Acid-Base Chemistry

Tentative Test Date

Monday, April 19th

Today’s Objectives• Distinguish among the three definitions of

acids and bases.• Calculate [H+], [OH-], and pH.

Definitions of Acids and Bases

ACIDS BASES

Arrhenius Proton Producer in

Aqueous Solutions

Hydroxide Producer in

Aqueous Solutions

Brønsted-Lowry

Proton Donor Proton Acceptor

Lewis Electron Acceptor

Electron Donor

More

Genera

l

Proton = H+

Identifying Acids and Bases

For the following neutralization reactions, identify the acid and the base.

1) HF + H2O F- + H3O+

2) NH3 + H2O NH4+ + OH –

Since water can act as both an acid and a base, it is called amphoteric.

Acids = Lose H+ Bases = Gain H+

Acid Base

AcidBase

Auto-Ionization of Water

H2O(l) + H2O(l) H3O+(aq) + OH-(aq)

hydroniumion

hydroxideion

watermolecule

Auto-Ionization of Water

H2O(l) + H2O(l) H3O+(aq) + OH-(aq)

Kw = [H3O+][OH-]

Kw = Water Dissociation Constant

At 25oC, Kw equals 1.0 x 10-14.

[H3O+] = Concentration of H3O+

[OH-] = Concentration of OH-

Example Problem

At 25oC, the concentration of hydroxide ions in an aqueous solution is 4.5 x 10-6M. Determine the concentration of hydronium ions or [H3O+].

Kw = [H3O+][OH-]

[H3O+] = = 2.2x10-9M

[OH-][OH-]

MOH

Kw6

14

105.4

100.1

][

“Simplified” Equation

H2O(l) + H2O(l) H3O+(aq) + OH-(aq)

H2O(l) H+(aq) + OH-(aq)

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

H+ and H3O+ are interchangeable in the equation.

Simplified

Now You Try

A solution’s hydrogen ion concentration is 9.01 x 10-10 M. What is the

concentration of hydroxide ions in this solution?

OHMMH

KOH w 5

10

14

1011.11001.9

100.1

][

pH

pH = - log [H+]

More Acidic More Basic

Simple pH Calculations

Ex. #1: What is the pH of a solution with a hydronium ion concentration of 1.0 x10-

4M?

pH = -log[H3O+] = - log (1.0x10-4 M) = 4

Ex. #2: Calculate the pH of a solution if [H+] equals 1.0x10-9M.

pH = -log[H+] = - log (1.0x10-9 M) = 9

Which is more acidic: Example #1 or 2?Example #1 is more acidic since it has a

lower pH.

pH with Log Tables

Ex. #3: Calculate pH of 4.58 x 10-3 M H+.

pH = - log [H+]

pH = - log (4.58 x 10-3 M)

pH = - [log (4.58) + log (10-3)]

pH = -[0.661 + -3] = - [ -2.339]

pH = 2.339

pH with Log Tables

Ex. #3: Calculate pH of a solution if the hydronium ion

concentration is 1.79 x 10-8 M.

pH = - log [H3O+]

pH = - log (1.79 x 10-8 M)

pH = - [log (1.79) + log (10-8)]

pH = -[0.253 + -8] = - [ -7.647] = 7.647

Homework

Lab notebooks due tomorrow!!

Worksheet on Naming and Writing Formulas for Acids and Bases

Book Problems pg. 582 # 6 & 7 pg. 586 # 8 & 9

Acid-Base Chemistry

Tentative Test Date

Monday, April 19th

Today’s Objective• Calculate [H+], [OH-], pH and pOH.

Review from Yesterday

Calculate the pH of a solution whose hydronium ion concentration is

2.71 x 10 -7 M.

pH = - log (2.71 x 10 -7 M) pH = -[log (2.71) + log (10 -7)]

pH = - [0.433 + -7] pH = 6.567

Combining the Problems

Using your calculator, calculate the pH of a solution whose hydroxide ion concentration is

1.3 x 10-3M.Step 1: Find [H+].

Step 2: Calculate pH.pH = -log (7.69 x 10-12 M)

pH = - [log (7.69) + log (10-12)]pH = -[0.886 + -12]

pH = 11.114

MMOH

KH w 12

3

14

1069.7103.1

100.1

][][

Relationships between pH, [H3O+], and [OH-]

As pH increases… The solution becomes more (acidic or

basic). The [H3O+] (increases or decreases). The [OH-] (increases or decreases).

[H+] from pH

pH = -log[H+]

-pH = log[H+]

-1 -1

10 10

HpH10

You will either need to be able to derive this formula or memorize this formula.

[H+] from pH

Ex #1: Calculate the hydronium ion concentration for a solution with a pH

of 6.719.

[H+] = 10–pH = 10–6.719

Using the Log Tables,

10-6.72 = 10-7 x 100.281

1.91 x 10-7 M

From Log

Tables

[H+] from pH

Ex #2: Calculate the hydronium ion concentration for a solution with a pH

of 10.5.

[H+] = 10–pH = 10–10.5

Using Your Calculator, press the following buttons for most

calculators.

2nd Log – 10.5

3.16 x 10-11 M

pOH

pOH = -log [OH-]

Using your log tables, calculate the pOH of a solution that contains 0.0807 M OH-.

0.0807 M = 8.07 x 10-2M

pOH = - log (8.07x10-2M) = -[log (8.07) + log(10-

2)]

pOH = -[0.907 + -2] = 1.093

Relating pH and pOH

Kw = [H+][OH-]

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

-log(1.0x10-14) = -log([H+][OH-])

-log(1.0x10-14) = -log[H+] + -log[OH-]

14 = pH + pOH

A Summary of Calculations

Important FormulasKw = [H+][OH-] pH + pOH =

14pH = -log[H+] [H+] = 10-pH

pOH = -log[OH-] [OH-] = 10-

pOHTypes of Solution

pH [H+] vs [OH-]

Acidic Less than 7 [H+] > [OH-]

Neutral 7.000… [H+] = [OH-]

Basic Greater than 7 [H+] < [OH-]