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

Aim: To estimate volumetrically the amount of Calcium carbonate present in the eggshell

Extracts from this document...


Chemistry Lab Report BACK TITRATION Aim: To estimate volumetrically the amount of Calcium carbonate present in the eggshell and analysing the sources of errors to evaluate the results. Hypothesis: Back Titration is used instead of conventional acid-base titration when one of the reactants is not very soluble in water. Variables: Independent variable: Potassium hydrogen phthalate, Volume of HCl, Weight of Eggshell Dependent variable: Concentration & volume consumed of NaOH. Concentration of Calcium Carbonate in the eggshell Controlled variables: Phenolphthalein solution, Ethanol Procedure Apparatus: 1 Boiled Egg Titration Setup Analytical Balance Watch Glass The experiment is divided in to three parts. Part A: (Standardisation of 1 M NaOH) We used the watch glass to take 2g of Potassium hydrogen phthalate and then recorded its exact mass up to 2 decimal figures. Then we transferred the salt into a 250ml conical flask and added 50ml of distilled water; swirled the flask to dissolve the salt by taking care that nothing remains in excess. We added 2-drops of phenolphthalein indicator to the solution in the conical flask and then we started doing titration by keeping NaOH in the burette. Then we kept of doing that until we reached the end point of it which was the change of colour from colourless to (pink), we recorded the Volume of NaOH consumed and calculated the molarity of the NaOH, Part B: (Standardisation of 2 M HCl) ...read more.


we had the minimum count of 0.1 which was divided by 2 to 0.05 but the uncertainty was occurring at two points so I kept it as 0.05 X 2= 0.1 Part B- Reaction involved in part B is- NaOH + HCl → NaCl + H2O Mole Ratio 1 1 The Strength of HCl (C) = (vol. of NaOH X Strength of NaOH)/ Vol. of HCl The Strength of HCl (C) = 17.9/1000±0.1 (dm3) X 0.93 (mol dm-3) = 1.66 (mol dm-3) 10/1000(dm3) The uncertainty is kept as 0.1 for the volumes and Concentration as we had the minimum count of 0.1 which was divided by 2 to 0.05 but the uncertainty was occurring at two points so I kept it as 0.05 X 2= 0.1 Part C- Reaction involved in part C is- CaCO3 + 2HCl → CaCl2 + CO2 + H2O Mole Ratio 1 2 Moles of NaOH = Concentration of NaOH X Volume of NaOH Moles of NaOH= 0.93 (mol dm-3) X 14.8/1000(dm3) = 0.014 (moles) Moles of NaOH = Moles of HCl(excess) 0.014 (moles) = 0.014 (moles) Moles of HCl reacted with CaCO3 = Moles of HCl initial – moles of HCl in ( (vol. X cocn.) (excess) Moles of HCl reacted with CaCO3 = [60/1000(dm3) X 1.66] – 0.014 = 0.09 Moles of HCl reacted with CaCO3 = 0.09 moles ½ moles of CaCO3 = Moles of HCl ½ moles of CaCO3 = 0.09/2 = 0.05 Mass of CaCO3 = moles of CaCO3 X Molar mass of CaCO3 Mass of CaCO3 = 0.05(gm) ...read more.


The overall quality of data is good but, the readings are précised but not exactly accurate that is because of huge alteration between the volumes of NaOH, yes there is another error of human reflection to the change in colour of liquid from colourless to pink which is an random error, The equipment used is Analytical Balance, Watch Glass, Beaker 100(ml), pipette bulb, Conical Flask- 250(ml), Burette, Funnel, and Stand. The reaction between the eggshell and the HCl, Ethanol took a lot of time so we were supposed to calculate the readings for the Part C the eggshell one at the first place. That saved our time and gave us time to complete the full experiment within the given time. The readings of data were perfect and I don’t think that I need much changes. The only thing that can be changed is using a steriliser which will reduce the time of reaction, and secondly I feel that using the same flask for three readings and cleaning it with distilled water was not accurate. There no such important issues to be dealt in the experiment as the apparatus used were accurate but we had some random errors due to human’s indulgence in calculating the readings, that was solved by using uncertainty values. The reading taken is once so we cannot be completely sure about the reading if we could have done it twice then we might have got the average readings which may have been a better judgement. Submitted By- LOVISH MEHNDIRATTA Class- 11th F ...read more.

The above preview is unformatted text

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

Found what you're looking for?

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

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 International Baccalaureate Chemistry essays

  1. Determination of potassium hydrogen carbonate into potassium carbonate

    X 100 = � 0.5% for trial 2: (0.1/14.300) X 100 = � 0.4 % The overall uncertainty for K2CO3 for the temperature change is the summing up of all the uncertainties. 0.5 + 0.4 = � 0.9% ( this has been used above ) This is the same method for the calculation of the uncertainties for KHCO3, For trial 1: (0.1/21.000)

  2. How much calcium carbonate is in an eggshell

    The volume of base added was used to calculate the amount of acid that had not reacted with the eggshell. The amount of acid that had reacted with the eggshell was calculated by subtraction. This amount of acid was used to calculate the amount of calcium carbonate present in the

  1. FInding the percentage purity of CACO3 in egg shell

    control the variable would be to prepare the solution and perform the experiment at the same temperature. 1. Mass of the egg shells The mass of the egg shells that is taken is controlled by the person performing the experiment.

  2. Calculate % of caco3 in white egg shell

    Add two drops of phenolphthalein to the solution. 4. In a burette add 0.100moldm-3 of sodium hydroxide. 5. Now titrate the solution. Add sodium hydroxide solution at a very slow rate, at the same time keep shaking the conical flask so that the phenolphthalein that shows the pink colour is quickly dissipated.

  1. Chemistry Titration Acid Base Lab

    Additionally this residue could have affected the pH of the sodium hydroxide solution that was to be put in the beaker for the purpose of this titration lab. However, there were some droplets of water left on the sides of the beaker after it was cleansed which would have decreased

  2. Enthalpy of Neutralisation Between HCl and NaOH

    ( = %n + %q) ( = ± 2.89) (% 100 = ±5.56%) Conclusion: The literature value for ΔH in the neutralization of NaOH and HCl is ΔH=-57.1 kJmol-1 The precision error, or random error attributed to the precision of instruments is ±5.56%, which is highish for a precision error.

  1. Research question: how to convert NaOH to NaCl by two different routes , and ...

    + 2M HCl (aq) 1MNaCl (aq) 50.00cm3 Uncertainties: value uncertainties Volume of HCl 50.00cm3 0.03 cm3 = 3x10-5 dm3 Volume of H2O 50.0 cm3 0.5cm3 = 5x10-4 dm3 Mass of NaOH 4.0 g 0.1 g Temperature Different original temperatures 0.1 ºC 1. to find the required amount of NaOH needed for each reaction :

  2. Acid Base titration to determine the percentage by mass of calcium carbonate in an ...

    For this procedure it cannot be used directly to titrate the CaCO3 because it is very slow when the reaction is close to reach the point of equivalence, so instead, the determination is achieved by adding an excess of acid, which was the HCl, to dissolve all of the CaCO3

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