Transcript Document

Acids & Bases
Acids:
acids are sour tasting
Arrhenius acid: Any substance that, when dissolved
in water, increases the concentration of hydronium
ion (H3O+)
Bronsted-Lowry acid: A proton donor
Lewis acid: An electron acceptor
Bases:
bases are bitter tasting and slippery
Arrhenius base: Any substance that, when dissolved
in water, increases the concentration of hydroxide
ion (OH-)
Bronsted-Lowery base: A proton acceptor
Lewis acid: An electron donor
Lone Hydrogen ions do not exist by themselves in
solution. H+ is always bound to a water molecule
to form a hydronium ion
Brønsted-Lowry Theory of Acids & Bases
Conjugate Acid-Base Pairs
General Equation
Brønsted-Lowry Theory of Acids & Bases
Brønsted-Lowry Theory of Acids & Bases
Brønsted-Lowry Theory of Acids & Bases
Notice that water is both an
acid & a base = amphoteric
Reversible reaction
ELECTROLYTES
Electrolytes are species which conducts electricity when
dissolved in water. Acids, Bases, and Salts are all
electrolytes.
Salts and strong Acids or Bases form Strong
Electrolytes. Salt and strong acids (and bases) are fully
dissociated therefore all of the ions present are
available to conduct electricity.
HCl(s) + H2O  H3O+ + ClWeak Acids and Weak Bases for Weak Electrolytes.
Weaks electrolytes are partially dissociated therefore
not all species in solution are ions, some of the
molecular form is present. Weak electrolytes have
less ions avalible to conduct electricity.
NH3 + H2O  NH4+ + OH-
Acids & Bases
STRONG
_ completely ionized
_ strong electrolyte
_ ionic/very polar bonds
bonds
Strong Acids:
HClO4
H2SO4
HI
HBr
HCl
HNO3
vs
WEAK
_ partially ionized
_ weak electrolyte
_ some covalent
Strong Bases:
LiOH
NaOH
KOH
Ca(OH)2
Sr(OH)2
Ba(OH)2
Acids & Bases
• One ionizable proton:
HCl → H+ + Cl• Two ionizable protons:
Combined:
H2SO4 → H+ + HSO4+ + SO 2H
SO
→
2H
2
4
4
+
2HSO4 → H + SO4
• Three ionizable protons:
H3PO4 → H+ + H2PO4–
Combined:
H2PO4- → H+ + HPO42H3PO4 → 3H+ + PO43HPO42- → H+ + PO4-3
Acids & Bases
For the following identify the acid and the base as strong or weak .
a. Al(OH)3 + HCl 
Weak base
Strong acid
b. Ba(OH)2 +
Strong base
HC2H3O2 
Weak acid
c. KOH + H2SO4 
Strong base Strong acid
d. NH3 + H2O 
Weak base Weak acid
Acids & Bases
For the following predict the product. To check your answer left click
on the mouse. Draw a mechanism detailing the proton movement.
a. Al(OH)3 + 3 HCl  AlCl3 + 3 H2O
b. Ba(OH)2 + 2 HC2H3O2  Ba(C2H3O2)2 + 2 H2O
c. 2KOH + H2SO4  K2SO4 + 2 H2O
d. NH3 + H2O  NH4+ + OH-
Conjugate Acid-Base Pairs
Conjugate Acid-Base Pairs
Acids & Bases
For the following Identify the conjugate acid and the conjugate base.
The conjugate refers to the acid or base produced in an acid/base
reaction. The acid reactant produces its conjugate base (CB).
a. Al(OH)3 + 3 HCl  AlCl3 + 3 H2O
CB
CA
b. Ba(OH)2 + 2 HC2H3O2  Ba(C2H3O2)2 + 2 H2O
CB
c. 2 KOH + H2SO4  K2SO4 + 2 H2O
CB
d. NH3 + H2O  NH4+ + OHCA
CB
CA
CA
TITRATION
Titration of a strong acid with a strong base
ENDPOINT = POINT OF NEUTRALIZATION =
EQUIVALENCE POINT
At the end point for the titration of a strong acid with a strong
base, the moles of acid (H+) equals the moles of base (OH-) to
produce the neutral species water (H2O). If the mole ratio in
the balanced chemical equation is 1:1 then the following
equation can be used.
MOLES OF ACID = MOLES OF BASE
nacid = nbase
Since M=n/V
MAVA = MBVB
TITRATION
MAVA = MBVB
1. Suppose 75.00 mL of hydrochloric acid was required to
neutralize 22.50 mLof 0.52 M NaOH. What is the molarity of
the acid?
HCl + NaOH  H2O + NaCl
Ma Va = Mb Vb rearranges to Ma = Mb Vb / Va
so Ma = (0.52 M) (22.50 mL) / (75.00 mL)
= 0.16 M
Now you try:
2. If 37.12 mL of 0.843 M HNO3 neutralized 40.50 mL of KOH,
what is the molarity of the base?
Mb = 0.773 mol/L
Molarity and Titration
TITRATION
Titration of a strong acid with a strong base
ENDPOINT = POINT OF NEUTRALIZATION =
EQUIVALENCE POINT
At the end point for the titration of a strong acid with a strong
base, the moles of acid (H+) equals the moles of base (OH-)
to produce the neutral species water (H2O). If the mole
ratio in the balanced chemical equation is NOT 1:1 then
you must rely on the mole relationship and handle the
problem like any other stoichiometry problem.
MOLES OF ACID = MOLES OF BASE
nacid = nbase
TITRATION
1. If 37.12 mL of 0.543 M LiOH neutralized 40.50 mL
of H2SO4, what is the molarity of the acid?
2 LiOH + H2SO4  Li2SO4 + 2 H2O
First calculate the moles of base:
0.03712 L LiOH (0.543 mol/1 L) = 0.0202 mol LiOH
Next calculate the moles of acid:
0.0202 mol LiOH (1 mol H2SO4 / 2 mol LiOH)= 0.0101 mol
H2SO4
Last calculate the Molarity:
Ma = n/V = 0.010 mol H2SO4 / 0.4050 L = 0.248 M
2. If 20.42 mL of Ba(OH)2 solution was used to
titrate29.26 mL of 0.430 M HCl, what is the molarity
of the barium hydroxide solution?
Mb = 0.308 mol/L
Molarity and Titration
• A student finds that 23.54 mL of a 0.122 M
NaOH solution is required to titrate a 30.00-mL
sample of hydr acid solution. What is the
molarity of the acid?
• A student finds that 37.80 mL of a 0.4052 M
NaHCO3 solution is required to titrate a 20.00mL sample of sulfuric acid solution. What is the
molarity of the acid?
• The reaction equation is:
H2SO4 + 2 NaHCO3 → Na2SO4 + 2 H2O + 2 CO2
Water Equilibrium
Water Equilibrium
Kw = [H+] [OH-] = 1.0 x 10-14
Equilibrium constant for water
 Water or water solutions in which [H+] = [OH-] = 10-7 M
are neutral solutions.
 A solution in which [H+] > [OH-] is acidic
 A solution in which [H+] < [OH-] is basic
pH
A measure of the hydronium ion
• The scale for measuring the hydronium ion concentration
[H3O+] in any solution must be able to cover a large range. A
logarithmic scale covers factors of 10. The “p” in pH stands for
log.
• A solution with a pH of 1 has [H3O+] of 0.1 mol/L or 10-1
• A solution with a pH of 3 has [H3O+] of 0.001 mol/L or 10-3
• A solution with a pH of 7 has [H3O+] of 0.0000001 mol/L or 10-7
pH = - log [H3O+]
The pH scale
The pH scale ranges from 1 to 10-14 mol/L or from 1
to 14.
pH = - log [H3O+]
1 2 3 4 5 6 7 8 9 10 11 12 13 14
acid
neutral
base
Manipulating pH
Algebraic manipulation of:
pH = - log [H3O+]
allows for:
[H3O+] = 10-pH
If pH is a measure of the hydronium ion
concentration then the same equations could be
used to describe the hydroxide (base)
concentration.
[OH-] = 10-pOH
pOH = - log [OH-]
thus:
pH + pOH = 14 ; the entire pH range!
PRACTICE PROBLEM #25
1. How many milliliters of 1.25 M LiOH must be added to neutralize
34.7 mL of 0.389 M HNO3?
10.8 mL
2. What mass of Sr(OH)2 will be required to neutralize 19.54 mL of
0.00850 M HBr solution?
0.0101 g
3. How many mL of 0.998 M H2SO4 must be added to neutralize 47.9
mL of 1.233 M KOH?
29.6 mL
4. What is the molar concentration of hydronium ion in a solution of
pH 8.25?
-9
5.623 x 10 M
5. What is the pH of a solution that has a molar concentration of
hydronium ion of 9.15 x 10-5?
pH = 4.0
6. What is the pOH of a solution that has a molar concentration of
hydronium ion of 8.55 x 10-10? pOH = 4.9
GROUP STUDY PROBLEM #25
______1. How many milliliters of 0.75 M KOH must be added to neutralize
50.0 mL of 2.50 M HCl?
______2. What mass of Ca(OH)2 will be required to neutralize 100 mL of
0.170 M HCl solution?
______3. How many mL of 0.554 M H2SO4 must be added to neutralize
25.0 mL of 0.9855 M NaOH?
______ 4. What is the molar concentration of hydronium ion in a solution
of pH 2.45?
______ 5. What is the pH of a solution that has a molar concentration of
hydronium ion of 3.75 x 10-9?
______ 6. What is the pOH of a solution that has a molar concentration of
hydronium ion of 4.99 x 10-4?