Chapter 16: Acid-Base Equilibria

Acid-Base Equilibria

I. Arrhenius Acid-Base Definition

A. Acids: proton generators in water (H⁺ are the acidic species)

Examples: HCl, H₂SO₄ e.g.: HCl <--> H⁺ + Cl^-

B. Bases: Hydroxide ion generators in water (OH^- are the basic species)

Examples: NaOH, NH3 e.g.: NH3 + H2O <--> NH4⁺ + OH^-

C. Unexplainables

What about carbonate acting as a base?

CO3^2- + 2 HC2H3O2 --> CO2 + H2O + 2 C2H3O2^-

=> carbonate is a base, but no hydroxide is generated.

II. Brönsted-Lowry Acid-Base Definition

A. Acids: proton donors

Examples: HCl, H2SO4

e.g.:

HCl + H2O <--> H3O⁺ + Cl^-

acid base C.A. C.B.

C.A. = conjugate acid ; C.B. = conjugate base

B. Bases: proton acceptors

Examples: NaOH, NH3

e.g.:

NH3 + H2O <--> NH4⁺ + OH^-

base acid C.A. C.B.

C. Water: special case!

It is amphoteric, since it can act as an acid or as a base.

Self ionization: H2O <--> H⁺ + OH^-

III. Definition of extent of ionization

A. Self ionization of water: define equilibrium constant K_w

K_w = [H⁺][OH^-]/[H2O] = [H⁺][OH^-] because the concentration of water is so great (as the solvent) that it can be considered constant (change in[H2O]<<<<[H2O]). Also, water is a pure liquid, and can be omitted from K.

B. Acids
1. Keq of acids in water: HA + H2O <--> H3O⁺ + A^-

K_a = [H3O⁺][A^-]/[H2O][HA] ; K_a (approximately)= [H⁺][A^-]/[HA]

simplifications: H3O⁺= H⁺ and [H2O] = Constant

The size of K_a indicates the extent of ionization of HA and thus its strength. Units of Ka, Kw (and Kb) are consistent throughout the discussion, so are not explicitly shown.

2. Strong Acids (K_a>>>>1): HCl, H2SO4, HNO3, HClO4

The ionization proceeds essentially to completion, therefore [HA] is very small and hard to measure; Ka is very hard to determine.

3. Weak acids (K_a<1): HSO4^-, HF, HC2H3O2, HCN, HNO2, HClO

The ionization is far from complete. All components are measureable and K_a can be determined.

4. General Rules
a. A strong acid has a weak conjugate base (as compared to H2O)

b. A weak acid has a strong conjugate base (as compared to H2O)

IV. Convenient measure of [H⁺] in aqueous solution => the pH scale

A. Definition: pH = -log[H⁺]

e.g.: for a solution of [H⁺] = 1.0 x 10^-6 M

pH = -log(1.00 x 10^-6 M) = 6.00

1. Rule of Significant Figures With Logarithms: The number of significant figuresin the concentration should equal the number of significant digits in the logarithm.

B. pH scale

Neutral Solution: pure water

K_w = [H⁺][OH^-]= 1.0 x 10^-14

[H⁺] = [OH^-] = 1.0 x 10^-7

pH = -log(1.0 x 10^-7) = 7.00 => midpoint of the scale

Max. basicity => pH = 14

Max. acidity => pH = 0

C. Can also define complement of pH: pOH = -log[OH^-]

e.g. a solution where [H⁺] = 1.0 x 10^-6M ; pH = 6.00 and

K_w = [H⁺][OH^-] = 1.0 x 10^-14= (1.0 x 10^-6)[OH^-]

[OH^-] = 1.0 x 10^-14/1.0 x 10^-6= 1.0 x 10^-8 M ; pOH = 8.00

D. General Relationship between pH and pOH

pK_w = -log(1.0 x 10^-14) = 14.00

pH + pOH = pK_w = 14.00

V. Calculation of pH of acidic solutions

A. Strong acids
1. 0.20 M HCl

Reactions:

(1) HCl <--> H⁺ + Cl^- ; K_a >>>1 (HCl ionizes completely)

(2) H2O <--> H⁺ + OH^- ; Kw = 1.0 x 10-14

K_a >>>K_w so max. [H⁺] = [HCl]_o because HCl ionizes completely

pH = -log(0.20) = 0.70

2. 2.0 x 10^-9 M HCl

Reactions:

(1) HCl <--> H⁺ + Cl^- ; [H⁺] = [HCl]_o = 2.0 x 10^-9 M

(2) H2O <--> H⁺ + OH^- ; [H⁺] = 1.0 x 10^-7 M

Reaction 2 dominates because [H⁺] from HCl << [H⁺] from water

=> the solution is too dilute to significantly affect [H⁺]

B. Weak Acids

Treat as standard equilibrium problems.

1. 0.20 M solution of acetic acid (HC2H3O2)

Reactions:

(1) HC2H3O2 <--> H⁺ + C2H3O2^- ; K_a = 1.8 x 10-5

(2) H2O <--> H⁺ + OH^- ; K_w = 1.0 x 10-14

K_a > K_w therefore reaction 1 dominates

HC2H3O2 <--> H⁺+ C2H3O2^-

initial 0.20 0 0 M

change -x +x +x M

eq. 0.20 - x x x M

K_a = [H⁺][C2H3O2^-]/[HC2H3O2]
1.8 x 10^-5 = (x)(x)/(0.20 - x) ; assume that 0.20 - x = 0.20
x² = (0.20)(1.8 x 10^-5)
x = 0.0019 M = [H⁺] = [C2H3O2^-]
pH = -log(0.0019) = 2.72

2. Check asumption with 5% rule

5% rule: can assume Y - x = Y if (x/Y)100 is less than 5%

(x/[HC2H3O2]_o)100 = (0.0019/0.20)100 = 0.95 % YES!

C. Compare strong and weak acids and resultant pH's

	Acid solution	pH	[H⁺] (M)
strong	0.20 M HCl	0.70	0.20
weak	0.20 M HC2H3O2	2.72	0.0019

There is a difference of over two pH units when ionization is essentially complete (HCl) and when one out of every 100 molecules ionizes (HC2H3O2).

D. Percent Dissociation (Percent Ionization)
1. Definition and calculation

% = ([H⁺]eq/[HA]_o)100

e.g. for 0.20 M HCl
% = (0.20/0.20) = 100%

e.g. for 0.20 M HC2H3O2
% = (0.0019/0.20) = 0.95%

2. Effect of concentration or dilution on %

[HC2H3O2]_o	pH	[H⁺]_eq	%
1.00	2.37	0.0042	0.42
0.20	2.72	0.0019	0.95
0.0010	3.87	0.00013	13

The percent dissociation increases as the solution becomes more dilute.

Relate to LeChatlier's Principle

HC2H3O2 + H2O <--> C2H3O2^- + H3O⁺

Get increasing relative amounts of water to acetic acid as [HC2H3O2] decreases, therefore shifts equilibrium to the right.

VI. Calculation of pH of basic solutions

A. Strong bases: treat analogously to strong acids

Generic case: B + H2O --> BH⁺ + OH^-

K_b = [BH⁺][OH^-]/[B][H2O] => K_b (approximately)= [BH⁺][OH^-]/[B]

1. 0.20 M NaOH

Reactions:

(1) NaOH --> Na⁺ + OH^- ; K_b >>>1 (NaOH ionizes completely)

(2) H2O <--> H⁺ + OH^- ; K_w = 1.0 x 10^-14

K_a >>>K_w so max. [OH^-] = [NaOH]_o because NaOH ionizes completely

pOH = -log(0.20) = 0.70
pH = 14.00 - 0.70 = 13.30

2. 2.0 x 10^-9 M NaOH

Reactions:

(1) NaOH --> Na⁺ + OH^- ; [OH^-] = [NaOH]o = 2.0 x 10^-9 M

(2) H2O <--> H⁺ + OH^- ; [OH^-] = 1.0 x 10^-7 M

Reaction 2 dominates because [OH^-] from NaOH << [OH^-] from water.

=> the solution is too dilute to significantly affect [OH^-]

B. Weak bases

Treat as standard equilibrium problems.

1. 0.20 M solution of ammonia (NH3)

Reactions:

(1) NH3 + H2O <--> NH4⁺ + OH^- ; K_b = 1.8 x 10^-5

(2) H2O <--> H⁺ + OH^- ; K_w = 1.0 x 10^-14

K_b > K_w , therefore reaction 1 dominates

NH3 + H2O <--> NH4⁺+ OH^-

initial 0.20 C 0 0 M

change -x -x +x +x M

eq. 0.20 - x C x x M

K_b = [NH4⁺][OH^-]/[NH3] = (x)(x)/(0.20-x) = 1.8 x 10^-5
x² = (0.20)(1.8 x 10^-5)
x = 0.0019 M = [NH4⁺] = [OH^-]
pOH = -log(0.0019) = 2.72
pH = 14.00 - 2.72 = 11.28

5% rule: (0.0019/0.20)100 = 0.95% YES!

D. Rule about percent dissociation holds for weak bases as well.

VII. Polyprotic acids

A. Sulfuric acid (H2SO4)

Equilibria in solution:

(1) H2SO4 <--> H⁺ + HSO4^- ; K_a >>>> 1 (complete ionization)

(2) HSO4^- <--> H⁺ + SO4^2- ; K_a = 1.2 x 10^-2

=> bisulfate is a weak acid: why?

Consider it as an electrostatic problem. Removing a positive charge from a negative species is more difficult than removing a positive charge from a neutral species.

B. Phosphoric acid (H3PO4)

Equilibria in solution:

(1) H3PO4 <--> H⁺ + H2PO4^- ; K_a = 7.5 x 10^-3

(2) H2PO4^- <--> H⁺ + HPO4^2- ; K_a = 6.2 x 10^-8

(3) HPO4^2- <--> H⁺ + PO4^3- ; K_a = 4.8 x 10^-13

C. General rules
1. Each successive ionization of a polyprotic acid is more difficult and the ionization constant is smaller.
2. Simplification: Consider only the first ionization when solving problems at this level.

VIII. Salt solutions and pH

A. Strong acid-strong base

e.g. NaCl (aq)

1. Ask from what acid did the anion derive.

Ans. Cl- from HCl (strong acid)

Cl- is the conjugate base (C.B.) of HCl

2. Ask from what base did the cation derive.

Ans. Na⁺ from NaOH (strong base)

Na⁺ is the conjugate acid (C.A.) of NaOH

3. Equation of salt formation: (thought experiment)

NaOH +	HCl	--> H2O +	Na⁺ +	Cl^-
SB	SA		C.A.	C.B.

4. The salt of a strong acid and a strong base is composed of a weak C.A., C.B. pair, therefore the solution will be neutral (pH = 7.00).
B. Weak acid-strong base

e.g. NaC2H3O2

1. C2H3O2^- derives from acetic acid (HC2H3O2), a weak acid
2. Na⁺ derives from NaOH, a strong base

NaOH +	HC2H3O2	--> H2O +	Na⁺+	C2H3O2^-
SB	WA		CA	CB

3. Sodium cations have no affinity for H⁺ or OH^- (CA of a SB), but acetate does have a significant affinity for H⁺ (CB of a WA).

Where will acetate get protons? from water

C2H3O2^- + H2O <--> HC2H3O2 + OH^- occurs in solution

=> the solution will be basic

a. Determining K_b of the conjugate base:

K_b = [HC2H3O2][OH^-]/[C2H3O2^-]

Known eq. constants: K_a = [H⁺][C2H3O2^-]/[HC2H3O2] ; K_w = [H⁺][OH^-]

Can we combine K_a and K_w to get K_b? Yes.

K_b = K_w/K_a = 1.0 x 10^-14/1.8 x 10^-5 = 5.6 x 10^-10

b. Calculating a specific pH:

e.g. 0.20 M NaC2H3O2

C2H3O2^- +	H2O <-->	HC2H3O2 +	OH^-	occurs in solution
0.20	C	0	0	M initial
-x	-x	+x	+x	M change
0.20-x	C	x	x	M final

K_b = 5.6 x 10^-10 = [HC2H3O2][OH^-]/[C2H3O2^-]
= (x)(x)/0.20-x = x²/0.20
x = 1.06 x 10^-5 M = [OH^-]
pOH = -log(1.06 x 10^-5) = 4.98
pH = 14.00 - 4.98 = 9.02

C. Strong acid-weak base

e.g. NH4Cl

1. Cl^- derives from HCl, a strong acid
2. NH4⁺ derives from NH3, a weak base

NH3 +	HCl	-->	NH4⁺ +	Cl^-
WB	SA		C.A.	C.B.

3. Chloride anions have no affinity for H⁺ or OH^- (CB of a SA), but ammonium ions have only a moderate affinity fro protons (CA of a WB). Ammonium ions will partially ionize to produce H⁺ in solution.

NH4⁺ --> H⁺ + NH3 occurs in solution

=> the solution will be acidic

a. Determining K_a of the conjugate acid:

Follow the same reasoning used for K_b of a conjugate base.
K_a = [H⁺][NH3]/[NH4⁺]
K_a = K_w/K_b = 1.0 x 10^-14/1.8 x 10^-5 = 5.6 x 10^-10

b. Calculating a specific pH:

e.g. 0.10 M NH4Cl

NH4⁺	-->	H⁺ +	NH3	occurs in solution
0.10		0	0	M initial
-x		+x	+x	M change
0.10-x		x	x	M final

K_a = 5.6 x 10^-10
= [H⁺][NH3]/[NH4⁺]
= (x)(x)/0.10-x = x²/0.10
x = 7.5 x 10^-6 M = [H⁺]
pH = -log(7.5 x 10^-6) = 5.13

D. Weak acid-weak base

e.g. ammonium acetate (NH4C2H3O2)

1. C2H3O2^- derives from acetic acid, a weak acid
2. NH4⁺ derives from ammonia, a weak base

NH3 +	HC2H3O2	-->	NH4⁺ +	C2H3O2^-
WB	WA		C.A.	C.B.

3. Both the conjugate acid and the conjugate base will contribute to the final pH of the solution.

NH4⁺ --> H⁺ + NH3 occurs in solution, and
C2H3O2^- + H2O <--> HC2H3O2 + OH^- occurs in solution

a. Determining the pH of a solution:

Must compare the K_a and K_b of the CA and CB.

1. K_a > K_b ; [H⁺] > [OH^-] and the solution will be acidic
2. K_a < K_b ; [H⁺] < [OH^-] and the solution will be basic
3. K_a = K_b ; [H⁺] = [OH^-] and the solution will be neutral

b. General Trends: Can compare K_a and K_bdirectly since they are related to CA and CB K's via K_w (K_b = K_w/K_a)
1. As K_a of a weak acid gets larger, K_bof the CB gets smaller
2. As K_b of a weak base gets larger, K_aof the CA gets smaller
3. Relationship:

acid:base	CA:CB	pH
K_a > K_b	K_a > K_b	< 7.00
K_a < K_b	K_a < K_b	> 7.00
K_a = K_b	K_a = K_b	= 7.00

IX. Salts of Polyprotic Acids

These can be treated like other salts.

e.g. Phosphoric acid and its salts: H3PO4; NaH2PO4; Na2HPO4; Na3PO4

Compare equilibrium constants for the possible processes:

Eq. Name Value Equilibrium

K_a1 7.5 x 10^-3 H3PO4 <--> H⁺ + H2PO4^-

K_a2 6.2 x 10^-8 H2PO4- <--> H⁺ + HPO4^2-

K_a3 4.8 x 10^-13 HPO4^2- <--> H⁺ + PO4^-3

K_b1 0.021 PO4^-3 + H2O <--> HPO4^2- + OH^-

K_b2 1.6 x 10^-7 HPO4^2- + H2O <--> H2PO4^- + OH^-

K_b3 1.3 x 10^-12 H2PO4^- + H2O <--> H3PO4 + OH^-

Choose the equilibrium with the largest K for the salt given.

Ex. 1: What is the pH of a 0.20 M solution of Na3PO4?

Dominant equilibrium: PO4^-3 + H2O <--> HPO4^2- + OH^- ; K_b1 = 0.021

PO4^-3 + H2O <--> HPO4^2- + OH^-

initial 0.20 C 0 0 M

change -x -x +x +x M

eq. 0.20 - x C x x M

Because K_b1 is greater than 10^-3, you CANNOT use the 5% rule to simplify the math!

0.021 = (x)(x)/(0.20-x)

x² + 0.021x - 0.0042 = 0

x = [-0.021 + (0.0004 + 0.0168)^1/2]/2

x = (0.1313-0.021)/2 = 0.055 M = [H⁺]

pH = 14.00 - (-log(0.055)) = 14.00-1.26 = 12.74

Na3PO4 is a basic salt.

Ex. 2: What is the pH of a 0.20 M NaH2PO4 solution?

Possible equilibria: H2PO4- <--> H⁺ + HPO4^2- ; K_a2 = 6.2 x 10^-8

H2PO4^- + H2O <--> H3PO4 + OH^- ; K_b3 = 1.3 x 10-12

The first one (K_a2) is dominant; K_a2 > K_b3

H2PO4^- <--> HPO4^2- + H⁺

initial 0.20 0 0 M

change -x +x +x M

eq. 0.20 - x x x M

6.2 x 10^-8 = (x)(x)/(0.20-x)

x² = (0.20)(6.2 x 10^-8) = 1.24 x 10^-8

x = 1.1 x 10^-4 M = [H⁺]

pH = -log(1.1 x 10-4) = 3.95

NaH2PO4 is an acidic salt.

Ex. 3: What is the pH of a 0.20 M Na2HPO4 solution?

Possible equilibria: HPO4^2- <--> H⁺ + PO4^-3 ; K_a3 = 4.8 x 10^-13

HPO4^2- + H2O <--> H2PO4^- + OH^- ; K_b2 = 1.6 x 10^-7

The second one (K_b2) is dominant; K_b2 > K_a3

HPO4^2- + H2O <--> H2PO4^- + OH^-

initial 0.20 C 0 0 M

change -x -x +x +x M

eq. 0.20 - x C x x M

1.6 x 10^-7= (x)(x)/(0.20-x)

x² = (0.20)(1.6 x 10^-7) = 3.2 x 10^-8

x = 1.78 x 10^-4 M = [OH^-]

pH = 14.00 - (-log(1.78 x 10^-4)) = 14.00-3.75 = 10.25

Na2HPO4 is a basic salt.

X. Lewis acid-base model

A more general description of acid-base behavior.

A. Definition

Acid: an electron pair acceptor

Base: an electron pair donor

This definition does not depend on the action of protons to explain behavior.

B. Determining Lewis acids and bases
1. Draw Lewis dot structures of acid and base
2. Show interacxtion of acid and base with Lewis dot structures
3. ID the electron pair donor (LB) and the electron pair acceptor (LA)
C. Examples
1. Water: H⁺ + OH^- <--> H2O

2. Complex ions: Cu(NH3)4²⁺

XI. Acid strength and molecular structure

A. Hydrides of elements
1. Two important factors determining relative acid strength
a. electronegativity of the element
b. size of the element
2. Trends observed: second row of the Periodic table

LiH	BeH2	BH3	CH4	NH3	H2O	HF	Ne
SB	SB	LB	N	WB	Amph.	WA	NR

third row of the Periodic table

NaH	MgH2	AlH3	SiH4	PH3	H2S	HCl	Ar
SB	SB	LB	N	WB	WA	SA	NR

As the electronegativity increases, the bond first becomes more covalent (start as ionic with M⁺ and H^-), then becomes more polar and ionizable to H⁺ and X^-.

3. Trends down a group

HF	HCl	HBr	HI
WA	SA	SA	SA

As the size increases, the ionizability of the X-H bond increases, and the acid strength increases.

A large atom is diffuse and polarizable; the small hydrogen atom is a 'pea in pudding'. The large atom can better stabilize a negative charge by spreading it out.

A small atom is hard and not as polarizable, like a billiard ball. The smaller atom has a harder time stabilizing a negative charge.

B. Oxyacids
1. Formula: (H-O)n-Z

third row of the Periodic table

NaOH	Mg(OH)2	Al(OH)3	H2SiO3	H3PO4	H2SO4	HClO4	Ar
SB	SB	WB	N	WA	SA	SA	NR

2. Compare two ends of the Periodic table
a. NaOH: The Na-O bond is more easily ionized than the O-H bond (larger electronegativity difference), therefore get Na⁺ and OH^- in solution.
b. H2SO4: The O-H bond(s) are more easily ionized than the S-O bond(s), therefore get H⁺ and HSO4^- in solution.
3. Trends down a group
HOCl > HOBr > HOI (electronegativity decreases Cl > Br > I)
4. Addition of oxygen atoms to the acid (groups of related oxyacids)
HClO4 > HClO3 > HClO2 > HClO

Each oxygen is more electronegative than chlorine, so will draw electron density from Cl and make it more electron poor. This increases the ionizability of the O-H bond by further stabilizing the anion.

5. General trends
a. As the electronegativity of Z increases, the acidity of the oxyacid increases.
b. As oxygen atoms are added, the acidity of the oxyacid increases.

	HC2H3O2	<-->	H⁺+	C2H3O2^-
initial	0.20		0	0	M
change	-x		+x	+x	M
eq.	0.20 - x		x	x	M

	NH3 +	H2O <-->	NH4⁺+	OH^-
initial	0.20	C	0	0	M
change	-x	-x	+x	+x	M
eq.	0.20 - x	C	x	x	M

Eq. Name	Value	Equilibrium
K_a1	7.5 x 10^-3	H3PO4 <--> H⁺ + H2PO4^-
K_a2	6.2 x 10^-8	H2PO4- <--> H⁺ + HPO4^2-
K_a3	4.8 x 10^-13	HPO4^2- <--> H⁺ + PO4^-3
K_b1	0.021	PO4^-3 + H2O <--> HPO4^2- + OH^-
K_b2	1.6 x 10^-7	HPO4^2- + H2O <--> H2PO4^- + OH^-
K_b3	1.3 x 10^-12	H2PO4^- + H2O <--> H3PO4 + OH^-

	PO4^-3 +	H2O <-->	HPO4^2- +	OH^-
initial	0.20	C	0	0	M
change	-x	-x	+x	+x	M
eq.	0.20 - x	C	x	x	M

	H2PO4^-	<-->	HPO4^2- +	H⁺
initial	0.20		0	0	M
change	-x		+x	+x	M
eq.	0.20 - x		x	x	M

	HPO4^2- +	H2O <-->	H2PO4^- +	OH^-
initial	0.20	C	0	0	M
change	-x	-x	+x	+x	M
eq.	0.20 - x	C	x	x	M