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

Determination of the concentration of lime water solution

Extracts from this document...


Victioria Nicholls Chemistry Coursework Title: Determination of the concentration of lime water solution Objective: The aim of this experiment is to determine the concentration of an unknown limewater solution in g dm-3 as accurately as possible, which contains 1g dm-3 of calcium hydroxide. This will be done by titration with hydrochloric acid. The Reaction occurring is: Ca(OH)2(l) + 2HCl(l) � CaCl2(l) + H2O(l) Apparatus: Metal stand Burette Conical flask 25cm3 Pipette 250cm3 volumetric flask Lime-water solution Hydrochloric acid Methyl-red indicator Diagram Method: 1.Firstly, determination of the concentration of hydrochloric acid to be used is to be carried out, as the 2.00 mol dm-3 solution of hydrochloric acid supplied is too concentrated. This was done by diluting the 2.00 mol dm-3 solution with water to the values 1.00, 0.50, 0.10 and 0.02mol dm-3 of hydrochloric acid. The was done using preliminary results: Concentration of HCl (mol dm-3) Volume of HCl required to neutralise the lime water (cm3) ...read more.


This is when the solution becomes neutral i.e. pH7. (See method on how to titrate accurately - see appendix 2) 7. So after finding a suitable dilution of hydrochloric acid to use which was 0.02 mol dm3, the burette was filled with 50cm3 of hydrochloric acid, using a measuring cylinder that measures to 1cm3. We had to make sure that there were no air bubbles present in the burette jet, this would have affected the volume of liquid in this method. To avoid this we kept the burette jet running until we could see clear that there were no air bubbles present in the burette to affect the accuracy of our results. 8. We needed an indicator which has an end point is below pH 7 because there is a sharp change below pH7 and a gradual change above pH 7. We needed a sharp change. The colours, ranges and end points vary considerably as can be seen in the table below: Name of dye Colour at low pH pH range End point (Pkind) ...read more.


Results Titration Start (cm3) Finish (cm3) Volume (cm3) Rough 50 19.0 31.00 1 50 19.3 30.70 2 50 19.4 30.60 3 50 19.5 30.50 Average Volume = 30.60cm3 Number of moles of HCl used = volume � concentration 1000 = 30.60 � 0.02 1000 = 0.0006mol dm-3 The reaction occurring is: Ca(OH)2 + 2HCl � CaCl2 + H2O As 2 : 1 HCl Ca(OH)2 0.0006 = 0.0003 moles of Ca(OH)2 2 0.0003 moles in 25cm3 0.0003 � 40 = 0.012 moles ArCa(OH)2 = 74 Moles = Mass � Mass = Moles � Ar Ar = 0.012 � 74 = 0.1g/dm-3 Therefore, the concentration of limewater is 0.1g/dm-3 Safety Substance Risk Precaution - Hydrochloric Acid - CORROSIVE - Can cause severe burns. Can be very dangerous to eyes and skin. Solutions equal to over 0.5m should be labelled corrosive. Those over 0.5m should be relabelled irritant. - Wear gloves and goggles at all times Instructions to follow before starting experiment: - Move all obstructions e.g. stools and bags - Clear work area of unnecessary equipment - Beware of spillage's - Take care with glass equipment - if breakage occurs report immediately ...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

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. The Determination of an Equilibrium Constant.

    Number of moles of CH3COOH produced = 1.65 - 1.3125 = 0.3375 mole According to the equation CH3COOH + C2H5OH CH3COOC2H5 + H2O The amount of C2H5OH produced equals the amount of CH3COOH reacted ?Number of moles of C2H5OH at equilibrium = 0.3375 mole The amount of CH3COOC2H5 (H2O)

  2. Finding out how much acid there is in a solution.

    might not have been shaken well; therefore the substances not mixed properly resulting in different concentrations of sodium carbonate for each titration. * When transferring 25cm3 of sodium carbonate solution into the conical flask using the pipette filler, all of the solution may not have been transferred properly, therefore leaving a few drops in the pipette filler.

  1. Chemistry concentration of lime water.

    The Limewater used will be a relatively weak base so it will be appropriate to use methyl orange as it has an end point on the ph scale between 8-5. Whereas phenolphthalein indicator has an end point much higher up in the ph scale.

  2. To determine the concentration of lime water

    in g dm?3: Formula mass (Ca(OH)2)= 40.1+(16+1)*2 = 74.1g Concentration of Ca(OH)2= 74.1* 1.13*10?2 g dm?3 = 0.83733 g dm?3 = 0.84 g dm?3? I conclude that my results are quite accurate as they are near to the 1g dm?3 estimate given to us.

  1. Determining the concentration of lime water

    * I will then add the deionised water which will be provided to the volumetric flask until it reaches the marked line on the volumetric flask. I will then shake the flask additionally in order to make sure that the Nitric acid has been diluted with the deionise water.

  2. Determine the concentration of lime water.

    After this it the results must be noted in a table as I have indicated in the table below. Titration 1 Titration 2 Titration 3 Final Reading 6.70 cm3 12.30 cm3 17.90 cm3 Initial Reading 1.00 cm3 6.70 cm3 12.30 cm3 Titre 5.70 cm3 5.60 cm3 5.60 cm3 Average titre

  1. Chemistry concentration of lime water.

    An accurate way of doing this is to use a volumetric flask and pipette. I want to dilute the acid to 0.1 moles, so I pipette 25.0cm3 of HCl from the original bottle and place it in the 500cm3 volumetric flask.

  2. To determine the concentration of a lime water solution

    Of moles = (volume/1000) x concentration = (100/1000) x 0.1 = 0.01 mol * The number of moles of calcium hydroxide is half the amount than that of the HCl so therefore: 0.01mol x 0.5 = 0.005 moles of Ca(OH)2 * The concentration therefore of the calcium hydroxide is: (0.005 x 1000)

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