BACK TITRATION
NUR SYAKINA BINTI ISHAK
NUR RAIHAN BINTI MOHD AZMI
NURUL SYAZWANI BT SHAFIEE
NUR MIRA NABILAH BINTI JAMALUDIN
NURUL ASYIQIN BINTI KHAIROLL ANNUAR
WAN NURULIYANA HAFIEZAH BINTI WAN ISMAIL
NURFATHINISSA BINTI ROSLAN
Alternative technique to
direct titration
In a simple acid-base
titration,a base (reagent) is
added in a known quantity
– greater than the amount
required for acid
neutralization.
The remaining base is
titrated with a standard
acid.
Acid and base reacts
completely.
The system has gone from
being ACID, past the
equivalent point to the
BASIC (excess base), and
back to the equivalent point
again. The final titration to
the equivalent point is called
BACK TITRATION.
Back titration
reaction is slow.
add NaOH in excess
allow the reaction to reach
completion
titrate the excess NaOH
with a standard solution of
HCl.
The system has gone from
being ACID , past the
equivalence point to the
BASIC side (excess base),
and then back to the
equivalence point.
The final titration to the
equivalence point is called
a BACK TITRATION.
EXAMPLE OF BACK TITRATION
The titration of insoluble acid organic acid with NaOH
SET UP EXPERIMENT
React with the excess volume of reactant which has
been left over after completing reaction with the
analyte from the normal titration
The substance or solution of unknown concentration
of excess intermediate reactant is made to react with
known volume and concentration of intermediate
reactant solution in back titration
Properties of back-titration
Throughout back titration, the reaction can reach the
completion quickly as the excess reactant that react with
the NaOH (as example) heated, and is much easier to
measure
Back titration also an indirect titration procedure
the proportion consumed in the reaction of back titration
being obtained by difference
PURPOSE OF BACK TITRATION
Back titration is
designed to
• 1: The analyte may be in solid
resolve some of
form
the problems
• 2: The analyte may contain
encountered
impurities which may interfere
with forward or
with direct titration. Consider the
case of contaminated chalk. We
direct titration.
can filter out the impurities before
Possible reasons
the excess reactant is titrated and
for devising
thus avoid this situation.
back titration
technique are :
Back titration is
designed to
resolve some of
the problems
encountered with
forward or direct
titration. Possible
reasons for
devising back
titration technique
are :
• 3: The analyte reacts slowly with
titrant in direct or forward titration.
The reaction with the intermediate
reactant can be speeded up and
reaction can be completed say by
heating.
• 4: Weak acid – weak base reactions
can be subjected to back titration for
analysis of solution of unknown
concentration. Recall that weak acidweak weak titration does not yield a
well defined change in pH, which can
be detected using an indicator.
useful if the
endpoint of the
reverse titration
is easier to
identify than
the endpoint of
the normal
titration
useful when
trying to work
out the amount
of an acid or
base in a nonsoluble solid.
ADVANTAGES OF BACK
TITRATION
Needs skill and
practise for
effective results
Instruments have
to be properly
calibrated since
it will give
affected the final
result.
Reactivity of the
elements to be
titrated should be
well researched
since this may
affect the end
point.
Time consuming
if done manually
DISADVANTAGES OF BACK
TITRATION
Example 1
150.0 mL of 0.2105 M nitric acid was added in excess
to 1.3415 g calcium carbonate. The excess acid was
back titrated with 0.1055 M sodium hydroxide. It
required 75.5 mL of the base to reach the end point.
Calculate the percentage (w/w) of calcium carbonate in
the sample.
CALCULATION
1.EXTRACT INFORMATION
• HNO3
V=150.0 mL
M=0.2105 M
• CACO3
Mass= 1.3415 g
• NAOH
M=0.1055 M
V=75.5 mL
2. Write balanced equation
2HNO3 + CaCO3  Ca(NO3)2 + CO2 + H2O ------ 1
HNO3 + NaOH  NaNO3 + H2O ------- 2
2 mole of HNO3 react with 1 mole of CaCO3
1 mole of HNO3 react with 1 mole of NaOH
3. Calculate no of mole
Initial amount HNO3:
No of mole of acid = 0.2105 x 150
= 31.575 mole acid.
Excess acid
No of mole of excess acid = 0.1055 x 75.5
= 7.965 mmole acid
mole of acid reacted with CaCO3 = ( 31.575 – 7.965 )
= 23.61 mole acid
4. Mole ratio
2 mole of HNO3 react with 1 mole of CaCO3
Thus,23.61 mole of HNO3 react with ½(23.61) mole of
CaCO3
mole of CaCO3 = ½ x mole acid
= ½ x 23.61
= 11.805 mole CaCO3.
5. Find mass
Gram CaCO3 = mole x molar mass
= 11.805 x 10-3 x 100
= 1.1805 g.
6.Find percentage
weight CaCO3
% CaCO3 = weight of sample  100
=
1.1805
1.3415
 100
= 87.99 % (w/w)
A 0.500g sample containing Na2CO3 is analyzed by adding
50.0ml of 0.100M HCL, a slight excess, boiling to remove
CO2, and then back-titrating the excess acid with 0.100M
NaOH . If 5.6ml NaOH is required for the back titration,
what is the percent Na2CO3 in the sample?
Molar mass for Na2CO3 = 106
Answer: 47.1%
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