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%