Announcements & Agenda (02/23/07): Announcements & Agenda (02/23/07) You should be reading Ch 10 this weekend!
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Acid & base strength: Quantitative (8.4-8.5)
the pH scale
Acid & base reactions
Last Time: Bronsted-Lowry Acids & Bases: Last Time: Bronsted-Lowry Acids & Bases acids donate a proton (H+)
bases accept a proton (H+)
Last Time: Strengths of Acids/Bases - Ionization: A strong acid/base completely ionizes (100%) in aqueous solutions.
HCl(g) + H2O(l) H3O+ (aq) + Cl− (aq)
COMPARE TO STRONG ELECTROLYTES
A weak acid/base dissociates only slightly in water to form a few ions in aqueous solutions.
H2CO3(aq) + H2O(l) H3O+(aq) + HCO3− (aq)
COMPARE TO WEAK ELECTROLYTES Last Time: Strengths of Acids/Bases - Ionization
Strong Acids (Know These): make up six (just a few) of all the acids.
have weak conjugate bases (the product formed after the proton is transferred). Strong Acids (Know These)
Strong Bases: Strong Bases are formed from metals of Groups 1A (1) and 2A (2).
include LiOH, NaOH, KOH, and Ca(OH)2.
dissociate completely in water.
KOH(s) K+(aq) + OH−(aq)
Ionization of Water: A Basis for Understanding pH (H+ concentrations): In water occasionally,
H+ is transferred from 1 H2O molecule to another.
one water acts an acid, the another acts as a base.
H2O + H2O H3O+ + OH−
.. .. .. ..
:O: H + H:O: H:O:H + + :O:H−
.. .. .. ..
H H H
water water hydronium hydroxide ion (+) ion (-) Ionization of Water: A Basis for Understanding pH (H+ concentrations)
Pure Water is Neutral (NOT ACIDIC OR BASIC): Pure Water is Neutral (NOT ACIDIC OR BASIC) the ionization of water molecules produces small, but equal quantities of H3O+ and OH− ions.
molar concentrations are indicated in brackets as [H3O+] and [OH−].
[H3O+] = 1.0 x 10−7 M
[OH−] = 1.0 x 10−7 M
Acidic Solutions: Acidic Solutions Adding an acid to pure water:
increases the [H3O+].
causes the [H3O+] to exceed 1.0 x 10-7 M.
decreases the [OH−].
Basic Solutions: Basic Solutions Adding a base to pure water:
increases the [OH−].
causes the [OH−] to exceed 1.0 x 10− 7M.
decreases the [H3O+].
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
Ion Product of Water, Kw: The ion product constant, Kw, for water
is the product of the concentrations of the hydronium and hydroxide ions.
Kw = [ H3O+] [ OH− ]
can be obtained from the concentrations in pure water.
Kw = [ H3O+] [ OH− ]
Kw = [1.0 x 10− 7 M] x [ 1.0 x 10− 7 M]
= 1.0 x 10− 14
Ion Product of Water, Kw
[H3O+] and [OH−] in Solutions: [H3O+] and [OH−] in Solutions IMPORTANT: Kw is always 1.0 x 10−14.
Calculating [H3O+]: Calculating [H3O+] What is the [H3O+] of a solution if [OH−] is 5.0 x 10-8 M?
STEP 1: Write the Kw for water.
Kw = [H3O+ ][OH− ] = 1.0 x 10−14
STEP 2: Rearrange the Kw expression.
[H3O+] = 1.0 x 10-14
[OH−]
STEP 3: Substitute [OH−].
[H3O+] = 1.0 x 10-14 = 2.0 x 10-7 M
5.0 x 10- 8
If lemon juice has [H3O+] of 2 x 10−3 M, what is the [OH−] of the solution?: If lemon juice has [H3O+] of 2 x 10−3 M, what is the [OH−] of the solution? 1) 2 x 10−11 M
2) 5 x 10−11 M
3) 5 x 10−12 M
Solution: 3) 5 x 10−12 M
Rearrange the Kw to solve for [OH- ]
Kw = [H3O+ ][OH− ] = 1.0 x 10−14
[OH− ] = 1.0 x 10 -14 = 5 x 10−12 M
2 x 10 - 3
Solution
pH Scale: pH Scale The pH of a solution
is used to indicate the acidity of a solution.
has values that usually range from 0 to 14.
is acidic when the values are less than 7.
is neutral with a pH of 7.
is basic when the values are > 7.
NOTE: pH is a logarithmic scale!!!
pH of Everyday Substances: pH of Everyday Substances
Testing the pH of Solutions: Testing the pH of Solutions The pH of solutions can be determined using
a) pH meter
b) pH paper
c) indicators that have specific colors at different pH values.
Calculating pH: pH is the negative log of the hydronium ion concentration.
pH = - log [H3O+]
Example: For a solution with [H3O+] = 1 x 10−4
pH = −log [1 x 10−4 ]
pH = - [-4.0]
pH = 4.0
Note: The number of decimal places in the pH equals
the significant figures in the coefficient of [H3O+].
4.0 1 SF in 1 x 10-4 Calculating pH
Learning Check: A. The [H3O+] of tomato juice is 2 x 10−4 M.
What is the pH of the solution?
1) 4.0 2) 3.7 3) 10.3
B. The [OH−] of a solution is 1.0 x 10−3 M.
What is the pH of the solution?
1) 3.00 2) 11.00 3) -11.00
Learning Check
If an area received 1 inch of rain with a pH of 4, how much more neutral rain would be needed to have a final pH of 6?: If an area received 1 inch of rain with a pH of 4, how much more neutral rain would be needed to have a final pH of 6? Approximately 2 inches
Approximately 9 inches
Approximately 20 inches
Approximately 100 inches
[H3O+], [OH-], and pH Values: [H3O+], [OH-], and pH Values
Calculating [H3O+] from pH: Calculating [H3O+] from pH The [H3O+] can be expressed by using the pH as the negative power of 10.
[H3O+] = 1 x 10 -pH
For pH = 3.0, the [H3O+] = 1 x 10 -3
On a calculator
1. Enter the pH value 3.0
2. Change sign -3.0
3. Use the inverse log key (or 10x) to obtain
the [H30+]. = 1 x 10 -3 M
Neutralization Rxns of Acids & Bases: In a neutralization reaction:
a base such as NaOH reacts with an acid such as HCl.
HCl + H2O H3O+ + Cl−
NaOH Na+ + OH−
the H3O+ from the acid and the OH− from the base form water.
H3O+ + OH− 2 H2O
Neutralization Rxns of Acids & Bases
Bases Used in Some Antacids: Bases Used in Some Antacids Antacids are used to neutralize stomach acid (HCl).
Neutralization Equations: In the equation for neutralization, an acid and a base produce a salt and water.
acid base salt water
HCl + NaOH NaCl + H2O
2HCl + Ca(OH)2 CaCl2 + 2H2O Neutralization Equations Balance these like any other reaction!
Solving Problems…: Solving Problems… What is the molarity of an HCl solution if 18.5 mL of a 0.225 M NaOH are required to neutralize 10.0 mL HCl?
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
Method:
Get into moles with “known”:
Given: 18.5 mL of 0.225 M NaOH
Do a moles-to-moles conversion
Get out of moles with “unknown”:
Slide27: 18.5 mL NaOH x 1 L NaOH x 0.225 mole NaOH
1000 mL NaOH 1 L NaOH
x 1 mole HCl = 0.00416 mole HCl
1 mole NaOH
MHCl = 0.00416 mole HCl = 0.416 M HCl
0.0100 L HCl
Calculate the mL of 2.00 M H2SO4 required to neutralize 50.0 mL of 1.00 M KOH. H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l): Calculate the mL of 2.00 M H2SO4 required to neutralize 50.0 mL of 1.00 M KOH. H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l) 1) 12.5 mL
2) 50.0 mL
3) 200. mL
Solution: Solution 1) 12.5 mL
0.0500 L KOH x 1.00 mole KOH x 1 mole H2SO4 x
1 L KOH 2 mole KOH
1 L H2SO4 x 1000 mL = 12.5 mL
2.00 mole H2SO4 1 L H2SO4
Two More Acid/Base Reactions…: Two More Acid/Base Reactions… 1. Acids react with metals
such as K, Na, Ca, Mg, Al, Zn, Fe, and Sn.
to produce hydrogen gas and the salt of the metal.
Molecular equations:
2K(s) + 2HCl(aq) 2KCl(aq) + H2(g)
Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)
Acids and Carbonates: Acids and Carbonates Acids react
with carbonates & hydrogen carbonates
to produce carbon dioxide gas, a salt, & water.
2HCl(aq) + CaCO3(s) CO2(g) + CaCl2(aq) + H2O(l)
HCl(aq) + NaHCO3(s) CO2(g) + NaCl (aq) + H2O(l)