Acid-Base Equilibria andSolubility Equilibria

Chapter 17

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산 - 염기 평형과 용해도 평형

1. 균일 대 불균일 용액 평형2. 완충용액3. 산 - 염기 적정 - 당량점4. 산 - 염기 지시약 - 종말점5. 용해도 평형6. 공통이온 효과와 용해도7. 착이온 평형과 용해도8. 정성분석에 대한 용해도곱 원리의 응용

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)

CH3COOH (aq) H+ (aq) + CH3COO- (aq)

common ion

Consider mixture of salt NaA and weak acid HA.

HA (aq) H+ (aq) + A- (aq)

NaA (s) Na+ (aq) + A- (aq)Ka =

[H+][A-][HA]

[H+] =Ka [HA]

[A-]

-log [H+] = -log Ka - log[HA]

[A-]

-log [H+] = -log Ka + log [A-][HA]

pH = pKa + log [A-][HA]

pKa = -log Ka

Henderson-Hasselbalch equation

pH = pKa + log[conjugate base]

[acid]

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?

Common ion effect

0.30 – x 0.30

0.52 + x 0.52

pH = pKa + log [HCOO-][HCOOH]

HCOOH pKa = 3.77

pH = 3.77 + log[0.52][0.30]

= 4.01

HCOOH (aq) H+ (aq) + HCOO- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.30 0.00

-x +x

0.30 - x

0.52

+x

x 0.52 + x

Mixture of weak acid and conjugate base!

A buffer solution ( 완충용액 ) is a solution of:

1. A weak acid or a weak base and

2. 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.

Add strong acid

H+ (aq) + CH3COO- (aq) CH3COOH (aq)

Add strong base

OH- (aq) + CH3COOH (aq) CH3COO- (aq) + H2O (l)

Consider an equal molar mixture of CH3COOH and CH3COONa

Pure water

0.005/0.005 AcH/Ac buffer

20 mL acetic acid/acetate(5/5 mM) buffer and water,With 0.1M HCl or 0.1 M NaOH

With 0.1M HCl 0.1 M NaOH

Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na2CO3/NaHCO3

(a) KF is a weak acid and F- is its conjugate base buffer solution

(b) HBr is a strong acidnot a buffer solution

(c) CO32- is a weak base and HCO3

- is it conjugate acidbuffer solution

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 0.050 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 = 9.25 + 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

pH = 9.25 + log[0.25][0.28]

[NH4+] =

0.0280.10

final volume = 80.0 mL + 20.0 mL = 100 mL

[NH3] = 0.0250.10

Chemistry In Action: Maintaining the pH of Blood

Titrations( 적정 , 滴定 )

Equivalence point( 당량점 ) – the point at which the reaction is complete

Indicator – substance that changes color at (or near) the equivalence point

Slowly add known titrant to unknown solution(titrand)UNTIL

the indicator changes color

물방울목적 : 당량점을 찾아내는 것

방법 : 종말점을 이용한다

End point( 종말점 ) – the point at which the indicator changes color

Strong Acid-Strong Base Titrations

NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq)

OH- (aq) + H+ (aq) H2O (l)equivalence point

•

phenolphthalein

phenolphthalein

methyl orange

methyl orange

Weak Acid-Strong Base Titrations

CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)

CH3COOH (aq) + OH- (aq) CH3COO- (aq) + H2O (l)

CH3COO- (aq) + H2O (l) OH- (aq) + CH3COOH (aq)

At equivalence point (pH > 7):

18.919 6.00019.297 6.19519.545 6.39019.707 6.58519.812 6.78019.880 6.97519.923 7.17019.951 7.36519.969 7.56019.980 7.75519.988 7.95019.992 8.14519.996 8.34019.998 8.53520.000 8.73020.002 8.92520.004 9.12020.008 9.31420.013 9.50920.020 9.70420.032 9.89920.050 10.09420.079 10.28920.123 10.48420.193 10.67920.304 10.87420.478 11.06920.754 11.264 equivalence point

•

phenolphthaleinphenolphthalein

methyl orange

methyl orange

Strong Acid-Weak Base Titrations

HCl (aq) + NH3 (aq) NH4Cl (aq)

NH4+ (aq) + H2O (l) NH3 (aq) + H+ (aq)

At equivalence point (pH < 7):

H+ (aq) + NH3 (aq) NH4+ (aq)

equivalence point

•

phenolphthalein

phenolphthalein

methyl orangemethyl orange

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)

0.01

0.0 0.0 0.01

NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.05 0.00

-x +x

0.05 - x

0.00

+x

x x

[NO2-] =

0.010.200 = 0.05 MFinal 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

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)

Initial (M)

Change (M)

Eq (M)

0.05 0.00

-x +x+y

0.05 - x

0.00

+x

x+y x

Kb =[OH-][HNO2]

[NO2-]

=x(x+y)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

H2O (l) OH- (aq) + H+(aq)

0.00

y

Kw = [OH-][H+] = y(x+y) = 1.0x 10-14

Acid-Base Indicators

HIn (aq) H+ (aq) + In- (aq)

10[HIn]

[In-]Color of acid (HIn) predominates

10[HIn]

[In-]Color of conjugate base (In-) predominates

Red cabbage juice and pH

pH

Phenolphthalein

HIn In-

Indicator Low pH colorTransition pH range

High pH color

Thymol blue (first transition) red 1.2–2.8 yellow

Thymol blue (second transition) yellow 8.0–9.6 blue

Methyl yellow red 2.9–4.0 yellow

Bromophenol blue yellow 3.0–4.6 purple

Congo red blue-violet 3.0–5.0 red

Methyl orange red 3.1–4.4 orange

Bromocresol green yellow 3.8–5.4 blue-green

Methyl red red 4.4–6.2 yellow

Bromocresol purple yellow 5.2–6.8 purple

Bromothymol blue yellow 6.0–7.6 blue

Phenol red yellow 6.8–8.4 red

Naphtholphthalein colorless to reddish 7.3–8.7 greenish to blue

Cresol Red yellow 7.2–8.8 reddish-purple

Phenolphthalein colorless 8.3–10.0 fuchsia

Thymolphthalein colorless 9.3–10.5 blue

Alizarine Yellow R yellow 10.2–12.0 red

Litmus red 4.5-8.3 blue

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

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) 2Ag+ (aq) + CO32- (aq) Ksp = [Ag+]2[CO3

2-]

Ca3(PO4)2 (s) 3Ca2+ (aq) + 2PO43- (aq) Ksp = [Ca2+]3[PO3

3-]2

Dissolution of an ionic solid in aqueous solution:

Q = Ksp Saturated solution

Q < Ksp Unsaturated solution No precipitate

Q > Ksp Supersaturated solution Precipitate will form

4.2

Na+ K+ NH4+ Al3+ H+ Mg2+ Ca2+ Sr2+ Ba2+ Cr3+ Fe3+ Fe2+ Co2+ Ni2+ Cu2+ Zn2+ Cd2+ Hg2+ Hg2

2+ Ag+ Pb2+ As3+ Sb3+ Bi3+

CH3CO2- S S S S S S S S S S U S S S S S S S I I S U U I

Br- S S S S S S S S S S S S S S S S S I Ia Ib I D D D

Cl- S S S S S S S S S S S S S S S S S S Ib Ib I D S D

I- S S S S S S S S S I U S S S U S S Ia Ia I Ia S D I

ClO3- S S S S S S S S S U U U S I S S S S S S S U U U

NO3- S S S S S S S S S S S S S S S S S S D S S U U D

CrO42- S S S U S S S Ia Ia U S I I U S Ia I Ia Ia Ia Ia U U U

SO42- S S S S S S Ia I Ib S S S S S S S S D I I I U D D

SO32- S S S U S S I I I I U I I I U I I U U I I U U U

C2O42- S S I I S I I I I S S I I I I I I I I I I U I D

PO43- S S S Ia S Ia Ia Ia Ia I Ia Ia I I Ia Ia I U U Ia Ia U U I

CO32- S S S U S Ia I Ia I U U I I I I Ia I I Ia Ia Ia U U U

SiO32- S S U I I Ia Ia Ia S U U I I U U Ia I U U U Ia U U I

O2- D D U Ib H2O Ia I I S I Ia Ia I I Ia I I I Ia I I I I I

OH- S S U Ia H2O Ia Ia I S I Ia Ia I I Ia Ia I U U U Ia U U D

S2- S S S D S D Ia I D I I Ia I I Ia Ia I I I Ia Ia I D I

Ia Soluble in AcidsIb Slightly Soluble in Acids

H2O Produces water

S Soluble I InsolubleD=Decomposes in

waterU=Compound does not exist or is

unstable

Solubility of Ionic Compounds in Water

Salt Ksp Salt Ksp 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 1.2 x 10-5 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 1.2 x 10-10

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

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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.

What is the solubility of silver chloride in g/L ?Ksp = 1.6x10-10, M(AgCl) = 143.35

0.00AgCl (s) Ag+ (aq) + Cl- (aq)

Ksp = [Ag+][Cl-]Initial (M)

Change (M)

Equilibrium (M)

+s

0.00

+s

s s

Ksp = s2

s = Ksps = 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 soln143.35 g AgCl

1 mol AgClx = 1.9 x 10-3 g/L

Ksp = 1.6 x 10-10

If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 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 = 0.100 M [OH-]0 = 4.0 x 10-4 M

Ksp = [Ca2+][OH-]2 = 8.0 x 10-6

Q = [Ca2+]0[OH-]02 = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8

Q < Ksp No precipitate will form

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 10-11 M

[Ag+] = Ksp

[Br-]1.6 x 10-10

0.020= = 8.0 x 10-9 M

3.9 x 10-11 M < [Ag+] < 8.0 x 10-9 M

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) 0.0010 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-] = 0.0010 M

AgBr (s) Ag+ (aq) + Br- (aq)

[Ag+] = s

[Br-] = 0.0010 + s 0.0010

Ksp = 0.0010 x s

s = 7.7 x 10-10

16.8

pH and Solubility

• 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)

Ksp = [Mg2+][OH-]2 = 1.2 x 10-11

Ksp = (s)(2s)2 = 4s3

4s3 = 1.2 x 10-11

s = 1.4 x 10-4 M

[OH-] = 2s = 2.8 x 10-4 M

pOH = 3.55 pH = 10.45

At pH less than 10.45

Lower [OH-]

OH- (aq) + H+ (aq) H2O (l)

Increase solubility of Mg(OH)2

At pH greater than 10.45

Raise [OH-]

Decrease solubility of Mg(OH)2

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-

Kf =[CoCl4 ]

[Co2+][Cl-]4

2-

The formation constant or stability constant (Kf) is the equilibrium constant for the complex ion formation.

Co(H2O)62+ CoCl4

2-

Kf

stability of complex

Qualitative Analysis of

Cations

Flame Test for Cations

lithium sodium potassium copper

Chemistry In Action: How an Eggshell is Formed

Ca2+ (aq) + CO32- (aq) CaCO3 (s)

H2CO3 (aq) H+ (aq) + HCO3- (aq)

HCO3- (aq) H+ (aq) + CO3

2- (aq)

CO2 (g) + H2O (l) H2CO3 (aq)carbonic

anhydrase