Acid-Base Titration

Acid-Base Titration
Learning Target:
Explain how neutralization reactions
are used in acid-base titrations.
A neutralization reaction is a reaction in
which an acid and a base in an aqueous
solution react to produce a salt and water.
A salt is an ionic compound made up of a
cation from a base and an anion from an
Neutralization is a double-replacement
Acid Base Titration
Titration is a method for determining the
concentration of a solution by reacting a known
volume of that solution with a solution of known
In a titration procedure, a measured volume of
an acid or base of unknown concentration is
placed in a beaker or flask, and initial pH
The standard solution is filled in a
A couple drops of acid base
indicator are added to the flask.
The standard solution is slowly
added to the unknown solution in
the flask.
As the two solutions are mixed the
acid and the base are neutralized.
Standard Solution or titrant– An acid/ base
solution whose concentration is known.
 End point: The point at which an indicator
changes color.
 Equivalence point: The point at which
moles of H+ ion from the acid equals
moles of OH– ion from the base. An
abrupt change in pH occurs at the
equivalence point.
Titration of Strong acid & strong base
Since the salt produced is
neutral, the solution at the
equivalence point has a pH of
the pH starts off low and
increases as you add more
sodium hydroxide solution.
Again, the pH doesn't change
very much until you get close
to the equivalence point.
Then it surges upwards very
Titration of weak acid and strong base
Salt is basic so,
equivalence point
comes at a pH > 7.
The start of the graph
shows a relatively
rapid rise in pH but
this slows down
Titration of weak base & strong acid
Salt formed is acidic,
hence, equivalence point
comes at a pH < 7.
At the very beginning of
the curve, the pH starts
by falling quite quickly as
the acid is added, but the
curve very soon gets less
Chemical dyes
whose color are
affected by acidic
and basic solutions are
called acid-base indicators.
A 43.0 mL of sodium hydroxide was titrated
against 32.0 mL of 0.100 M hydrochloric
acid. What is the molarity of sodium
hydroxide solution?
Step 1: Find the moles of hydrochloric acid:
Mol = M x L
Mol = 0.100 mol x .032 L
HCl = .0032 mol
Step 2: Write a balanced equation, and find
the molar ratio between the acid and the
HCl + NaOH  NaCl + H2O
HCl : NaOH
1 mol : 1mol
 NaoH = 0.0032 Mol
Answer ….
Step 3: Find the concentration of sodium
M (NaOH) = mol/ L
 M = 0.0032 mol
.043 L
NaOH = 0.0744 M
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