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