Transcript 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.
Acid and base reacts completely.
The remaining base is titrated with a standard acid.
The system has gone from being
, past the equivalent point to the (excess base), and back to the equivalent point again. The final titration to the equivalent point is called
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
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 resolve some of the problems encountered with forward or direct titration. Possible reasons for devising back titration technique are : • 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.
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.
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
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.
• 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 2HNO 3 + CaCO 3 Ca(NO 3 ) 2 HNO 3 + NaOH NaNO 3 + CO 2 + H 2 O ------ 1 + H 2 O ------- 2 2 mole of HNO 3 react with 1 mole of CaCO 3 1 mole of HNO 3 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 CaCO 3 = ( 31.575 – 7.965 ) = 23.61 mole acid 4. Mole ratio 2 mole of HNO 3 react with 1 mole of CaCO 3 Thus,23.61 mole of HNO 3 CaCO 3 mole of CaCO 3 react with ½(23.61) mole of = ½ x mole acid = ½ x 23.61
= 11.805 mole CaCO 3 .
5. Find mass Gram CaCO 3 = mole x molar mass = 11.805 x 10 -3 x 100 = 1.1805 g.
6.Find percentage % CaCO 3 weight CaCO 3 = 100 1.1805
100 = 87.99 % (w/w)
A 0.500g sample containing Na 2 CO 3 is analyzed by adding 50.0ml of 0.100M HCL, a slight excess, boiling to remove CO 2, 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 Na 2 CO 3 in the sample?
Molar mass for Na 2 CO 3 = 106 Answer: 47.1%