Transcript Ch. 14 - Liquids - Clayton Valley Charter High School

Ch. 19 – Acids & Bases
III. Titration
(p. 612 – 616)
A. Neutralization
 Chemical
reaction between an acid and
a base
 Products are a salt (ionic compound)
and water
A. Neutralization
ACID + BASE  SALT + WATER
HCl + NaOH  NaCl + H2O
strong
strong
neutral
HC2H3O2 + NaOH  NaC2H3O2 + H2O
weak
strong
basic
• Salts can be neutral, acidic, or basic
• Neutralization does not mean pH = 7
A. Neutralization
ACID + BASE  SALT + WATER
• Double replacement reaction
• Can predict products – ex. antacids in
your stomach
2 HCl + Ca(OH)2  CaCl2 + 2H2O
strong
strong
neutral
B. Titration
 Titration
standard solution
• Analytical method
in which a
standard solution
is used to
determine the
concentration of an
unknown solution
unknown solution
B. Titration
 Stoichiometry
• Provides basis for titration
• Examples:
HCl + NaOH  NaCl + H2O
• 1 mol HCl will neutralize 1 mol NaOH
2HCl + Mg(OH)2  MgCl2 + 2H2O
• 2 mol HCl will neutralize 1 mol Mg(OH)2
B. Titration
 Equivalence
point
• Point at which equal number
of moles of H3O+ and OHare in solution
• Determined by…
• indicator color change
• dramatic change in pH
B. Titration
Strong acid, strong base
Weak acid, strong base
B. Titration
 Endpoint
• Point at which indicator color changes
• Should be close to equivalence point if
correct indicator is chosen
B. Titration
+
O
OH
moles H3 = moles
MAVAnA = MBVBnB
M: Molarity
V: volume
n: # of H+ ions in the acid
or OH- ions in the base
B. Titration
1.
50.0 mL of an unknown solution of NaOH are
titrated with 0.100 M HCl. Find the molarity of
the NaOH solution if 52.5 mL of acid are
required to reach the equivalence point.
H3O+
OH-
M = 0.100M
M=?
V = 52.5 mL
n=1
V = 50.0 mL
n=1
MAVAnA = MBVBnB
(0.100)(52.5mL)(1)
=M(50.0mL)(1)
M = 0.105M NaOH
B. Titration
2.
42.5 mL of 1.3M KOH are required to
neutralize 50.0 mL of H2SO4. Find the
molarity of H2SO4.
H3O+
OH-
M=?
M = 1.3M
V = 50.0 mL
n=2
V = 42.5 mL
n=1
MAVAnA = MBVBnB
M(50.0mL)(2)
=(1.3M)(42.5mL)(1)
M = 0.55M H2SO4