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BACK TITRATION

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. Back titration.

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BACK TITRATION

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  1. 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

  2. Back titration

  3. EXAMPLE OF BACK TITRATIONThe titration of insoluble acid organic acid with NaOH

  4. SET UP EXPERIMENT

  5. Properties of back-titration

  6. PURPOSE OF BACK TITRATION

  7. ADVANTAGES OF BACK TITRATION

  8. DISADVANTAGES OF BACK TITRATION

  9. 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

  10. 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

  11. 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 HNO3react with 1 mole of NaOH

  12. 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

  13. 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.

  14. 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)

  15. 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% Try this

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