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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-]