• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Back Titration Lab Report. In my experiment, I hoped to find the amount of calcium carbonate in some mineral limestone using the back titration method

Extracts from this document...

Introduction

Back Titration Lab Report Aim: Determining the percentage purity of calcium carbonate in a sample of limestone. Introduction: In my experiment, I hoped to find the amount of calcium carbonate in some mineral limestone using the back titration method The equation of the reaction is as follows: 2HCl + CaCO3 � CaCl2 + CO2 + H2O As not all the acid will be used up in the above reaction, I plan to obtain the amount of acid not used up and consequently the amount of calcium carbonate in the limestone, by titrating it with known sodium hydroxide solution. The equation of the reaction is as follows: HCl + NaOH � NaCl + H2O Apparatus: The equipment and reagents that I used are as follows: 250 cm3 beaker Electronic balance (� 0.01 g) ...read more.

Middle

I repeated the procedure to obtain 3 sets of readings and will take the average value as my result. Data processing and presentation: With all the above processes being done, I wrote down the values which I obtained in the raw data table below. Raw data Raw Data Measure Volume of acid used / cm3 (� 0.05 cm3) 1 41.6 2 41.4 3 41.5 I then processed my data to find the average volume of acid used. Vavg = V1+V2+V3 3 Vavg = 41.6 + 41.4 + 3 Vavg = 41.5 (� 0.05 cm3) Now I used the various formulas related to the mole concept to find the amount of calcium carbonate in the sample of limestone. The equations of the reactions are as follows: 1) 2HCl + CaCO3 � CaCl2 + CO2 + H2O 2) ...read more.

Conclusion

1 2 = 4.075 x 10-3 moles Mass of CaCO3 in 1.5g of limestone = Moles x RMM (RMM of CaCO3 = 100) = 4.075 x 10-3 x 100 = 0.41g Percentage purity of calcium carbonate = 0.41 x 100 1.5 = 27.3 � 0.35 % Conclusion: A possible source of error in this experiment is the determination of the end-point, which is characterised by the solution just turning orange. This is because a slightly greater volume of acid may have been used than required to produce the pink colour. To try to reduce the effects of this error I would like to carry out a large number of titrations and their average used in the calculation. Another possible way to reduce error is by using more accurate measuring instruments like a more precise burette so as to reduce the uncertainty of the measurements. ?? ?? ?? ?? ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Aqueous Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Here's what a teacher thought of this essay

4 star(s)

This is a four star piece of work with excellent scientific knowledge of molar calculations and demonstrated great skill in their work. A clear, concise piece of work but they could have put more into the introduction and conclusion.

Marked by teacher Patricia McHugh 01/12/2012

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Aqueous Chemistry essays

  1. Marked by a teacher

    ANALYSIS OF ASPIRIN BY BACK TITRATION

    4 star(s)

    Aspirin is a medicine commonly found in households around the world. It also is one of the least expensive and most useful drugs in the market. A Chemist named Felix Hoffmann first synthesized aspirin, otherwise known as acetylsalicylic acid, in 1897 from salicylic acid.

  2. Titration Experiment

    Readings Results 1st reading 23.5 cm3 2nd reading 23.3 cm3 3rd reading 23.2 cm3 Average reading 23.3 cm3 Calculations Reading 1= 21.9 cm3 reading 2=22.1 cm3 reading=21.8cm3 Calculation of average titre: 21.9+22.1+21.8= 65.8 cm3 65.8 3 Volume of alkali used=25 cm3 Concentration of alkali used=0.1 mol Volume of acid used=

  1. To investigate the rate of reaction between different concentrations of hydrochloric acid with metal ...

    right hand as well, I can then pour the acid and time it at the same time and can quickly put on the stopper without losing too much CO2. 3. Put the CaCO3 first, before the acid and use a paper funnel so that the Calcium carbonate bit does not get stuck on the side.

  2. A Colorimetric Determination of Manganese In Steel

    Calibration Graph. Concentration of Acidified Potassium Permanganate (mol/l) Absorbance 4.0x10-5 0.095 8.0x10-5 0.170 1.2x10-4 0.260 1.6x10-4 0.340 2.0x10-4 0.480 2.4x10-4 0.550 2.8x10-4 0.650 Unknown 1 0.260 Unknown 2 0.335 From the results it can be concluded that the concentration of the unknown 1 solution is 1.16mol/l and the concentration of the unknown 2 solution is 1.34mol/l.

  1. Indigestion Tablets Investigation

    "2NaHCO3 --> Na2CO3 + H2O + CO2" (page 169) This sodium carbonate would then react with the acid to form sodium chloride, water and carbon dioxide. 2HCl + Na2CO3 --> 2NaCl + H2O + CO2 Alternatively the sodium bicarbonate could react directly with the acid - HCl + NaHCO3 --> NaCl + H2O + CO3 The other active ingredient is magnesium Trisilicate.

  2. Investigating the effects of varying pH levels on the germination of cress seeds

    be detrimental to the experiment as a whole, as all of the samples will be subject to identical conditions, whatever the change. After completing the experiment, it was clear that the only samples which were successful in any germination were those of both acids with either 0% concentration or 20% concentration i.e.

  1. To investigate the effect of concentration on the temperature rise, heat evolved and heat ...

    because HCl and NaOH in a particular volume will give same amount of moles of H+ ions and OH- ions. So when the temperature reached maximum it is obvious that all the H+ ions and OH- ions have bonded, and no more H+ ions left in the mixture to react.

  2. To prepare antifebrin using phenylammonium chloride C6H5NH3CL and Ethanoic anhydride (CH3CO)2O.

    + (CH3CO)2O CH3CONHC6H5 + CH3COOH + HCL Molar ratio: 1 : 1 The molar mass of Phenylammonium chloride (C6H5NH3CL) is: 129.5 g mol-1 The amount of Phenylammonium chloride used 1.0g Therefore the moles of Phenylammonium chloride is calculated using: Mass = 1__ = 0.00722 mol-1 Molar mass 129.5 1 mol of Phenylammonium chloride gives 1 mol of antifebrin.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work