1. Acid-Base Equilibria and Solubility Equilibria
Chapter 17
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2. 산-염기 평형과 용해도 평형 균일 대 불균일 용액 평형 완충용액 산-염기 적정 - 당량점 산-염기 지시약 - 종말점
공통이온 효과와 용해도
착이온 평형과 용해도
정성분석에 대한 용해도곱 원리의 응용

3. CH3COONa (s) Na+ (aq) + CH3COO- (aq)
The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.
The presence of a common ion suppresses the ionization of a weak acid or a weak base.
Consider mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid).
CH3COONa (s) Na+ (aq) + CH3COO- (aq)
common ion
CH3COOH (aq) H+ (aq) + CH3COO- (aq)

4. Henderson-Hasselbalch equation
Consider mixture of salt NaA and weak acid HA.
NaA (s) Na+ (aq) + A- (aq)
Ka =
[H+][A-]
[HA]
HA (aq) H+ (aq) + A- (aq)
[H+] =
Ka [HA]
[A-]
Henderson-Hasselbalch equation
-log [H+] = -log Ka - log
[HA]
[A-]
pH = pKa + log
[conjugate base]
[acid]
-log [H+] = -log Ka + log
[A-]
[HA]
pH = pKa + log
[A-]
[HA]
pKa = -log Ka

5. What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?
HCOOH (aq) H+ (aq) + HCOO- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.30
0.00 -x +x x
0.52 x x
Mixture of weak acid and conjugate base!
Common ion effect
0.30 – x  0.30
x  0.52
pH = pKa + log
[HCOO-]
[HCOOH]
HCOOH pKa = 3.77
pH = log
[0.52]
[0.30]
= 4.01

6. A buffer solution (완충용액) is a solution of:
A weak acid or a weak base and
The salt of the weak acid or weak base
Both must be present!
A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.
Consider an equal molar mixture of CH3COOH and CH3COONa
Add strong acid
H+ (aq) + CH3COO- (aq) CH3COOH (aq)
Add strong base
OH- (aq) + CH3COOH (aq) CH3COO- (aq) + H2O (l)

7. 20 mL acetic acid/acetate(5/5 mM) buffer and water,
With 0.1M HCl or 0.1 M NaOH
Pure water
0.005/0.005 AcH/Ac buffer
With 0.1M HCl
0.1 M NaOH

8. Which of the following are buffer systems? (a) KF/HF
(b) KBr/HBr, (c) Na2CO3/NaHCO3
KF is a weak acid and F- is its conjugate base
buffer solution
(b) HBr is a strong acid
not a buffer solution
(c) CO32- is a weak base and HCO3- is it conjugate acid
buffer solution

9. Calculate the pH of the 0. 30 M NH3/0. 36 M NH4Cl buffer system
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of M NaOH to 80.0 mL of the buffer solution?
NH4+ (aq) H+ (aq) + NH3 (aq)
= 9.20
pH = pKa + log
[NH3]
[NH4+]
pKa = 9.25
pH = log
[0.30]
[0.36]
= 9.17
NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)
start (moles)
end (moles)
0.029
0.001
0.024
0.028
0.0
0.025
[0.25]
[0.28]
[NH4+] =
0.10
final volume = 80.0 mL mL = 100 mL
[NH3] =

10. Chemistry In Action: Maintaining the pH of Blood

11. Titrations(적정, 滴定) 물방울 목적:당량점을 찾아내는 것 Equivalence point(당량점)
– the point at which the reaction is complete
방법:종말점을 이용한다
End point(종말점)
– the point at which the indicator changes color
Indicator – substance that changes color at (or near) the
equivalence point
Slowly add known titrant to unknown solution(titrand)
UNTIL
the indicator changes color

12. Strong Acid-Strong Base Titrations
NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq)
OH- (aq) + H+ (aq) H2O (l)
equivalence point
phenolphthalein •
methyl orange

13. Weak Acid-Strong Base Titrations
CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)
CH3COOH (aq) + OH- (aq) CH3COO- (aq) + H2O (l)
At equivalence point (pH > 7):
CH3COO- (aq) + H2O (l) OH- (aq) + CH3COOH (aq)
18.919
6.000
19.297
6.195
19.545
6.390
19.707
6.585
19.812
6.780
19.880
6.975
19.923
7.170
19.951
7.365
19.969
7.560
19.980
7.755
19.988
7.950
19.992
8.145
19.996
8.340
19.998
8.535
20.000
8.730
20.002
8.925
20.004
9.120
20.008
9.314
20.013
9.509
20.020
9.704
20.032
9.899
20.050
10.094
20.079
10.289
20.123
10.484
20.193
10.679
20.304
10.874
20.478
11.069
20.754
11.264
phenolphthalein •
methyl orange
equivalence point

14. Strong Acid-Weak Base Titrations
HCl (aq) + NH3 (aq) NH4Cl (aq)
H+ (aq) + NH3 (aq) NH4+ (aq)
phenolphthalein
At equivalence point (pH < 7):
NH4+ (aq) + H2O (l) NH3 (aq) + H+ (aq) •
methyl orange
equivalence point

15. Exactly 100 mL of 0. 10 M HNO2 are titrated with a 0
Exactly 100 mL of 0.10 M HNO2 are titrated with a 0.10 M NaOH solution. What is the pH at the equivalence point ?
0.01
HNO2 (aq) + OH- (aq) NO2- (aq) + H2O (l)
start (moles)
end (moles)
0.0
NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.05
0.00 -x +x x x
[NO2-] =
0.200
= 0.05 M
Final volume = 200 mL
Kb =
[OH-][HNO2]
[NO2-] = x2
0.05-x
= 2.2 x 10-11
0.05 – x  0.05
x  1.05 x 10-6
= [OH-]
pOH = 5.98
pH = 14 – pOH = 8.02

16. Exactly 100 mL of 0. 10 M HNO2 are titrated with a 0
Exactly 100 mL of 0.10 M HNO2 are titrated with a 0.10 M NaOH solution. What is the pH at the equivalence point ?
NO2- (aq) + H2O (l) OH- (aq) HNO2 (aq)
H2O (l) OH- (aq) + H+(aq)
Initial (M)
Change (M)
Eq (M)
0.05
0.00
0.00
0.00 -x
+x+y +x x
x+y y x
Kb =
[OH-][HNO2]
[NO2-] =
x(x+y)
0.05-x
= 2.2 x 10-11
pOH = 5.98
Kw =
[OH-][H+] =
y(x+y)
= 1.0x 10-14
pH = 14 – pOH = 8.02
0.05 – x  0.05
x  1.05 x 10-6
= [OH-]

17. Acid-Base Indicators HIn (aq) H+ (aq) + In- (aq) [HIn]  10
Color of acid (HIn) predominates
 10
[HIn]
[In-]
Color of conjugate base (In-) predominates
Red cabbage juice and pH pH

18. Phenolphthalein
HIn
In-

19. Indicator Low pH color Transition pH range High pH color
Thymol blue (first transition)
red
1.2–2.8
yellow
Thymol blue (second transition)
8.0–9.6
blue
Methyl yellow
2.9–4.0
Bromophenol blue
3.0–4.6
purple
Congo red
blue-violet
3.0–5.0
Methyl orange
3.1–4.4
orange
Bromocresol green
3.8–5.4
blue-green
Methyl red
4.4–6.2
Bromocresol purple
5.2–6.8
Bromothymol blue
6.0–7.6
Phenol red
6.8–8.4
Naphtholphthalein
colorless to reddish
7.3–8.7
greenish to blue
Cresol Red
7.2–8.8
reddish-purple
Phenolphthalein
colorless
8.3–10.0
fuchsia
Thymolphthalein
9.3–10.5
Alizarine Yellow R
10.2–12.0
Litmus

20. Which indicator(s) would you use for a titration of HNO2 with KOH ?
Weak acid titrated with strong base.
At equivalence point, will have conjugate base of weak acid.
At equivalence point, pH > 7
Use cresol red or phenolphthalein

21. Solubility Equilibria
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = [Ag+][Cl-]
Ksp is the solubility product constant
MgF2 (s) Mg2+ (aq) + 2F- (aq)
Ksp = [Mg2+][F-]2
Ag2CO3 (s) Ag+ (aq) + CO32- (aq)
Ksp = [Ag+]2[CO32-]
Ca3(PO4)2 (s) Ca2+ (aq) + 2PO43- (aq)
Ksp = [Ca2+]3[PO33-]2
Dissolution of an ionic solid in aqueous solution:
Q < Ksp
Unsaturated solution
No precipitate
Q = Ksp
Saturated solution
Q > Ksp
Supersaturated solution
Precipitate will form

22. 4.2

23. U=Compound does not exist or is unstable Slightly Soluble in Acids
Solubility of Ionic Compounds in Water S
Soluble I
Insoluble
D=Decomposes in water
U=Compound does not exist or is unstable
Na+ K+
NH4+
Al3+ H+
Mg2+
Ca2+
Sr2+
Ba2+
Cr3+
Fe3+
Fe2+
Co2+
Ni2+
Cu2+
Zn2+
Cd2+
Hg2+
Hg22+
Ag+
Pb2+
As3+
Sb3+
Bi3+
CH3CO2- S U I
Br- Ia Ib D
Cl- I-
ClO3-
NO3-
CrO42-
SO42-
SO32-
C2O42-
PO43-
CO32-
SiO32-
O2-
H2O
OH-
S2- Ia
Soluble in Acids Ib
Slightly Soluble in Acids
H2O
Produces water

24. Solubility Product Constant (Ksp) Values at 25 oC
Salt
Ksp
Bromides
Carbonates
Oxalates
PbBr2
6.6 x 10-6
MgCO3
6.8 x 10-6
MgC2O4
4.8 x 10-6
CuBr
6.3 x 10-9
NiCO3
1.3 x 10-7
FeC2O4
2 x 10-7
AgBr
5.4 x 10-13
CaCO3
5.0 x 10-9
NiC2O4
1 x 10-7
Hg2Br2
6.4 x 10-23
SrCO3
5.6 x 10-10
SrC2O4
5 x 10-8
Chlorides
MnCO3
2.2 x 10-11
CuC2O4
3 x 10-8
PbCl2
1.2 x 10-5
CuCO3
2.5 x 10-10
BaC2O4
1.6 x 10-7
CuCl
1.7 x 10-7
CoCO3
1.0 x 10-10
CdC2O4
1.4 x 10-8
AgCl
1.8 x 10-10
FeCO3
2.1 x 10-11
ZnC2O4
1.4 x 10-9
Hg2Cl2
1.4 x 10-18
ZnCO3
1.2 x 10-10
CaC2O4
2.3 x 10-9
Fluorides
Ag2CO3
8.1 x 10-12
Ag2C2O4
3.5 x 10-11
BaF2
1.8 x 10-7
CdCO3
6.2 x 10-12
PbC2O4
4.8 x 10-12
MgF2
7.4 x 10-11
PbCO3
7.4 x 10-14
Hg2C2O4
1.8 x 10-13
SrF2
2.5 x 10-9
Hydroxides
MnC2O4
1 x 10-15
CaF2
1.5 x 10-10
Ba(OH)2
5.0 x 10-3
Phosphates
Iodides
Sr(OH)2
6.4 x 10-3
Ag3PO4
8.9 x 10-17
PbI2
8.5 x 10-9
Ca(OH)2
4.7 x 10-6
AlPO4
9.8 x 10-21
CuI
1.1 x 10-12
Mg(OH)2
5.6 x 10-12
Mn3(PO4)2
1 x 10-22
AgI
8.5 x 10-17
Mn(OH)2
2.1 x 10-13
Ba3(PO4)2
3 x 10-23
Hg2I2
4.5 x 10-29
Cd(OH)2
5.3 x 10-15
BiPO4
1.3 x 10-23
Sulfates
Pb(OH)2
1.2 x 10-15
Sr3(PO4)2
4 x 10-28
CaSO4
7.1 x 10-5
Fe(OH)2
4.9 x 10-17
Pb3(PO4)2
7.9 x 10-43
Ag2SO4
Ni(OH)2
5.5 x 10-16
Chromates
Hg2SO4
6.8 x 10-7
Co(OH)2
1.1 x 10-15
CaCrO4
7.1 x 10-4
SrSO4
3.5 x 10-7
Zn(OH)2
4.1 x 10-17
SrCrO4
2.2 x 10-5
PbSO4
1.8 x 10-8
Cu(OH)2
1.6 x 10-19
Hg2CrO4
2.0 x 10-9
BaSO4
1.1 x 10-10
Hg(OH)2
3.1 x 10-26
BaCrO4
Acetates
Sn(OH)2
5.4 x 10-27
Ag2CrO4
2.0 x 10-12
Ag(CH3COO)
4.4 x 10-3
Cr(OH)3
6.7 x 10-31
PbCrO4
2.8 x 10-13
Hg2(CH3COO)2
4 x 10-10
Al(OH)3
1.9 x 10-33
Fe(OH)3
2.6 x 10-39
Solubility Product Constant (Ksp) Values at 25 oC

25. Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a saturated solution.
Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution.

26.  What is the solubility of silver chloride in g/L ?
Ksp = 1.6x10-10, M(AgCl) =
0.00
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = [Ag+][Cl-]
Initial (M)
Change (M)
Equilibrium (M) +s s
Ksp = s2
s = Ksp 
s = 1.3 x 10-5
[Ag+] = 1.3 x 10-5 M
[Cl-] = 1.3 x 10-5 M
Solubility of AgCl =
1.3 x 10-5 mol AgCl
1 L soln
g AgCl
1 mol AgCl x
= 1.9 x 10-3 g/L
Ksp = 1.6 x 10-10

28. The ions present in solution are Na+, OH-, Ca2+, Cl-.
If 2.00 mL of M NaOH are added to 1.00 L of M CaCl2, will a precipitate form? Ksp = 8.0x10-6
The ions present in solution are Na+, OH-, Ca2+, Cl-.
Only possible precipitate is Ca(OH)2 (solubility rules).
Is Q > Ksp for Ca(OH)2?
[Ca2+]0 = M
[OH-]0 = 4.0 x 10-4 M
Ksp = [Ca2+][OH-]2 = 8.0 x 10-6
Q = [Ca2+]0[OH-]0 2
= 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8
Q < Ksp
No precipitate will form

29. What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br- and Cl- at a concentration of 0.02 M? Ksp = 7.7x10-13
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = [Ag+][Cl-]
Ksp = 1.6 x 10-10
AgBr (s) Ag+ (aq) + Br- (aq)
Ksp = 7.7 x 10-13
Ksp = [Ag+][Br-]
[Ag+] =
Ksp
[Br-]
7.7 x 10-13
0.020 =
= 3.9 x M
1.6 x 10-10
= 8.0 x 10-9 M
3.9 x M < [Ag+] < 8.0 x 10-9 M

30. The Common Ion Effect and Solubility
The presence of a common ion decreases the solubility of the salt.
What is the molar solubility of AgBr in (a) pure water and (b) M NaBr? Ksp = 7.7x10-13
AgBr (s) Ag+ (aq) + Br- (aq)
Ksp = 7.7 x 10-13
s2 = Ksp
s = 8.8 x 10-7
NaBr (s) Na+ (aq) + Br- (aq)
[Br-] = M
[Ag+] = s
[Br-] = s 
Ksp = x s
s = 7.7 x 10-10
16.8

31. pH and Solubility Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq)
The presence of a common ion decreases the solubility.
Insoluble bases dissolve in acidic solutions
Insoluble acids dissolve in basic solutions
Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq)
At pH less than 10.45
Ksp = [Mg2+][OH-]2 = 1.2 x 10-11
Lower [OH-]
Ksp = (s)(2s)2 = 4s3
OH- (aq) + H+ (aq) H2O (l)
4s3 = 1.2 x 10-11
Increase solubility of Mg(OH)2
s = 1.4 x 10-4 M
[OH-] = 2s = 2.8 x 10-4 M
At pH greater than 10.45
pOH = pH = 10.45
Raise [OH-]
Decrease solubility of Mg(OH)2

32. Complex Ion Equilibria and Solubility
A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions.
Co2+ (aq) + 4Cl- (aq) CoCl4 (aq) 2-
The formation constant or stability constant (Kf) is the equilibrium constant for the complex ion formation.
Kf =
[CoCl4 ]
[Co2+][Cl-]4 2-
Co(H2O)6 2+
CoCl4 2-
stability of complex Kf

35. Qualitative Analysis of Cations

36. Flame Test for Cations
lithium
sodium
potassium
copper

37. Chemistry In Action: How an Eggshell is Formed
Ca2+ (aq) + CO32- (aq) CaCO3 (s)
CO2 (g) + H2O (l) H2CO3 (aq)
carbonic
anhydrase
H2CO3 (aq) H+ (aq) + HCO3- (aq)
HCO3- (aq) H+ (aq) + CO32- (aq)