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1

Aqueous Equilibria: Acids and Bases

Ch. 16

What is an acid? What is a base? There are actually multiple definitions

Arrhenius: Dealt with species in aqueous solutions. Most basic definition of acis-base.

Brønsted-Lowrey: Acid-Base need not be in aqueous solutions . Acid and bases are part of a related conjugate pair.

Lewis: No need for hydrogen in definitions

•Acid: Proton (H+) donor •Base: Proton (H+) acceptor

•Acid: Electron (e- ) acceptor •Base: Electron (e- ) donor

•Acid: increases H+ in water. •Base: increases OH- in water

Ag+ + 2 :NH3 → [H3N:Ag:NH3]+

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Weak vs. Strong Acids/Bases What does “strong” as in strong acids and bases mean?

•Highly Concentrated?

•Dangerous?

Strong just mean 100% dissociation That’s all.

H Cl

HCl in the gas phase

H +

-

-

-

-

HCl in the aqueous phase

H Cl

H Cl

H Cl

Cl

Cl

Cl Cl

HCl(g) HCl(aq)

H2O

K’s large HCl is a strong acid

H +

H +

H +

HCl = 1.3x106

Weak vs. Strong Acids/Bases

Weak just means much less than 100% dissociation.

HF in the gas phase

HF in the aqueous phase

H F

H2O H F

H F H F

H F

H F

H F

F -

Most HF molecules are actually NOT dissociated

HF(g) HF(aq) K’s are very small HF is a weak acid

H F

H F H +

HF = 6.6x10-4

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Weak vs. Strong Acids/Bases

Just like with acids, strong bases are those with100% dissociation.

NaOH in the solid phase

NaOH in the aqueous phase

NaOH(s) NaOH(aq)

H2O

K’s large NaOH is a strong base

H O Na

H O Na

H O Na

H O Na

H O Na

H O Na

Na

+

H O

-

Na

+

H O -

H O -

H O - Na

+

Na

+

Weak vs. Strong Acids/Bases Weak just means much less than 100% dissociation! For NH3, it itself is not

dissociated, however, it takes an H+ from water to create OH-

NH3 in the liquid phase

NH+ in the aqueous phase

H2O

Most NH3 molecules are actually NOT converted to NH4

+

NH3(l) NH4+(aq) K’s are very small NH3 is a weak base

N

H O -

H +

N

NH3(l)+H2O(l) → NH4+(aq) + OH-(aq)

NH3 = 1.8x10-5

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Strong vs. Weak acids

If you know the strong acids/bases, assume the rest are weak

• HCl(aq) • HBr(aq) • HI(aq) • HNO3 • HClO3 • HClO4 • H2SO4

**

Strong Acids Strong Bases

• (soluble metal hydroxides)

• LiOH • NaOH • KOH • Ca(OH)2

(kinda-ish)

• Sr(OH)2 • Ba(OH)2

100% ionization

Examples of weak acids

Examples of weak bases

• HF • CH3COOH (acetic) • HCOOH (formic) • NH4

+ (ammonium) • H3PO4 • HClO2 • HClO • (small, highly-charged

metal ions)

• Al3+

• (nitrogen-containing bases)

• NH2R R = anything • NH3 (ammonia) • (“Insoluble” hydroxides)

<100% ionization

A common rule for oxo-acids: If there are at least 2 more O’s than H+’s in the molecule, it is strong

**Only the 1st H+ removed is “strong”

Auto ionization of water – Kw

At any time in a sample of pure water, some water molecules, very few, will dissociate by themselves

The amount (concentration) that dissociates is quantified by Kw

Kw = [H+][OH-]= 1x10-14

Only 1.8x10-6% of H2O molecules within a container of water do this!!

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The pH scale A simple way of expressing concentration of H+ ions in solution.

pH = -log[H+]

0.00000000000001

0.0000000000001

0.000000000001

0.00000000001

0.0000000001

0.000000001

0.00000001

0.0000001

0.000001

0.00001

0.0001

0.001

0.01

0.1

1 0.00000000000001

0.0000000000001

0.000000000001

0.00000000001

0.0000000001

0.000000001

0.00000001

0.0000001

0.000001

0.00001

0.0001

0.001

0.01

0.1

1

Notice the relationship:

[H+][OH-] = 1.0x10-14 = Kw

[H+] in M [OH-] in M

Also notice the logarithmic nature

of pH vs. [H+]

For example, what is the

difference in [H+] between the pH

of 7 and 4?

pH of 4 has 103, (or 1,000x) more

H+ ions in solution than pH 7.

pH Common

solutions

[H+] = 10-pH

pH and [H+]

Determine the pH of a solution with 1.55x10-6 M H+. Also, what will be the

concentration of OH-?

pH = -log(1.55x10-6M) = 5.810

[H+][OH-] = 1.0x10-14 = Kw

(1.55x10-6M)[OH-] = 1.0x10-14

M1.02x100.977M

1.00x10][H 14

14

Determine the pH of a solution with 0.977 M OH-. Also, what will be the concentration of

H+?

pH = -log(1.02x10-14M) = 13.991

M6.452x10M1.55x10

1.00x10][OH 9

6-

14-

Determine the H+ and OH- concentrations for a solution with a pH of 7.02

pH = -log[H+]

[H+] = 10-pH M9.8x1010][H 87.01 M1.0x10M9.8x10

1.00x10][OH 7

8-

14-

Notice that the closer to pH = 7, the closer the [H+] and [OH-]

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pH and pOH

[H+][OH-] = 1.0x10-14 = Kw

Although most people know about pH, we can also calculate something called pOH

take the -log

pH + pOH = 14

So…let’s say we have a solution with a [H+] of 2.9x10-5 M. Determine the concentration of [OH], find pH and pOH.

(2.9x10-5M)[OH-] = 1.00x10-14

M2.9x10

1.0x105

14

pOH = -log(3.4x10-10M) = 9.47

pH = -log(2.9x10-5M) = 4.54

pH = 14.00 – pOH = 4.53

pOH = 14 – 4.54 = 9.46

[OH-] = 10-4.54 = 3.5x10-10M

[OH-] = = 3.4x10-10M

Difference in last digit due to rounding (keep an extra sig fig in calculations)

Qualitatively: •The larger the [H+], to lower the [OH-] •The smaller the pH, the larger the pOH

or

pH of a strong acid/strong base Remember that when a strong acid or a strong base is placed in water, we assume

that it is 100% dissociated.

So, for HCl, all of the “H” becomes dissociated as H+. If we know the concentration of HCl, we

also know the concentration of H+.

What is the pH of a 0.050M solution of HCl?

Since HCl is 100% dissociated, the [H+] is also 0.050M

pH = -log[H+]

pH = -log(0.050) = 1.30

And for NaOH, all of the “OH” becomes dissociated as OH-. If we know the concentration of NaOH, we

also know the concentration of OH-.

What is the pH of a 0.050M solution of NaOH?

Since NaOH is 100% dissociated, the [OH-] is also 0.050M

pOH = -log[OH-]

pOH = -log(0.050) = 1.30

pH = 14 - pOH

pH = 14.00 -1.30 = 12.70

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Conjugate Acid/Base pairs

HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)

Conjugate acid/base pairs are acids and bases that differ by only an H+

acid base

Conjugate acid/base pair

acid base

Conjugate acid/base pair

The stronger the acid, the weaker

its conjugate base.

HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)

stronger

acid

stronger

base

weaker

acid weaker

base

Conjugate Acid/Base pairs

HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)

stronger

acid

stronger

base

weaker

acid weaker

base

Reaction proceeds in direction that results in a weaker acid and weaker base produced

NO3- (aq) + H2O(l) → OH-(aq) + HNO3(aq)

stronger

acid

stronger

base

weaker

acid weaker

base

The reaction will not proceed in the forward direction

NH4+ (aq) + CH3COO-(aq) ⇌ NH3(aq) + CH3COOH(aq)

stronger

acid

stronger

base

weaker

acid

weaker

base

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The Acid and Base Equilibrium constants: Ka and Kb

[HA]

]][AO[H 3

a

K

The strength of a acid or base can be related to the an equilibrium constant

HA(aq) + H2O(l) → H3O+(aq) + A-(aq)

[B]

]][BH[OH

-

b

K

B(aq) + H2O(l) → OH-(aq) + BH+(aq)

pKa = -log(Ka) pKb = -log(Kb) Ka = 10-pKa Kb = 10-pKb

The larger the Ka, and thus smaller the pKa, the stronger the acid

The larger the Kb, and thus smaller the pKb, the stronger the base

A certain acid has a Ka of 1.2x10-7. Calculate it’s pKa.

A certain acid has a pKa of 9.23. Calculate it’s Ka.

Based on pKa or Ka values, which acid is stronger?

CH3COOH Ka = 1.6x10-5 CH3COO- Kb = 6.2x10-10

How are conjugate acid/base pairs related?

Kb*Ka = Kw = 1x10-14

For conjugate acid/base pairs

pKb +pKa = pKw = 14

We can also take the negative log of

Ka, Kb, and Kw.

CH3COOH pKa CH3COO- pKb

Conjugate Acid/Base pair relationship – Ka and Kb

NH3 Kb = 1.9x10-5

NH3 pKb

NH4+

Ka = 5.4x10-10

H3PO4 Ka = 7.6x10-3

H3PO4 pKa

H2PO4- Kb = 1.3x10-12

H2PO4- pKb

-

2.12 11.88

4.79 9.21

9.27 4.73 NH4+ pKa

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Polyprotic Acids While binary acids only have one H+ to donate, some oxoacids have multiple protons. They

can be diprotic or triprotic.

100% dissociation

⇌

HSO4- H2SO4 SO4

2- pKa2 = 1.99 pKa1 ≈ -3

<100% dissociation

- 2-

H3PO4 H2PO4- HPO4

2- PO43-

⇌ ⇌ ⇌

pKa1 = 2.12 pKa2 = 7.20 pKa3 = 12.38

- 2- 3-

Only the first H+ is considered

strong.

Each subsequent H+

dissociates less.

Acidic, Basic, and Neutral Salts Remember: in chemistry, a salt is just an ionic compound containing an anion and

cation, but is neutral overall.

The anion and cation in this case can be viewed as conjugate acid/bases of something else…

Rules to remember: •The child of a strong acid or base is negligible in strength.

•The child of weak acid or base is also weak

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Acidic, Basic, and Neutral Salts Whether a salt is neutral, acidic, or basic will depend on the specific anion and

anion

Neutral salts

Cation Anion (conjugate base)

From strong base …of a strong acid

Acidic salts Conjugate acid of

weak base …of a strong acid

Basic salts From strong base …of a weak acid

Acidic salts Small/highly-charged metal on

…of a strong acid

?????? Conjugate acid weak base or small/highly-

charged metal ion

…of a weak acid

Ka ≈ Kb

Ka > Kb

Ka < Kb Basic salts

Acidic salts

Neutral salts

NaCl

NH4Cl

AlCl3

NaClO

NH4CH3COO

NH4NO2

NH4CN

“from” NaOH

“from” HCl

“from” HCl

“from” NH3

“from” HCl

“from“NaOH

“from” HClO

“from” NH3

“from” CH3COOH

“from” NH3

“from” HNO2

“from” NH3

“from” HCN

Salt

Metal ions as acids Certain metal ions act as acids, increasing the H+ concentration in water.

When these ions dissolve in water, water ions “coordinate” with the ions

creating a “complex”.

The metal ions pulls enough electron density from one water molecule outside the complex to allow it to dissociate into H3O+ while an OH- stays

bound to the complex. Thus, increasing H3O+ concentration.

Al3+ Al3+

H2O outside the complex

H3O+ outside the complex

Acidity increases with higher charge and smaller size of ion

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Determining the pH of a Weak Acid Solution

Determine the pH of a 0.123M aqueous solution of HClO. Ka = 2.95x10-8.

I

E C

nitial

hange

quilibirum

0.123M 0M

-x +x

0.123M - x x

HClO(aq)+ H2O(l)⇌ H3O+(aq) + ClO-(aq)

0M

+x

x

Write the balanced equation for the acid in H2O…producing H3O+.

Just like any other equilibrium problem, plug in what we know.

Since H2O is a pure liquid (and does not change appreciably over time), we don’t need

to deal with it

[HClO]

]][ClOO[H3a

K

x-0.123

(x)(x)2.95x10 8

0.123

x2.95x10

28

29 x3.63x10

x = 6.02x10-5

[H3O+] = x = 6.02x10-5M

pH = -log(6.02x10-5M)

pH = 4.220

0.0487% x1000.123

6.02x10-5

Less than 5%, good assumption 5-

2-5

6.02x10-0.123

)(6.02x10

Determining the pH of a Weak Base Solution

Determine the pH of a 0.555M aqueous solution of NH3. pKb = 4.75.

I

E C

nitial

hange

quilibirum

0.555M 0M

-x +x

0.555M - x x

NH3(aq)+ H2O(l)⇌ OH-(aq) + NH4

+(aq)

0M

+x

x

Write the balanced equation for the base in H2O…producing OH- and the conjugate acid, NH4

+ .

][NH

]][NH[OH

3

4b

K

x-0.555

(x)(x)1.78x10 5-

0.555

x1.78x10

25

26 x9.88x10

x = 3.14x10-3

[OH-] = x = 3.14x10-3M

pOH = -log(3.14x10-3M)

pOH = 2.503 pH = 14.00-2.503 = 11.497

0.566% x1000.555

3.14x10-3

Less than 5%, good assumption 4-

2-3

3.14x10-0.555

)(3.14x10

First, we need to deal with pKb.

54.75b 1.78x10 10 K

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Calculating Ka from pH If 1.55M solution of an acid has a pH of 5.15, determine the Ka for the acid (assume it is monoprotic).

I

E C

nitial

hange

quilibirum

1.55M 0M

-x +x

1.55M - x x

HA(aq)+ H2O(l) ⇌ H3O+(aq) + A-(aq)

0M

+x

x

[H+] = 10-5.15 = 7.1x10-6M

Since we know pH, we also know [H+] at equilibrium

7.1x10-6M 7.1x10-6M

7.1x10-6 M = x

1.55M – 7.1x10-6M

[HA]

]][AO[H3a

K

)1.7x10-(1.55

))(7.1x10(7.1x106-

-6-6

a K

11a 3.3x10K

“Short-Cuts”

initial

2

cbase] or acid [weak

xK

This assumes that Kc is small enough where x will also be small. Therefore we

leave it out.

If you complete enough ice tables, you will start noticing trends.

For example, placing a weak acid or base in water will give you the following expression using an ice table:

This also assumes that there are no products of ionization in the water

before the weak base or acid is added.

If you are ever unsure of a short cut…always do it the long way!!

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Percent ionization

x100[HA]

]O[Hacid anof ionization %

initial

mequilibriu3

x100[B]

][OHbase aof ionization %

initial

mequilibriu-

Percent ionization is the amount of a weak acid or base that has become H3O+ or OH-.

Determine the percent ionization if 0.76M aqueous solution of an unknown acid mixed with water

resulted in a an [H3O+] of 9.05x10-4M

0.12%x1000.76M

M9.05x10 ionization %

4

Determine the percent ionization if 5.00M aqueous solution of an unknown acid mixed with water

resulted in a pH of 3.12

0.015%x1005.00M

M7.6x10 ionization %

4

[H3O+] = 10-pH = 10-3.12 = 7.6x10-4 M

Determine the percent ionization of a 0.100M aqueous solution of NH3. pKb = 4.75.

initial

2

cbase] or acid [weak

xKLet’s use the “short-cut”

Kb = 10-4.75 = 1.8x10-5

0.100

x1.8x10

25

62 1.8x10x

x = 1.3x10-3 = [OH-]

1.3%x1000.100M

M1.3x10 ionization %

3

Percent Ionization and Concentration

initial

2

cbase] or acid [weak

xK

Weak acids and bases do not ionize to the same extent in every solution. The amount to

which it ionizes depends on, among other things, the initial concentration of the weak acid or

base.

Let’s use the “short-cut” Determine the % ionization of a weak acid with

a Ka of 1.0x10-5 with the initial concentrations of

1.28M, 0.040M, and 0.010M.

1.28

x1.0x10

25

]O[H 0.0036Mx 3

0.040

x1.0x10

25

0.00063Mx

0.010

x1.0x10

25

0.00032Mx

0

0.5

1

1.5

2

2.5

3

3.5

0 0.5 1 1.5

% io

niz

atio

n

Initial acid concentration (M)

0.28%x1001.28M

0.0036Mionization %

1.6%x1000.040M

0.00063Mionization %

3.2%x1000.010M

0.00032Mionization %

The more concentrated a weak base solution, the lower

the % ionization

0

0.0005

0.001

0.0015

0.002

0.0025

0.003

0.0035

0.004

0 0.2 0.4 0.6 0.8 1 1.2 1.4Hyd

ron

ium

co

nce

ntr

ati

on

at

equ

ilib

riu

m (M

)

Initial weak acid concentration (M)

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Binary Acid Strengths

H2O HF

HCl

HBr

HI

pKa =15.74 pKa = 3.17

H2S

H2Te

pKa ≈ -10

pKa ≈ -8

pKa ≈ -6 pKa = 6.9

pKa = 2.6

Acid Strength

Acid

Strength

Electronegativity

Bo

nd

Len

gth

Moving down a group, the bond

length gets longer. Longer bond

length = weaker X-H bond

pKa = 3.9

H2Se

-15

-10

-5

0

5

10

15

20

80 100 120 140 160 180

pK

a

Bond Length (pm)

H2O

H2S

H2Se H2Te

HCl

HF

HBr HI

Carboxylic Acid Strengths

4.79 2.86 1.35 0.77

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Oxoacid Strengths – Inductive effects

In oxoacids, The weaker the O-H bond, the stronger the acid

What can make the O-H bond weaker? Highly electronegative atoms withdraw

electron density from the O-H bond, making it weaker.

≈ -1 ≈ -8 7.53

Shift in electron

density away from O-H makes the

bond weaker (more acidic)

Less electron density at

the H

1.96

Carboxylic Acid Strengths

4.74 2.86 1.35 0.77

Explain the following trend in acid strength:

4.74

NaOH

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Oxoacid Strengths – Inductive effects

9.24 6.37 -1.4

≈ -8 ≈ -3 2.15 9.84

Acid Strength

Electronegativity of central atom Across each row, electron

density of being shifted

away from the O-H bond.

Stabilizes negative charge

after proton loss