Post on 30-Dec-2015
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BACK TITRATIONNUR 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
Back titration
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.
Acid and base reacts completely.
The remaining base is titrated with a standard
acid.
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.
EXAMPLE OF BACK TITRATIONThe titration of insoluble acid organic acid with NaOH
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.
SET UP EXPERIMENT
Properties of back-titration
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
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
• 1: The analyte may be in solid form
• 2: The analyte may contain impurities which may interfere with direct titration. Consider the case of contaminated chalk. We can filter out the impurities before the excess reactant is titrated and thus avoid this situation.
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 acid-weak weak titration does not yield a well defined change in pH, which can be detected using an indicator.
Back titration is designed to
resolve some of the problems
encountered with forward or direct titration. Possible
reasons for devising back
titration technique are :
ADVANTAGES OF BACK TITRATION
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 non-soluble solid.
DISADVANTAGES 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
CALCULATION
Example 1150.0 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.
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
% CaCO3 = 100
= 100
= 87.99 % (w/w)
sample ofweight
CaCOweight 3
1.3415
1.1805
Try this
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%