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Acids & Bases
1 © Prof. Zvi C. Koren 20.07.2010
Definitions
Arrhenius
Acid releases H+ in water: HCl(aq) H+ + Cl-
Base releases OH- in water: NaOH(aq) Na+ + OH-
Brønsted-Lowry (don’t need water)
Acid donates proton (H+)
Base accepts proton
HCl(g) + NH3(g) NH4+Cl-(s)
Lewis (don’t need protons)
Acid accepts an e-pair for sharing
Base donates an e-pair for sharing
BF3 + :NH3 F3B–NH3
coordinate covalent bond
2 © Prof. Zvi C. Koren 20.07.2010
Brønsted-Lowry
Conjugate Acid-Base Pairs
All Bronsted-Lowry acid-base reactions consist of 2 conjugate pairs
HF(aq) + NH3(aq) NH4+ + F-
acid base acid base
HCO3- + H2O H3O+ + CO3
2-
(hydronium ion)
acid base acid base
HCO3- + H2O OH- + H2CO3
base acid base acid
Conjugate acid-base pairs: HA/A-, B/BH+
difference within a pair is H+
amphotericamphiprotic
3 © Prof. Zvi C. Koren 20.07.2010
H2O + H2O H3O+ + OH-
or
H2O H+ + OH-
Water ionization constant Kw = [H+][OH-]
= 1.0 x 10-14 @ 25oC
In neutral solutions and pure water: [H+] = [OH-] = 1.0 x 10-7 M
In acidic solutions: [H+] > [OH-]
In basic solutions: [H+] < [OH-]but [H+][OH-] = 1.0 x 10-14
(Note: H3O+ and H+ are the same.)
The Acid-Base Properties of Water:
Auto-Ionization
4 © Prof. Zvi C. Koren 20.07.2010
Sorensen: “The potential (or power) of H”
pH –log[H+]
[H+] = 10–pH
Also:
pOH –log[OH–]
and also:
pK –logK
From before:
[H+][OH–] = Kw
log[H+] + log[OH–] = logKw
–log[H+] + –log[OH–] = –logKw
Note about pH and Significant Figures:
If [H+] = 1.0 x 10–9 M pH = 9.002 sig. figs. 2 sig. figs.
Quick calculation regarding pH:
Sig. Figs. begin after
the decimal point!
If [H+] = 5.0 x 10–9 M pH = 9 – log5.0
(Note: small “p” big “H”)
pH + pOH = pKw = 14.00@ 25oC
The pH Scale
5 © Prof. Zvi C. Koren 20.07.2010
For a neutral solution at 25 oC: pH = pOH = 7.00
For acidic solutions at 25 oC: pH < 7.00
For basic solutions, at 25 oC: pH > 7.00
Strong Acids
HClO4 perchloric acid
HClO3 chloric acid
HCl hydrochloric acid
H2SO4 sulfuric acid (1st proton)
HNO3 nitric acid
H3O+ hydronium ion
Strong Bases
Metal hydroxides:
MOH (from Group I) – very strong
M(OH)2 (from Group II) – strong
Others (ions):
CH3- methide ion
H- hydride ion
NH2- amide ion
C2H5O- ethoxide ion
OH- hydroxide ion
HClO4(aq) H+ + ClO4-
orHClO4(aq) + H2O(l) H3O
+ + ClO4-
Dissociates “completely”
to release H+
K >> 1
NaOH(aq) Na+ + OH-
Ca(OH)2(aq) Ca2+ + 2OH-
Dissociates “completely” to
release OH-: K >> 1
Strong Acids & Bases
(In oxyacids, the acidic H is bonded to O)6 © Prof. Zvi C. Koren 20.07.2010
Weak Acids & Bases
Weak AcidsHA(aq) + H2O H3O+ + A-
1 ][
]][[ constant ionization acid
HA
AHKa
Weak BasesB(aq) + H2O BH+ + OH-
1 ][
]][[ constant hydrolysis base
B
OHBHKb
HAc(aq) H+ + Ac- (HAc or HOAc is HC2H3O2, “acetic acid”)
NH3(aq) + H2O NH4+ + OH-
5-1.8x10 ][
]][[ )(
HAc
AcHHAcKa
5-
3
43 1.8x10
][
]][[ )(
NH
OHNHNHKb
HA(aq) H+ + A-
7 © Prof. Zvi C. Koren 20.07.2010
8 © Prof. Zvi C. Koren 20.07.2010
Compare the acid strengths of HNO3 vs. HNO2 and H2SO4 vs. H2SO3
Relative Strengths of Conjugate Acid-Base Pairs
HCl + H2O H3O+ + Cl-
ACID BASE acid base
Acid Base
HCl Cl-
H3O+ H2O
HAc + H2O H3O+ + Ac-
acid base ACID BASE
HAc Ac-
H2O OH-CH3NH2 + H2O CH3NH3
+ + OH-
base acid ACID BASE
HAc + CH3NH2 CH3NH3+ + Ac-
ACID BASE acid base
CH3NH3+ CH3NH2
9 © Prof. Zvi C. Koren 20.07.2010
Determining Ka & Kb values from pH measurements - 1
0.0300 M HCOOH (formic acid)
HCOOH(aq) H+ + COOH-
][
]][[
HCOOH
COOHHKa
Use “ICE” Box Method:
pH=2.66
[H+]=2.2x10-3M
I
C
E
HCOOH H+ COOH-M
0.0300-x x=2.2x10-3 x
0.0300 - -
-x +x +x
4
3
232
107.1102.20300.0
)102.2(
0300.0][
]][[
xx
x
x
x
HCOOH
COOHHKa
Ka of Weak Acid
10 © Prof. Zvi C. Koren 20.07.2010
Determining Ka & Kb values from pH measurements - 2
C6H5NH2(aq) + H2O C6H5NH3+ + OH-
x
x
NHHC
OHNHHCK b
15.0][
]][[
2
256
356
The pH of a 0.15-M solution of aniline (C6H5NH2) is 8.89.
Calculate the Kb of aniline:
0.15
0.15-x x x
Need to calculate x = [OH-] from pH (two methods):
pH [H+] [OH-] or pH pOH [OH-]
[OH-] = 7.8x10-6 M
Kb = 4.1x10-10
Kb of Weak Base
11 © Prof. Zvi C. Koren 20.07.2010
Calculation of pH of Strong Acids & Bases
0.10-M HNO3:
HNO3(aq) H+ + NO3-
0.10
0.10 0.10
pH = -log[H+] = 1.00
0.00020-M Ca(OH)2:
Ca(OH)2(aq) Ca2+ + 2OH-2.0x10-4
2.0x10-4 4.0x10-4
[OH-] = 4.0x10-4 M
[OH-] pOH pH
or
[OH-] [H+] pH
pH = 10.60
Strong Base:
Strong Acid:
12 © Prof. Zvi C. Koren 20.07.2010
Weak Acid
Calculate the pH of a 0.10-M HAc solution:
HAc(aq) H+ + Ac-,0.10
x x0.10-x
MxHxx
x
x
HAc
AcHKx a
322
5 103.1][ 10.010.0][
]][[108.1
pH=2.89
Weak Base
Calculate the pH of a 0.10-M NH3 solution:
NH3(aq) + H2O NH4+ + OH-,
0.10
xx0.10-x
MxOHxx
x
x
NH
OHNHKx b
322
3
45 103.1][ 10.010.0][
]][[108.1
[OH-] [H+] pH
pOH pH= 11.11
Calculation of pH of Weak Acids & Bases
13 © Prof. Zvi C. Koren 20.07.2010
+
Acid Base
Salt
Water
The Birth of a Salt: A Family Portrait
Koren’s Acid-Base Family Rule:The Strength of the Baby-Salt
will be according to the Stronger Parent
Acid-Base Properties of Salts – Qualitative
Hydrolysis of Salts
Reminder: The rxn between an acid and a base produces a SALT.
For example, HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
(Daddy) (Mommy) (Baby)
Types of Salts:
(Which are acidic, basic, or neutral?)
1. Salt from a strong acid & a strong base: NaCl
2. Salt from a weak acid & a strong base: NaAc
3. Salt from a strong acid & a weak base: NH4Cl
4. Salt from a weak acid & a weak base: NH4Ac
Koren’s Family Rule:
The nature of the baby is influenced by the stronger parent!
15 © Prof. Zvi C. Koren 20.07.2010
Hydrolysis of Salts
1. From st A + st B: NaCl
Dissociation of Dissolved Salt: NaCl(aq) Na+ + Cl-
Hydrolysis of Ions: Na+ + H2O NaOH(aq) + H+?
?
XX
2. From w A + st B: NaAc
Dissociation of Dissolved Salt: NaAc(aq) Na+ + Ac-
Hydrolysis of Ions: Na+ + H2O X
Cl- + H2O HCl(aq) + OH-
NaCl is a NEUTRAL salt.
Ac- + H2O HAc(aq) + OH-
NaAc is a BASIC salt.
16 © Prof. Zvi C. Koren 20.07.2010
Hydrolysis of Salts (continued)
3. From st A + w B: NH4Cl
Dissociation of Dissolved Salt: NH4Cl(aq) NH4+ + Cl-
Hydrolysis of Ions: Cl- + H2O X
4. From w A + w B: NH4Ac
Dissociation of Dissolved Salt: NH4Ac(aq) NH4+ + Ac-
Hydrolysis of Ions: NH4+ + H2O NH3(aq) + H3O+
Ac- + H2O HAc(aq) + OH-
NH4+ + H2O NH3(aq) + H3O+
NH4Cl is an ACIDIC salt.
NH4Ac will depend on which is stronger (or less weak)
17 © Prof. Zvi C. Koren 20.07.2010
Kb of Conjugate Base
1. A- + H2O HA + OH-, Kb(A-)
2. HA + H2O H3O+ + A-, Ka(HA)
3. H2O + H2O H3O+ + OH-, Kw= Ka(HA) • Kb(A-)
Must be conjugates
Also,
Kw = Kb(B) • Ka(BH+),
For example,
Kw = Kb(NH3) • Ka(NH4+)
Reminder:
Ka refers to rxn of “a + H2O” and NOT with anything else!
Kb refers to rxn of “b + H2O” and NOT with anything else!
18 © Prof. Zvi C. Koren 20.07.2010
Acid-Base Properties of Salts – Quantitative - 1
1. From st A + st B: NaCl
Calculate the pH of a 0.10-M NaCl solution:
Dissociation of Dissolved Salt: NaCl(aq) Na+ + Cl-
Hydrolysis of Ions: Na+ + H2O
XX
Cl- + H2O
only H2O produces significant amounts of H+ and OH- :
H2O H+ + OH-, Kw = [H+][OH-] = 1.0x10-14
And [H+] = [OH-] = 1.0x10-7 M pH = 7.00
19 © Prof. Zvi C. Koren 20.07.2010
Acid-Base Properties of Salts – Quantitative - 2
2. From w A + st B: NaAc
Calculate the pH of a 0.10-M NaAc solution:
Dissociation of Dissolved Salt: NaAc(aq) Na+ + Ac-
Hydrolysis of Ions: Na+ + H2O X
Ac- + H2O HAc(aq) + OH-
0.1
0.1 0.1
0.1
0.1-x x x
Kb of Ac-:10
5
14
106.5 1081
1001
)( )(
x
x.
x.
HAcK
KAcK
a
w
b
= 5.6x10-10 x = [OH-] = 7.5x10-6 M
pH = 8.88x
x
Ac
OHHAcK b
10.0][
]][[ 2
Is a solution containing HAc and NaAc, acidic, basic, or neutral?
20 © Prof. Zvi C. Koren 20.07.2010
Acid-Base Properties of Salts – Quantitative - 3
3. From st A + w B: NH4Cl
Calculate the pH of a 0.10-M NH4Cl solution:
Dissociation of Dissolved Salt: NH4Cl(aq) NH4+ + Cl-
Hydrolysis of Ions: Cl- + H2O X
NH4+ + H2O NH3 (aq) + H3O+
0.1
0.1 0.1
0.1
0.1-x x x
Ka of NH4+:
10
5
14
3
4 106.5 1081
1001
)( )(
xx.
x.
NHK
KNHK
b
w
a
= 5.6x10-10 x = [H3O
+ ]= 7.5x10-6 M
pH = 5.12
x
x
NH
OHNHKa
10.0][
]][[ 2
4
33
21 © Prof. Zvi C. Koren 20.07.2010
Polyprotic Acids & Bases
Acids:H2SO4 (sulfuric acid)
H2SO3 (sulfurous acid)
H2C2O4 (oxalic acid)
H2CO3 (carbonic acid)
H3PO4 (phosphoric acid)
others
Bases:SO3
2- (sulfite ion)
C2O42- (oxalate ion)
CO32- (carbonate ion)
PO43- (phosphate ion)
HPO42- (hydrogen phosphate ion)
others
H3PO4(aq) H+ + H2PO4- , Ka1 = 7.5x10-3
H2PO4- H+ + HPO4
2- , Ka2 = 6.2x10-8
HPO42- H+ + PO4
3- , Ka3 = 3.6x10-13
CO32- + H2O HCO3
- + OH-, Kb1 = Kb(CO32-) = Kw/Ka(HCO3
-) = Kw/K2(H2CO3)
= 2.1x10-4
HCO3- + H2O H2CO3(aq) + OH-, Kb2 = Kb(HCO3
-) = Kw/Ka(H2CO3) =
Kw/K1(H2CO3)
= 2.4x10-8
For “H2CO3”:
K1 = 4.2x10-7
K2 = 4.8x10-11
which is stronger?
22 © Prof. Zvi C. Koren 20.07.2010
Polyprotic Acids & Bases (continued)
Problem:
Calculate the pH of a 0.10-M solution of H3PO4:
H3PO4(aq) H+ + H2PO4- , Ka1 = 7.5x10-3
H2PO4- H+ + HPO4
2- , Ka2 = 6.2x10-8
HPO42- H+ + PO4
3- , Ka3 = 3.6x10-13
Because K1 >> K2 >> K3, [H+]1 >> [H+]2 >> [H+]3
We only need to consider the first ionization!!!
H3PO4(aq) H+ + H2PO4- , Ka1 = 7.5x10-
3
0.10
x0.10-x x= x2/(0.10-x) x2/0.10
x = [H+] = 2.7x10-2 M pH = 1.56
Answer:
23 © Prof. Zvi C. Koren 20.07.2010
Polyprotic Salts
Problem:
Calculate the equilibrium concentrations of all the species in a 0.10-M Na2CO3 solution.
Identify all the species.
Because K1 >> K2, [OH-]1 >> [OH-]2
For the calculation of [OH-]eq, we only need to consider the 1st ionization (so go back):
x0.10-x x
2.1x10-4 = Kb(CO32-) = x2/(0.10-x) x2/0.10 x = [OH-] = 4.6x10-3 M
[H+] = 2.18x10-12 MpH = 11.66
Answer:
Note that we have a salt:
Dissociation of dissolved salt: Na2CO3(aq) 2Na+ + CO32-
0.10
0.10 [Na+] = 0.20 M
CO32- + H2O HCO3
- + OH-
HCO3- + H2O H2CO3(aq) + OH-
0.10
[CO32-] = 0.10-x 0.10 Mfrom before
Hydrolysis of non-spectator ions:
(continued)24 © Prof. Zvi C. Koren 20.07.2010
Polyprotic Salts (continued)
Answer continued:
recall: CO32- + H2O HCO3
- + OH-
HCO3- + H2O H2CO3(aq) + OH-
[HCO3-]: produced in 1st rxn, but reacts in 2nd rxn. BUT, K1 >> K2.
[HCO3-] [OH-] = 4.6x10-3 M
Theoretically, which is greater, [HCO3-] or [OH-]?
[H2CO3]: only produced in 2nd rxn.
][
]][[104.2
3
322
8
HCO
OHCOHKx b
[H2CO3] = Kb2 = 2.4x10-8 M
(continued)25 © Prof. Zvi C. Koren 20.07.2010
Polyprotic Salts (continued)
Σqi = 0 Σn+C+ = Σn–C–
[Na+] + [H+] = 2·[CO32-] + [OH-] + [HCO3
-] 0.20 2.18x10-12 2·(0.10- 4.6x10-3) 4.6x10-3 4.6x10-3
Note the BALANCES in this reaction system:
Charge Balance & Material Balance
Charge Balance:
Total concentrations of (+) ions = Total concentrations of `(–) ions
Material Balance (Conservation) for “Carbonate”:Total amount of carbonate material is conserved,
from the beginning to the end "(הוא שיהיה, מה שהיה)"
[Na2CO3]0 = [CO32-]eq + [HCO3
-]eq + [H2CO3]
0.10 0.10 - 4.6x10-3 4.6x10-3 2.4x10-8
26 © Prof. Zvi C. Koren 20.07.2010
Polyprotic Salts (continued)
Is a solution of NaHSO3 basic or acidic?
NaHSO3(aq) Na+ + HSO3–
HSO3– + H2O H3O
+ + SO32–
HSO3– + H2O OH– + H2SO3(aq)
As an Acid:
As a Base:
Ka(HSO3–) = K2(H2SO3) = 6.2x10–8
Kb(HSO3–) =
13
2
14
32a
w 8.3x10 102.1
100.1
)SO(HK
K
x
x
HSO3– is much stronger as an acid than it is as a base
27 © Prof. Zvi C. Koren 20.07.2010
H+ + OH– H2O
H3N + BF3 H3N–BF3
NH3 + H2O NH4+ + OH–
CO2(g) + OH– HCO3
–
bicarbonate ion
Cu2+ + 2OH- Cu(OH)2(s)
Lewis Acids & Bases
28 © Prof. Zvi C. Koren 20.07.2010
Identify the Lewis acids and bases in the following reactions:
Note: A Lewis acid accepts a pair of e’s; a Lewis base donates the e’s