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

Titration of an Unknown Potassium Halide Against Silver Nitrate

Extracts from this document...

Introduction

Titration of an Unknown Potassium Halide Against Silver Nitrate Apparatus * 250cm3 volumetric flask + 0.3cm3 * 25cm3 pipette + 0.06cm3 * Filler * Conical flask * 50cm3 burette + 0.05cm3 * Retort stand and burette holder Chemicals * 1.035g + 0.001g unknown potassium halide * Distilled water * 0.1moldm-3 Potassium chromate indicator * 0.05moldm-3 Silver Nitrate Solution Data Collection Mass of solid white Unknown potassium Halide used: 1.035g +0.001g. This was made up to a 250cm3 colourless solution. 0.1moldm-3 yellow potassium chromate indicator was used in this titration. The end point was judged to be when a dark pink precipitate started to form. Rough Titration Titration 1 Titration 2 Titration 3 Initial Burette Reading + 0.05cm3 0.00cm3 18.80cm3 1.00cm3 18.40cm3 Final Burette Reading + 0.05cm3 18.80cm3 36.30cm3 18.40cm3 35.60cm3 Titre +0.1cm3 18.80cm3 17.50cm3 17.40cm3 17.20cm3 The average Titre = 17.50 + 17.40 + 17.20 = 17.36666...cm3 3 = 17.40 (4sf) ...read more.

Middle

Therefore 1 mole of KX = 1.035 . 8.70 x 10-3 = 118.9gmol-1 One mole of KX has an atomic mass of 118.9gmol-1. Atomic mass of one mole of KX = 118.9gmol-1 Atomic Mass of one mole of K = 39.10gmol-1 Therefore the atomic mass of one mole of X = 118.9 - 39.10 Atomic mass of X = 79.80gmol-1 Therefore X = Br The unknown potassium halide was Potassium Bromide. Percentage Uncertainty Percentage Uncertainty of Volumetric Flask = 0.3 . x 100 = 0.12% 250.0 Percentage Uncertainty of Pipette: 0.06 x 100 = 0.24% 25.0 Percentage Uncertainty of balance: 0.001 x 100 = 0.10% 1.035 Percentage Uncertainty of Burette: 0.1 x 100 = 0.57% 17.4 Overall Percentage Uncertainty of Equipment = 0.24 + 0.10 + 0.57 + 0.12 = 1.03% Effect of the total percentage uncertainty on the result [Atomic Mass KX = 118.9gmol-1] = 118.9 x 1.03 100 = 1.22gmol-1 (3sf) ...read more.

Conclusion

This may have resulted in the end point being overstepped, or not quite reached, which would have affected the accuracy of the results. Although there was a total percentage error on the equipment of 1.03%, this is negligible, as the greatest error on the results in this titration would be caused by personal judgement, as when the end point is judged to have occurred is dependent on the person carrying out the titration. The accuracy of the results obtained from this titration may be improved if the silver nitrate solution was held in a container that blocked out light. As a result of this, the silver nitrate would be less affected by the light, and consequently the end points of each titration would be more consistent, therefore improving the accuracy of the results. ?? ?? ?? ?? Emma Wellham 0502966 Titration of an Unknown Potassium Halide Against Silver Nitrate Page 1 of 3 ...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. Aim: To determine the concentration of chloride ions in sea water by titration ...

    Thus, the end-points achieved will be precise. Concentration and type of titrant. To find accurate and reliable results, the same silver nitrate solution must be used. Otherwise, it would result in different readings which cannot be relied upon for further calculations. Use F1 solution provided for every titration carried out.

  2. Testing for halide ions. The objective of the experiment was: ...

    � Ag2(NH3)+(aq) + 2Cl-(aq) * Potassium A15 + ammonia? Silver nitrate+ A15 ion 2AgA15(s) + NH3 (aq) � A152(NH3)+(aq) + 2Cl-(aq) * Potassium B15 + ammonia? Silver nitrate+ B15 ion 2AgB15(s) + NH3 (aq) � Ag2(NH3)+(aq) + 2B15-(aq) Evaluation: The experiment went very well. I have read about the testing the halogens but I did not remember.

  1. Determination of potassium hydrogen carbonate into potassium carbonate

    and of � 1.2 % as the uncertainty for measuring temperature. Therefore 1.7 + 1.2 = 2.9 % Therefore the total uncertainty of the enthalpy change is 2.9 + 3.0 = � 5.9 % By using Hess's law the enthalpy change for the thermal decomposition of potassium hydrogen carbonate into potassium carbonate can be found.

  2. IB questions and answers on Atomic Theory

    The following graph is produced when a pure sample of boron is passed through a mass spectrometer. Use the data to calculate the relative atomic mass for boron. 100 + 24.65 = 124.65 10B = (24.65/124.65) x 100 = 19.78% 11B = (100/124.65) x 100 = 80.22% (10 x 0.1978)

  1. Using Solubility Rules to Indentify Unknown Solutions

    Data Collection and Processing: Photograph of Lap Setup: Qualitative Observations and Raw Data Table: Five drops ionic solution � 1/2 drop Five drops ionic solution � 1/2 drop KCl MgCl2 Na2SO4 NaOH BaCl2 MgSO4 KCl NR NR NR NR NR MgCl2 NR Ppt - off white, thick consistency, vapor observed

  2. Determining the position of unknown element X in the Reactivity Series

    Take a strip of Element X and cut it into 7 equal pieces 2. Pour 5ml of CuSO4 into a test tube 3. Put a thermometer into one test CuSO4 and measure the temperature 4. Now put a piece of Element X into the test tube and measure ?H 5.

  1. Research into the production of Nitrate Fertillisers.

    Manufacturing process for Ammonium sulphate Dryer type Particulate Kg/Mg lb/Ton VOCb Kg/Mg lb/Ton Rotatory dryer (uncontrolled) 23 46 0.74 1.48 Wet scrubber 0.02c 0.04c 0.11 0.22 Fluidize- bed dryers Uncontrolled 109 218 0.74 1.48 Wet scrubber 0.14 0.28 0.11 0.22 Ammonium Nitrate.

  2. The balanced equation shows that two moles of silver nitrate react with one mole ...

    % uncertainty = ±0.111 Calculation of Ratio n(Cu) : n(Ag) 0.00522 : 0.0167 total % uncertainty = ±0.111+0.605 1:3.20 % uncertainty = ±0.716 1:3 % uncertainty = ±0.7 Conclusion This reaction is an example of a single replacement redox reaction, as the copper element replaces an element in a compound, silver nitrate producing silver and copper nitrate.

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