 −5 CH32COOH(aq) ++H O(l)←⎯⎯ H3O (aq) CH3CO2(aq) , Ka ==1.8×10 [CH3COOH]
+ – HAc H3O Ac
Step 1 Initial concentration 0.10 0 0
Step 2 Change in concentration –x +x +x
Step 3 Equilibrium concentration 0.10 – x x x
Step 4 Substitute the equilibrium concentrations into Ka.
x × xx2 K =×1.8 10−5 = ≈ Æ x =×1.3 10−3 (>0) a 0.10 − x 0.10
+ −3 From pH =−log[H3O ] , pH ≈−log(1.3×10 ) =2.89
+ From Percentage deprotonated =×()[H3O ]/[HA]initial 100%
with [HA]initial = 0.10 , 1.3×10−3 Percentage deprotonated =×100% =1.3% 0.10
Ex. 10.8 Calculating the Ka of a weak acid from the pH.
pH = 2.95 for 0.010 M mandelic acid, C6H5CH(OH)COOH(aq). Æ Ka = ?
+p− H +2− .95−−1 1 From [H3O ] = 10 , [H3O ] =⋅10 mol L =0.0011 mol ⋅L
13 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
+ − + Relations between equilibrium concentrations: [H3O ] = [A ] , [HA] =−[HA]initial [H3O ]
[H O++][A− ] [H O ]2(0.0011)2 33 −4 Ka == + = =1.4 ×10 [HA] ()[HA]initial − [H3O ] 0.010 − 0.0011
10.12 Solutions of Weak Bases concentration of BH++[BH ] Percentage protonated =×100% =×100% (13) initial concentration of B [B]initial
TOOLBOX 10.2 HOW TO CALCULATE THE pH OF A SOLUTION OF A WEAK BASE
B BH+ OH–
Step 1 Initial [B]initial 0 0 concentration
Step 2 Change in –x +x +x concentration Step 3 Equilibrium [B]initial – x x x concentration
Ex. 10.9 Calculating the pH and percentage protonation of a weak base.
Calculate the pH and percentage of protonation of a 0.20 M aqueous solution of methylamine.
[CH NH + ][OH− ] ZZX + − 33 −4 CH32NH()aq ++H2O(l)YZZ CH3NH3()aq OH(aq), Kb ==3.6 ×10 [CH32NH ]
+ – CH3NH2 CH3NH3 OH
Step 1 Initial concentration 0.20 0 0
Step 2 Change in concentration –x +x +x
Step 3 Equilibrium concentration 0.20 – x x x
x × xx2 Step 4 K =×3.6 10−4 = ≈ Æ x =×8.5 10−3 b 0.20 − x 0.20
From pOH =−log[OH − ] , pOH =−log(8.5×10−3 ) =2.07
Step 5 From pH =−pKw pOH , pH = 14.00 −=2.07 11.93
14 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
+ From Eq.13, with [BH ] = x and [B]initial = 0.20 ,
8.5×10−3 Percentage protonated =×100 % =4.2 % 0.20
10.13 The pH of Salt Solutions
Neutralization of 0.3 M HAc(aq) with 0.3 M NaOH(aq) Æ NaAc(aq) with pH 9
+ (1) All cations that are conjugate acids of weak bases produce acidic solutions. NH4
(2) Small, highly charged metal cations that can act as Lewis acids in water produce acidic solutions. Al3+, Fe3+
Fig. 10.18 Acidic solutions of hydrated cations.
H2O, 0.1M Al2(SO4)3(aq), 0.1M Ti2(SO4)3(aq), 0.1M HAc(aq)
Fig. 10.19 Hydrated metal cation acts as a Brønsted acid.. Æ
15 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
(3) Cations of Group 1, 2, and those of charge +1 from other groups, are such weak Lewis acid that the hydrated
ions do not act as acids.
– – (4) Very few anions that contain hydrogen produce acidic solutions. H2PO4 , HSO4
(5) All anions that are the conjugate bases of weak acids produce basic solutions.
− ZZX − HCO22(aq) ++H O(l)YZZ HCOOH(aq) OH (aq)
(6) Anions of strong acids are such weak bases that they have no effect on the pH of solutions.
Ex. 10.10 Calculating the pH of a salt solution with an acidic cation.
Estimate the pH of 0.15 M NH4Cl(aq). [H O+ ][NH ] + ZZX + 33 NH42(aq) ++H O(l)YZZ H3O (aq) NH3(aq) , Ka = + [NH4 ]
+ + NH4 H3O NH3
Step 1 Initial concentration 0.15 0 0
Step 2 Change in concentration –x +x +x
Step 3 Equilibrium concentration 0.15 – x x x
−14 Kw 1.0 ×10 −10 Ka == −5 =5.6 ×10 Kb 1.8×10 x× xx2 K =×5.6 10−10 = ≈ Æ x =×9.2 10−6 a 0.15 − x 0.15
16 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
+ From pH =−log[H3O ] ,
pH =−log(9.2×10−6 ) =5.04
Ex. 10.11 Calculating the pH of a salt solution with a basic anion.
Estimate the pH of 0.15 M Ca(CH3CO2)2(aq). [CH COOH][OH− ] − ZZX − 3 CH32CO (aq) ++H2O(l)YZZ CH3COOH(aq) OH (aq) , Kb = − [CH32CO ]
– – CH3CO2 CH3COOH OH
Step 1 Initial concentration 0.30 0 0
Step 2 Change in concentration –x +x +x
Step 3 Equilibrium concentration 0.30 – x x x
−14 Kw 1.0 ×10 −10 Kb == −5 =5.6 ×10 Ka 1.8×10 x× xx2 K =×5.6 10−10 = ≈ Æ x =×1.3 10−5 b 0.30 − x 0.30
From pOH =−log[OH− ],
pOH =−log(1.3×10−5 ) =4.89
pH =−pKw pOH =14.00 −4.89 =9.11
POLYPROTIC ACIDS AND BASES
10.14 The pH of a Polyprotic Acid Solution
+ − + − [H33O ][HCO ] −7 H23CO(aq)H++2O(l)R H3O(aq)HCO3(aq), Ka1 ==4.3×10 [H23CO ]
17 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
+ 2− −−+ 2 [H33O ][CO ] −11 HCO32(aq) ++H O(l) R H3O (aq) CO3(aq) , Ka2 ==− 5.6×10 [HCO3 ]
KKaa12>> >> Ka3>> ⋅⋅⋅
▶ The pH of the solution of 0.010 M HS24O(aq):
First deprotonation:
+ − H24SO(aq)H+ 2O(l)⎯⎯→ H3O(aq)H+ SO4(aq)
Initial 0.010 0 0 Equil. 0 0.010 0.010
Second deprotonation:
−−+ 2 HSO42(aq) ++H O(l) R H3O (aq) SO4(aq) , Ka2 = 0.012
Initial 0.010 0.010 0 Equil. 0.010 – x 0.010 + x x
[H O+ ][SO 2− ] (0.01+ xx) 34 −2 −3 Ka2 ==− =0.012 Æ x = −2.65×10 or 4.51× 0 [HSO4 ] 0.01− x
+1− ∴ [H3O ] =+0.010 x =0.014 mol ⋅L Æ pH1= .9
10.15 Solutions of Salts of Polyprotic Acids
Conjugate base of a polyprotic acid is amphiprotic(양쪽성양성자성):
−+ZZX 2−−15 HS (aq) ++H23O(l)YZZ H O (aq) S (aq) , Ka2 =×7.1 10 , p1Ka2 = 4.15
−−ZZX −8 HS()aq ++H22O(l)YZZ HS()aq OH(aq), KKbw11==/7Ka.7×10, p7Kb1 = .11
18 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
HS− very weak acid Æ basicity of S2− will dominate Æ pH > 7
HS− weak base Æ pH > 7
pH will be high if both pKa1 and pKa2 are relatively large. ( p7Ka1 = .89, pKa2 = 14.15)
▶ Empirical formula good for SK w/ Ka2 and S Ka1 (S: initial conc. of the salt)
1 pH =+2 (pKa1pKa2) (14)
Ex. 10.12 Estimating the pH of a solution of an amphiprotic salt. 구연산(枸櫞酸)
Estimate the pH of (a) 0.20 M NaH2PO4(aq); (b) 0.20 M Na2HC6H5O7(aq), a salt of citric acid, H3C6H5O7.
– (a) For the solution of phosphoric acid, H2PO4
−3 −8 Ka1 =×7.6 10 , Ka2 =×6.2 10 , p2Ka1 = .12, p7Ka2 = .21
−−14 8 −7 SK==0.20 wa/ K2 1.0×10 / 6.2 ×10 =1.6 ×10
−3 SK==0.20 a1 7.6 ×10
11 pH =+22()pKKaa12p =()2.12 +7.21 =4.66
2– (b) For the solution of citric acid, HC6H5O7
p5Ka2 = .95, p6Ka3 = .39
−−14 78− SK==0.20 wa/ K3 1.0 ×10 / 4.1×10 =2.4 ×10
−6 SK==0.20 a2 1.1×10
11 pH =+22()pKKaa23p =()5.95 +6.39 =6.17
10.16 The Concentrations of Solute Species
TOOLBOX 10.3 HOW TO CALCULATE THE CONCENTRATIONS OF ALL SPECIES IN A POLYPROTIC ACID SOLUTION
CONCEPTUAL BASIS
+ Assume that only the first deprotonation contributes significantly to [H3O ].
+ – Autoprotolysis of water does not contribute significantly to [H3O ] or [OH ].
PROCEDURE FOR A DIPROTIC ACID
19 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
– Step 1. Determine [HA ] and [H3O+] from the deprotonation equilibrium (Ka1) of H2A.
2– – – Step 2. Determine [A ] from the second deprotonation equilibrium (Ka2) of HA , using the values of [HA ] and
[H3O+] obtained from Step 1.
– Step 3. Calculate [OH ] by dividing Kw by [H3O+] obtained from Step 1.
PROCEDURE FOR A TRIPROTIC ACID Å one more step
Ex. 10.13 Calculating the concentrations of all solute species in a polyprotic acid solution.
Calculate the concentrations of all solute species in 0.10 M H3PO4(aq)..
Step 1. Find the concentration of H3PO4, assuming that only H3PO4 affects the pH significantly.
+ − + − [H32O ][H PO4] −3 H34PO(aq)H++2O(l)R H3O(aq)H2PO4(aq), Ka1 ==7.6 ×10 [H34PO ]
+ – H3PO4 H3O H2PO4
Initial concentration 0.10 0 0
Change in concentration -x +x +x
Equilibrium concentration 0.10 - x x x
+2− 2 −3 [H32O ][H PO4] x x Ka1 =×7.6 10 = = ≈ [H34PO ] 0.10− x 0.10
Æ x ≈ 0.028 no good approximation ! 28% w.r.t. 0.10
Go back to full quadratic equation: xx23+×(7.6 10−−) −7.6 ×10 4=0
x = 2.4 ×10−2 or −×3.2 10−2
− + − −2 [H24PO ] = [H3O ] Æ [H24PO ] =×2.4 10
−1 [H34PO ] = 0.10 −⋅0.024 mol L [H34PO ] =0.10−x Æ =⋅0.08 mol L−1
20 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
2– Step 2. Find the concentration of HPO4
+ 2− −−+ 2 [H34O ][HPO ] −8 H24PO(aq)H++2O(l)R H3O(aq)HPO4(aq), Ka2 ==− 6.2 ×10 [H24PO ]
– + 2– H2PO4 H3O HPO4
Initial concentration (from Step 1) 2.4×10–2 2.4×10–2 0
Change in concentration -x +x +x
Equilibrium concentration 2.4×10–2 - x 2.4×10–2 +x x
+ 2− 2.4 ×+10−−22xx 2.4 ×10 x −8 [H34O ][HPO ] ()( ) Kxa2 =×6.2 10 = − = −−22≈ = [H24PO ] 2.4 ×−10 x 2.4 ×10
Æ x =×6.2 10−8( 2.4 ×10−2) Good approximation !
2− 2− −81− From [HPO42] =≈x Ka , [HPO4 ] = 6.2 ×⋅10 mol L
3– Step 3. Find the concentration of PO4
+ 3− 23−−+ [H34O ][PO ] −13 HPO42(aq)H++O(l)R H3O(aq)PO4(aq), Ka3 ==2− 2.1×10 [HPO4 ]
2– + 3– HPO4 H3O PO4
Initial concentration (from Step 2) 6.2×10–8 2.4×10–2 0
Change in concentration -x +x +x
Equilibrium concentration 6.2×10–8 - x 2.4×10–2 +x x
+ 3− 2.4 ×+10−−22x xx2.4 ×10 −13 [H34O ][PO ] ( ) ( ) −19 Ka3 =×2.1 10 = 2− = −−88≈ Æ x ≈×5.4 10 [HPO4 ] 6.2 ×−10 x 6.2 ×10
3− −−19 1 ∴= [PO4 ] x ≈5.4 ×10 mol ⋅L
Step 4. Find the concentration of OH– .
1.0 ×10−14 From [OH−+] = K /[H O ], [OH−−] = mol⋅=L 14.2 ×10−13 mol ⋅L−1 w 3 2.4 ×10−2
21 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
10.17 Composition and pH
BOX 10.1 WHAT HAS THIS TO DO WITH … THE ENVIRONMENT?
Acid Rain
⎧ N(gg)+ O( )YZZZXZ 2 NO(g) ⎪ 22 ⎨ 2 NO(gg) +⎯O22( ) ⎯→2 NO (g) ⎪ + − ⎩3 NO22(gl) +⎯3 H O( ) ⎯→+2 H3O (aq) 2 NO3(aq) +NO(g)
SO22(gl) +⎯H O( ) ⎯→H2SO3(aq)
⎧⎪ 2 SO22()gg+⎯O() ⎯→2 SO3()g
⎨ + − ⎩⎪SO32(g) +⎯2 H O(l) ⎯→+H3O (aq) HSO4(aq)
Eastern gamma grass (protein-rich seeds)Æ
Precipitation across U.S. gradually become more acidic from west to east
22 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
◆ Concentrations of species present in a carbonic acid solution varying with pH. HOW DO WE DO THAT?
+ − + − [H33O ][HCO ] H23CO(aq)H++2O(l)R H3O(aq)HCO3(aq) Ka1 = [H23CO ]
+ 2− −+ 2−[H33O ][CO ] HCO32(aq) ++H O(l) R H3O (aq) CO3(aq) Ka2 = − [HCO3 ]
Fraction of each species X present:
− []X []X /[HCO3 ] fX()==−−22− −− [H23CO ]++[HCO3] [CO3] [H2CO3]/[HCO3]+1+[CO3]/[HCO3]
+ 2− [H23CO ] [H3O ] [CO32] Ka − = , − = + [HCO3 ] Ka1 [HCO3 ] [H3O ]
+ 2 + [H O ] − [H O ]K 2− KK f ()HCO = 3 , f HCO = 31a , f CO = a12a 23 H ()3 H ()3 H
++2 where HK=+[H33O] [HO] a1+Ka1Ka2
For any diprotic weak acid H2A,
+ 2 + [H3O ] − [H31O ]Ka 2− KKa1a2 f ()HA2 = , f ()HA = , f ()A = (15a) H H H
++2 where H=+[H33O] [HO]Ka1+Ka1Ka2 (15b)
At pH < pKa1, fully protonated form (H2CO3) is dominant.
2– At pH > pKa2, the fully deprotonated form (CO3 ) is
dominant.
− At intermediate pH range, f (HCO3 ) ≈ 1.
⎡ − ⎤ 1 f HCO3 at pH =+2 ()pKKaa12p ⎣ ( )⎦max
Fig. 10.20 The fractional composition of the species in carbonic acid as a function of pH.
23 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
Fig. 10.21 The fractional composition of the species in phosphoric acid (triprotic acid) as a function of pH.
AUTOPROTOLYSIS (자체양성자이전 반응) AND pH
10.18 Very Dilute Solutions of Strong Acids and bases
▶ Very dilute solution of HCl(aq)
−+− Charge balance: [OH ] =[H3O ]−[Cl ]
− −+ Material balance: [Cl ] = [HCl]initial ∴ [OH ] =−[H3O ] [HCl]initial
+− + + Autoprotolysis constant: Kw ==[H3O ][OH ] [H3O ]([H3O ]−[HCl]initial )
+2 + Or [H3O ] −[HCl]initial[H3O ]−Kw =0 (16)
▶ Very dilute solution of NaOH(aq)
−++ Charge balance: [OH ] =[H3O ]+[Na ]
+ −+ Material balance: [Na ]= [NaOH]initial ∴ [OH ] =+[H3O ] [NaOH]initial
+− + + Autoprotolysis constant: Kw ==[H3O ][OH ] [H3O ]([H3O ]+[NaOH]initial ) (17)
++2 Or [H3iO ] +−[NaOH] nitial[H3O ] Kw =0
24 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
Ex. 10.14 Finding the pH of a very dilute aqueous solution of a strong acid.
What is the pH of 8.0×10–8 M HCl(aq)?
−−81 + Let [HCl]initial =×8.0 10 mol⋅L and [H3O ] = x .
++2 From Eq.(16), [H3O ] −−[HCl]initial[H3O ] Kw =0
Æ xx28−×(8.0 10−−) −(1.0×10 14) =0
x = 1.5×10−7 or −×6.8 10−8
+ From pH =−log[H3O ] ,
pH =−log(1.5×10−7 ) =6.82
10.19 Very Dilute Solutions of Weak Acids
◆ Calculating the pH for very dilute solutions of a weak acid. HOW DO WE DO THAT?
– + – Four unknowns: HA, A , H3O , OH
Two equilibria:
[H O+ ][A− ] K = [H O+][OH−] K = 3 w 3 a [HA]
Charge balance:
+−− [H3O ] =+[OH ] [A ]
Material balance:
− [HA]initial =[HA]+[A ]
−+−+Kw [A][=−H33O][OH][=−HO] + [H3O ]
−+⎛⎞Kw [HA] =−[HA]initial [A ] =−[HA]initial ⎜⎟[H3O ]−+ [H O ] ⎝⎠3 + Kw =−[HA]initial [H3O ] ++ [H3O ]
25 2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 10
++⎛⎞Kw [H33O]⎜⎟[HO]− + ⎝⎠[H3O ] Ka = + Kw [HA]initial −+[H3O ] + [H3O ]
+−6 +−8 For many solutions of weak acids, [H3O ] ><10 (or pH 6) Æ Kw /[H3O ] < 10
+ 2 [H3O ] ∴ Ka + (18a) [HA]initial − [H3O ]
+−6 When the acid is so dilute that [H3O ] ≤10 (or 7 ++32 + [H3O ] +−KKaw[H3O ] ()+Ka[HA]initial [H3O ]−KaKw=0 (18b) Find the solutions of a cubic equation graphically (or numerically). Ex. 10.15 Estimating the pH of a dilute aqueous solution of a weak acid when the autoprotolysis of water must be considered. Estimate the pH of a 1.0 ×10−4 M aqueous phenol solution. + Let x = [H3O ]. −10 Value of Ka for phenol: Ka =×1.3 10 −−14 10 −4 −14 KKwa+=[HA]initial 1.0 ×10 +(1.3×10 ) ×(1.0×10 ) =2.3×10 −10 −−14 24 KKaw=×()1.3 10 ×(1.0 ×10 )=1.3×10 Eq.(18b) becomes xx31+×()1.3 10−−02−(2.3×10 14)x−1.3×10−24=0 Let xy=×10−7 and divide the resulting equation through by 10−21 . yy32+−0.0013 2.3y−0.0013 =0 Finding the positive root: y = 1.516 Æ x =×1.5 10−7 + pH =−log[H3O ] =−log x =−log()1.5×10−7 =6.82 26