Unit 8 Chapter16 B and L notes.ppt

Chapter 16

Some Properties of Acids

• Produce H+ (as H

3O+) ions in water (the hydronium ion is a hydrogen ion attached to a water molecule)

• Taste sour

• Corrode metals and produce H₂ gas

• Electrolytes (can be both strong or weak)

• React with bases to form a salt and water

• pH is less than 7

Some Properties of Bases

• Produce OH_{Produce OH}-- ions in water _{ions in water}

• Taste bitter, chalky_{Taste bitter, chalky}

• Are electrolytes (Strong and weak)_{Are electrolytes (Strong and weak)}

• Feel soapy, slippery_{Feel soapy, slippery}

• React with acids to form salts and water_{React with acids to form salts and water}

• _{pH greater than 7pH greater than 7}

Arrhenius acid is a substance that produces

H

or

(H

O

-

hydronium ion

)

Arrhenius base is a substance that produces

OH

Definition #1: Arrhenius

KOH



K

+

OH

-HCl



H

+ Cl

Use chemical formulas to help

identify each from Arrhenius:

1. HCl

Acid Base Salt

2. Fe(OH)

Acid Base Salt

3. CH

COOH Acid Base Salt

4. H

SO

Acid Base Salt

5. Na

O Acid Base Salt

6. KI Acid Base Salt

Definition #2: Brønsted – Lowry

Definition #2: Brønsted – Lowry

(Most important one!)

(Most important one!)

B.L. Acid: A substance that will donate a proton (H+) to another species in solution.

HF + H₂O  H₃O+ + F

Acid base conjugate acid conjugate base

B.L. Base: A substance that will accept a

proton (H+) from another species in a solution. NH₃ + H₂O  NH₄+ + OH

Base acid conjugate acid conjugate base

Conjugate Pairs

Conjugate Pairs

*Must be able to easily recognize pairs –

Your Turn:

Label the acid, base, conjugate acid, and

Label the acid, base, conjugate acid, and

conjugate base in each reaction:

conjugate base in each reaction:

2. CH

2. CH

COOH + H

COOH + H

O

O





CH

CH

COO

COO

+ H

+ H

3

O

O

1. H

1. H

CO

CO

+ H

+ H

O

O





  HCO

  HCO

+ H

+ H

3

O

O

1. H

1. H

CO

CO

+ H

+ H

O

O





  HCO

  HCO

+ H

+ H

3

O

O

3. H

3. H

O + HSO

O + HSO





  SO

  SO

42-2-

+ H

+ H

O

O

3. H

3. H

O + HSO

O + HSO





  SO

  SO

42-2-

+ H

+ H

O

O

Acid Acid Acid Acid Acid Acid Acid Acid Base Base Base Base Base Base Base Base Conj. Conj. Base Base Conj. Conj. Base Base Conj. Conj. Base Base Conj. Conj. Base Base Conj. Conj. Acid Acid Conj. Conj. Acid Acid Conj. Conj. Acid Acid Conj. Conj. Acid Acid Acid Acid Acid

Acid BaseBase_BaseBase Conj.Conj. Base

Base

Conj.

Conj.

Base

Base Conj._AcidConj._Acid

Conj.

Conj.

Acid

Definition #3 – Lewis

(Not required in AP Chem)

Lewis acid

Lewis acid - a substance that - a substance that accepts an electron pair

accepts an electron pair

Lewis base

Lewis base - a substance - a substance that donates an electron

that donates an electron

pair

Formation of

Formation of

hydronium ion

hydronium ion

is also an

is also an

excellent example.

excellent example.

Lewis Acids & Bases

Lewis Acids & Bases

•

Electron pair of the new O-H bond

Electron pair of the new O-H bond

originates on the Lewis base.

Strength of Acids and Bases

1. Strong acid: completely

transfers its proton(s)

(100% dissociation).

Conjugate base is very

weak (negligible) due to

equilibrium.

2. Strong base: completely

dissociates. Conjugate

acid is very weak

Strength of Acids and Bases

1. Weak acid: partially

dissociates.

Conjugate base is

also weak.

2. Weak base: partially

dissociates.

Strong Acid Weak Acid

Can determine which direction

equilibrium lies.

CH₃COOH(aq) + H₂O(l)  H₃O+(aq) + CH

3COO-(aq)

Reactant favored (Reverse)

HCl(g) + H₂O(l)  H₃O+(aq) + Cl-(aq)

Product favored (Forward) *Use strength chart to compare (or K_a values).

Memorize Strong Acids and

Bases:

Strong Acids

HNO

– nitric acid

HCl – hydrochloric acid

HBr – hydrobromic acid

HI – hydroiodic acid

H

SO

- sulfuric acid

HClO

– perchlroic acid

Strong Bases

LiOH – lithium hydroxide NaOH – sodium hydroxide KOH – potassium hydroxide *Ca(OH)2 – calcium hydroxide *Ba(OH)2 – barium hydroxide *Sr(OH)2 – strontium hydroxide

The

The

pH scale

pH scale

is a way

is a way

of classifying

of classifying

substances as acids

substances as acids

and bases. Instead of

and bases. Instead of

using very small

using very small

numbers, we use a

numbers, we use a

logarithmic scale

logarithmic scale

based on Molarity of

based on Molarity of

the H

the H

(or OH

(or OH

) ion.

) ion.

Under 7 = acid

Under 7 = acid

7 = neutral

7 = neutral

Over 7 = base

Over 7 = base

pH Equations

*Remember that these (used by themselves) are for strong acid or base ONLY!

• For Acids:

pH= -log [H+]

[H+] = 10 –pH

• _{For Bases:}

pOH = -log [OH-]

[OH-] = 10 –pOH

• _{To Convert scales:}

pH + pOH = 14 K_w = [H+][OH-]

*K_w = 1 x 10-14

Reminders: [ ] means concentration measured in molarity (M).

Pay attention to given info (acid or base?)

*Only used ALONE for STRONG acids or

Autoionization of Water (Equilibrium)

Autoionization of Water (Equilibrium)

• HH₂₂O can function as O can function as bothboth an ACID and a BASE an ACID and a BASE

• _{In pure water there can beIn pure water there can be AUTOIONIZATION,AUTOIONIZATION,} • There will be equal amounts of [HThere will be equal amounts of [ ₃O+] and [OH_and -]

Equilibrium constant for water = K

Equilibrium constant for water = K

K

K

= [H

= [H

O

O

] [OH

] [OH

] =

] =

1.00 x 10

1.00 x 10

at 25

at 25

C

C

2 H₂O(l) <--> H₃O+(aq) + OH-(aq)

*When writing the equilibrium

More About Water

More About Water

K

K_{w w} called called Ion-Product ConstantIon-Product Constant

K

K_ww = [H = [H₃₃OO++] [OH_{] [OH}--] = 1.00 x 10_{] = 1.00 x 10}-14-14 at 25 _{at 25} ooC_C

In a neutral solution [H

In a neutral solution [H₃₃OO++] = [OH_{] = [OH}--]_]

and so [H

Calculate:

Calculate the concentration of H

(aq)

from a

solution in which [OH

] = 0.010

M

. Predict

whether this solution is an acid or base.

*Reminder:

K

K

= [H

= [H

O

O

] [OH

] [OH

]

]

1.0 x 10

1.0 x 10

M

M

Basic [OH

Calculating the pH

pH = - log [H+]

(Remember that the [ ] mean Molarity)

1. What is the pH of a solution If [H+] = 1 X 10-10 M?

pH = - log 1 X 10-10

pH = - (- 10) pH = 10

2. What is the pH of a solution if [H+] = 1.8 X 10-5 M?

pH = - log 1.8 X 10-5

pH calculations – Solving for [H

pH calculations – Solving for [H

]

]

pH calculations – Solving for [H

pH calculations – Solving for [H

]

]

3.

3.

If the pH of a sample of soda is 3.12,

If the pH of a sample of soda is 3.12,

what is the [H

what is the [H

]?

]?

10

10

=

=

[H

[H

]

]

[H

[H

] = 10

] = 10

= 7.6 x 10

= 7.6 x 10

M

M

**You can use the antilog button on your calculator, You can use the antilog button on your calculator, usually “Shift” or “2

usually “Shift” or “2nd nd function” and then the log _{function” and then the log}

button

Calculating the pOH

•

Since acids and bases are opposites,

Since acids and bases are opposites,

pH and pOH are opposites!

pH and pOH are opposites!

•

pOH does not really exist, but it is

pOH does not really exist, but it is

useful for changing bases to pH.

useful for changing bases to pH.

•

pOH looks at the perspective of a base

pOH looks at the perspective of a base

pOH = - log [OH

pOH = - log [OH

] and [OH-] = 10

] and [OH-] = 10

•

Since pH and pOH are on opposite

Since pH and pOH are on opposite

ends

ends

Calculating the pOH

pOH = - log [OH-]

(Same equation as pH, just with base information)

4.

What is the pOH of a 0.10 M NaOH

What is the pOH of a 0.10 M NaOH

solution?

solution?

pOH = - log 0.1 pOH = - (- 1) pOH = 1

5. What is the pH of the solution above?

Multiple ways to solve

Multiple ways to solve

6.

6.

What is the pH of a solution if [OH-] = 0.0010

What is the pH of a solution if [OH-] = 0.0010

M?

M?

pOH = - log [0.0010]

pOH = - log [0.0010]

pOH = 3

pOH = 3

pH = 14 – 3 = 11

pH = 14 – 3 = 11

OR

OR

K

K

= [H

= [H

O

O

] [OH

] [OH

]

]

[H

[H

O

O

] = 1.0 x 10

] = 1.0 x 10

M

M

AP Sample Question

At 25°C, aqueous solutions with a pH of 8 have a hydroxide ion concentration, [OH-], of

A. 1 x 10-14 M

B. 1 x 10-8 M

C. 1 x 10-6 M

D. 1 M

AP Sample Question

Which of the following statements about

the pH of 0.010M HClO

is correct?

A

. pH=2.00, because [H+]=1.0×10

M.

B.

pH=2.00, because [H+]=2.0×10

M.

C.

pH>2.00, because HClO

is a strong acid.

AP Sample Question

Which of the following is a weak acid in aqueous solution?

A. HCl

B. HClO₄

C. HNO₃

D. H₂S

Equilibria Involving

Equilibria Involving

Weak Acids and Bases

Weak Acids and Bases

Consider acetic acid, CH

Consider acetic acid, CH₃₃COOH (HOAc)COOH (HOAc)

CH

CH33COOH + HCOOH + H22O O ↔↔ H H33OO++ + CH + CH 3

3COO COO -

AcidAcid Conj. base Conj. base

• (K is designated K(K is designated K_aa for ACID and K for ACID and K_bb for BASE) for BASE)

• K gives the ratio of ions (split up) to molecules _{K gives the ratio of ions (split up) to molecules}

(don

(don’’t split up) – just like Kt split up) – just like K_eqeq from the equilibrium from the equilibrium

constant

Ionization Constants for Acids/Bases

Ionization Constants for Acids/Bases

Acids

Acids ConjugateConjugate

Bases

Bases

Increase strength

Equilibrium Constants

Equilibrium Constants

for Weak Acids: in general:

for Weak Acids: in general:

Equilibrium Constants

Equilibrium Constants

for Weak Acids: in general:

for Weak Acids: in general:

Weak acid has K

Weak acid has K

< 1

< 1

Leads to small [H

Example: Equilibria Involving A

Example: Equilibria Involving A

Weak Acid

Weak Acid

Example: Equilibria Involving A

Example: Equilibria Involving A

Weak Acid

Weak Acid

You have 1.00 M

You have 1.00 M

. Calculate the

. Calculate the

equilibrium conc. of

equilibrium conc. of

, H

, H

O

O

,

,

3

3COO COO --

,

,

and the pH. The K

and the pH. The K

= 1.8 x 10

= 1.8 x 10

.

.

Step 1.

Step 1.

Define equilibrium conc. in ICE box

Define equilibrium conc. in ICE box

CH

CH₃₃COOH + HCOOH + H₂₂O O ↔↔ H H₃₃OO++ + CH _{+ CH} 3

3COO COO -

[HOAc]

[HOAc]

[H

[H

O

O

] + [OAc

] + [OAc

]

]

initial

initial

change

change

equilib

equilib

1.00 0 0 00 1.00

1.00 0 0 00 -x

-x +x +x +x+x -x

-x +x +x +x+x

1.00-x

1.00-x xx xx

1.00-x

Step 2

Step 2

.

.

Write K

Write K

expression

expression

You have 1.00 M HOAc. Calc. the equilibrium concs.

You have 1.00 M HOAc. Calc. the equilibrium concs.

of HOAc, H

of HOAc, H₃₃OO++, OAc_{, OAc}--, and the pH._{, and the pH.}

This is a quadratic. Solve using quadratic

This is a quadratic. Solve using quadratic

formula.

formula.

or you can make an approximation if x is very

or you can make an approximation if x is very

small! (Rule of thumb: 10

small! (Rule of thumb: 10-5-5 or smaller is ok) _{or smaller is ok)}

or you can make an approximation if x is very

or you can make an approximation if x is very

small! (Rule of thumb: 10

small! (Rule of thumb: 10-5-5 or smaller is ok) _{or smaller is ok)}

Equilibria Involving A Weak Acid

Equilibria Involving A Weak Acid

Equilibria Involving A Weak Acid

Equilibria Involving A Weak Acid

Equilibria Involving A Weak Acid

Equilibria Involving A Weak Acid

Equilibria Involving A Weak Acid

Step 3

Step 3

.

.

Solve K

Solve K

expression

expression

You have 1.00 M HOAc. Calc. the equilibrium concs.

You have 1.00 M HOAc. Calc. the equilibrium concs.

of HOAc, H

of HOAc, H₃₃OO++, OAc_{, OAc}--, and the pH._{, and the pH.}

First assume x is very small because

First assume x is very small because

K

K_aa is so small. is so small.

Now we can more easily solve this

Now we can more easily solve this

approximate expression.

Step 3

Step 3

.

.

Solve K

Solve K

approximate

approximate

expression

expression

You have 1.00 M HOAc. Calc. the equilibrium concs.

You have 1.00 M HOAc. Calc. the equilibrium concs.

of HOAc, H

of HOAc, H₃₃OO++, OAc_{, OAc}--, and the pH._{, and the pH.}

x =

x = [ [HH₃₃OO++] = [] = [_OAcOAc--] = 4.2 x 10] = 4.2 x 10-3-3 M M

pH = - log [

pH = - log [HH₃₃OO++] = -log (4.2 x 10] = -log (4.2 x 10-3-3) =) = 2.37 2.37

Equilibria Involving A Weak Acid

Equilibria Involving A Weak Acid

Equilibria Involving A Weak Acid

Now you try:

Now you try:

Now you try:

Now you try:

Calculate the pH of a 0.20

Calculate the pH of a 0.20 MM solution of HCN. solution of HCN. The K

The K_aa value of HCN is: K value of HCN is: K_aa = 4.9 x 10 = 4.9 x 10-10-10._.

HCN + H

HCN + H22O O ↔↔ CN CN-- + H + H 3

3OO++ Approximate solution

Approximate solution

Equilibrium Constants

Equilibrium Constants

for Weak Bases: in general:

for Weak Bases: in general:

Equilibrium Constants

Equilibrium Constants

for Weak Bases: in general:

for Weak Bases: in general:

Weak base has K

Weak base has K

< 1

< 1

Leads to small [OH

Eample: Equilibria Involving A

Eample: Equilibria Involving A

Weak Base

Weak Base

You have 0.010 M NH

You have 0.010 M NH₃₃. Calculate the pH. . Calculate the pH. K

K_bb = 1.8 x 10 = 1.8 x 10-5-5

NH

NH33 + H + H22O O ↔ ↔ NHNH44++ + OH + OH-

-Step 1.

Step 1. Define equilibrium conc. in ICE box.Define equilibrium conc. in ICE box.

[NH[NH₃₃] ] ↔↔ [NH [NH₄₄++] + [OH_{] + [OH}--]_]

initial

initial

change

change

equilib

equilib

0.010 00 00

0.010

0.010 00 00

-x

-x +x+x +x+x

-x

-x +x+x +x+x

0.010 - x

0.010 - x x x xx

0.010 - x

Equilibria Involving A Weak Base

Equilibria Involving A Weak Base

You have 0.010 M NH

You have 0.010 M NH₃₃. Calc. the pH.. Calc. the pH. NH

NH₃₃ + H + H₂₂O O  NH NH₄₄++ + OH _{+ OH}- -K

K_bb = 1.8 x 10 = 1.8 x 10-5-5 Step 2

Step 2.. Solve the equilibrium expressionSolve the equilibrium expression

Assume x is small, so

Assume x is small, so

x = [OH

x = [OH--] = [NH_{] = [NH} 4

4++] = 4.2 x 10] = 4.2 x 10-4-4 M M

and [NH

and [NH₃₃] = 0.010 - 4.2 x 10] = 0.010 - 4.2 x 10-4-4 ≈ 0.010 M _{≈ 0.010 M} The approximation is valid

Equilibria Involving A Weak Base

Equilibria Involving A Weak Base

You have 0.010 M NH

You have 0.010 M NH

. Calc. the pH.

. Calc. the pH.

NH

NH

+ H

+ H

O

O





NH

NH

+ OH

+ OH

-K

K

= 1.8 x 10

= 1.8 x 10

Step 3.

Step 3.

Calculate pH

Calculate pH

[OH

[OH

] = 4.2 x 10

] = 4.2 x 10

M

M

so pOH = - log [OH

so pOH = - log [OH

] = 3.37

] = 3.37

Because pH + pOH = 14,

Because pH + pOH = 14,

pH = 10.63

What else can I solve for?

A student prepared a 0.10 M solution of

formic acid, (HCOOH) and found its pH at

25°C to be 2.38. Calculate the K

of formic

acid.

You Try:

A 0.020

M

solution of niacin, one of the B

vitamins, has a pH of 3.26. What is its acid

dissociation constant?

AP Sample Question

At 25°C, aqueous solutions with a pH of 8 have a hydroxide ion concentration, [OH-], of

A. 1 x 10-14 M

B. 1 x 10-8 M

C. 1 x 10-6 M

AP Sample Question

Caffeine (C₈H₁₀N₄O₂) is a weak base with a K_b value of 4 x 10-4. The pH of a 0.01 M solution of caffeine

is in the range of

Frequently Missed Q’s:

13 15 24 23 4

MC Scores:

Polyprotic Acids?

Monoprotic

acids:

3COO- Ka

Diprotic

acids:

H

SO



H

+ HSO

HSO



H

+ SO

42-

K

Triprotic

acids:

H

PO



H

+ H

2

PO

K

H

PO



H

+ HPO

42-

K

HPO



H

+ PO

K

Polyprotic acids and pH?

The solubility of CO

in water at 25°C and 0.1

atm is 0.0037

M

. The common practice is to

assume that all the dissolved CO

is in the form

of carbonic acid (H

CO

), which is produced:

CO

(aq)

+ H

O

(l)

↔

↔

H

CO

(aq).

K

K

= 4.3 x 10

= 4.3 x 10

and K

and K

a2

= 5.6 x 10

= 5.6 x 10

What is the pH of 0.0037

What is the pH of 0.0037

M

M

solution of

solution of

H

H

CO

CO

? What is the [CO

? What is the [CO

]?

]?

Ionization Constants of Conjugate Acid-Base Pairs If you know K_a but need a K_b value:

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

A- _{(aq) + H}

2O (l) OH- (aq) + HA (aq)

K_a

K_b

H₂O (l) H+ _(aq) + OH- _(aq) K

w

K_aK_b = K_w

Weak Acid and Its Conjugate Base

K_a = Kw

K_b Kb =

Find the K

value?

Ammonia is a weak base, and is written in the

hydrolysis equation below:

NH

NH

+ H

+ H

O

O

↔

↔

NH

NH

+ OH

+ OH

K

K

b

= 1.8 x 10

= 1.8 x 10

Write this equation to represent an acid dissociation.

Write this equation to represent an acid dissociation.

Calculate the corresponding acid dissociation

Calculate the corresponding acid dissociation

constant.

constant.

K

Molecular Structure and Acid Strength

Z O H Z _O- + H+

- _+

The O-H bond will be more polar and easier to break if: • Z is very electronegative or

Molecular Structure and Acid Strength

1. Oxoacids having different central atoms (Z) that are from the same group and that have the same oxidation number.

Acid strength increases with increasing electronegativity of Z

H O Cl O O •• •• •• •• •• •• •• •• ••

H O Br O O •• •• •• •• •• •• •• •• ••

Cl is more electronegative than Br

Molecular Structure and Acid Strength

2. Oxoacids having the same central atom (Z) but different numbers of attached groups.

Acid strength increases as the oxidation number of Z increases.

Molecular Structure and Acid Strength

3. Small molecular diatomic acids.

Acid strength increases with DECREASING electronegativity of Z

Cl is more electronegative than I,therefore:

Each particle diagram shown is a representation of an aqueous solution of one of the acids listed in the table. The molarity of the acids in the solutions is the same. Based on the information,

which particle diagram best corresponds HIO?

The table above

provides the chemical structures for weak bases and their

ionization

constants, K_b. Based on the data, which of the following provides the best reason for the

trend in base strengths?

The different

Anion

Ending Acid Name

-ide

hydro

-

(stem)

-

ic acid

-ate

(stem)

-

ic acid

-ite

(stem)

-

ous acid

Quick Acid Nomenclature

No Oxygen

No Oxygen

w/Oxygen

w/Oxygen

An easy way to remember which goes with which…

An easy way to remember which goes with which…

“

Examples:

Name the following acids:

Nitric acid

Sulfuric Acid

Hydrofluoric Acid

Nitrous Acid

HNO₃

H₂SO₄

HF

percent ionization = Ionized acid concentration at equilibrium

Initial concentration of acid x 100%

For a monoprotic acid HA

Percent ionization = [H

+]

Percent Ionization Trend

• % ionization INCREASES with DECREASING acid concentration

• % ionization

DECREASES with INCREASING acid concentration

Calculate % ionization:

Calculate the percent ionization of a 0.020 M

solution of niacin. K

= 1.5 x 10

You Try:

If a solution of 0.10 M formic acid, (HCOOH)

is 4.2% ionized, calculate the value of K

and

find the pH of this solution.

AP Sample Question

HCOOH

(aq)

+ H

O

(l)



HCOO

(aq)

+ H

3

O

(aq)

The ionization of HCOOH in water is represented by

the equation above. In a 0.020 M HCOOH

(aq)

solution, the percent ionization of HCOOH is 9.5%.

Calculate the value of

K

for HCOOH.