http://wps.prenhall.com/wps/media/o
bjects/3312/3392202/blb1703.html
Sites:

http://chemed.chem.wisc.edu/chempaths/GenChemTextbook/Titrations-875.html
IB: http://ibchem.com/IB/ibnotes/full/aab_htm/18.5.htm
Simulation:
http://www.avogadro.co.uk/chemeqm/acidbase/titration/phcurves.htm
http://www.ausetute.com.au/titrcurv.html

A titration is a volumetric technique in which a
solution of one reactant (the titrant) is added to a
solution of a second reactant (the "analyte") until the
equivalence point is reached.

The equivalence point is the point at which titrant
has been added in exactly the right quantity to react
stoichiometrically with the analyte.

An indicator may be added which has an "endpoint"
(changes color) at the equivalence point, or the
equivalence point may be determined from a titration
curve.
 In
an acid-base titration, a solution
containing a known concentration of base
is slowly added to an acid (or the acid is
added to the base).
 Acid-base
indicators can be used to signal
the equivalence point of a titration (the
point at which stoichiometrically
equivalent quantities of acid and base
have been brought together).
The titrant is added to the
solution from a buret, and
the pH is continually
monitored using a pH meter.
To understand why titration
curves have certain
characteristic shapes, we will
examine the curves for three
kinds of titrations:
(1) strong acid-strong base;
(2) weak acid-strong base;
a
pH meter can be used to monitor the
progress of the reaction producing a pH
titration curve, a graph of the pH as a
function of the volume of the added titrant.

The shape of the titration curve makes it
possible to determine the equivalence point
in the titration.

The titration curve can also be used to
select suitable indicators and to determine
the Ka of the weak acid or the Kb of the weak
base being titrated.

Titration – the progressive transfer of a solution from a
buret (called the titrant) into a measured volume of
another solution (called the sample).

Equivalence point – the volume of titrant required to
neutralize the sample (# mol acid = # mol base).

Endpoint – the pH at the equivalence point of a
titration, where the indicator changes color.

Indicator – a chemical which is added to the sample that
changes colour at the equivalence point of a titration.

Buffering region – a horizontal region of the pH curve
where pH is not changing significantly.
 The
endpoint of a titration is NOT the same
thing as the equivalence point:
 The
equivalence point is a single point
defined by the reaction stoichiometry as the
point at which the base (or acid) added
exactly neutralizes the acid (or base) being
titrated.
 The endpoint is defined by the choice of
indicator as the point at which the colour
changes. Depending on how quickly the
colour changes, the endpoint can occur
almost instantaneously or be quite wide.
 If
an appropriate indicator is chosen such
that the endpoint of the titration occurs at
the equivalence point, then a colour change
in the solution being titrated can be used as
a signal that the equivalence point has been
reached.

Intuition may suggest that the endpoint of the titration will occur
at the equivalence point if we choose an indicator whose pKa is
equal to the pH of the equivalence point. If such an indicator was
chosen, the colour change would be half complete at the
equivalence point.

Thus for titrating a weak acid with a strong base where, at
equivalence point the solution is slightly acidic, an optimum
indicator should have a pKa > 7, for example Thymol Blue (pKa =
8.9) or Phenolphthalein (pKa = 9.4)

For titrating a weak base with a strong acid at equivalence point
encounters a slightly basic solution. Thus an optimum indicator
should have a pKa < 7, for example Bromocresol Green (pKa =
4.7) or Methyl Red (pKa = 5.1)

For titration involving strong base and a strong acid an optimum
indicator should have a pKa ~ 7, for example Bromothymol Blue
(pKa = 7.0) or Phenol Red (pKa = 7.9
 1.
Consider the titration when 0.100 M NaOH
solution have been added to 50.00 mL of
0.100 M HCl.

The initial pH—The pH of the solution before the
addition of any base is determined by the initial
concentration of the strong acid.
For a solution of 0.100 M HCl
[H+] = 0.100 M
pH = -log(0.100) = 1.00



Between the initial pH and the equivalence
point— ( adding 49 mL of NaOH)
As NaOH is added, the pH increases slowly at first and then
rapidly in the vicinity of the equivalence point.
The pH of the solution before the equivalence point is
determined by the concentration of acid that has not yet
been neutralized.
nH+ = 0.1 x 0.05 = 5 x 10-3 nOH- = 0.1 x 0.049 = 4.9 x 10-3
nH+ left =
5 x 10-3 – 4.9 x 10-3 = 1 x 10-4 mol H+
Vsolution = 50 + 49 = 99 mL = 0.099L
[H+ ] = 1x10-4 / 0.099 = 0.001M
=> pH = 3

The equivalence point—At the equivalence
point equal number of moles of the NaOH and
HCl have reacted, leaving only a solution of their
salt, NaCl. No calculation is required to deduce
that the pH is 7.00, because the cation of a
strong base (in this case Na+) and the anion of a
strong acid (in this case Cl–) do not hydrolyze and
therefore have no appreciable effect on pH.

After the equivalence point—The pH of the
solution after the equivalence point is determined
by the concentration of the excess NaOH in the
solution.(adding 51 mL of NaOH)
V=50 + 51 = 101 mL = 0.101 L
nH+ = 5 x 10-3 nOH- = 0.1 x 0.051 = 5.1 x 10-3
nOH- left = 5.1 x 10-3 – 5.0 x 10-3 = 1 x 10-4 mol OH[OH- ] = 1x10-4 / 0.101 = 9.9 x 10-4 M
pOH = 3.00
pH = 11.00
 3.
What is the pH when 48.00 ml of .100 M
NaOH solution have been added to 50.00 ml
of .100 M HCl solution?

As an example, consider the titration curve for the titration of 50.0 mL
of 0.100 M acetic acid, HC2H3O2, with 0.100 M NaOH.
The initial pH—This pH is just the pH of the 0.100 M HC2H3O2.
The calculated pH of
0.100 M HC2H3O2 is 2.89
(using Ka)


Between the initial pH and the equivalence point—To determine
pH in this range, we must consider the neutralization of the acid
Calculate the pH of the solution formed when 45.0 mL of 0.100 M
NaOH is added to 50.0 mL of 0.100 M HC2H3O2 (Ka = 1.8 10–5).
Consider the neutralization of the acid.

Prior to reaching the equivalence point, part of the HC2H3O2 is
neutralized to form C2H3O2–. Thus, the solution contains a mixture of
HC2H3O2 and C2H3O2–.

pH = 5.70

The equivalence point is reached after adding
50.0 mL of 0.100 M NaOH to the 50.0 mL of
0.100 M HC2H3O2. At this point the 5.00 10–3 mol
of NaOH completely reacts with the 5.00 10–3 mol
of HC2H3O2 to form 5.00 10–3 mol of their salt,
NaC2H3O2.

The Na+ ion of this salt has no significant effect
on the pH.

The C2H3O2– ion, however, is a weak base, and
the pH at the equivalence point is therefore
greater than 7.
 Finding
the concentration of the salt :
na = 0.005
nsalt = 0.005 = n C2H3O2V = 50 + 50 = 100 mL = 0.100 L
Because the salt is a weak base
Kb = Kw/Ka
[OH- ] = 5.3 x 10-6 pH = 8.72