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

Chemistry report - In this experiment, it is a redox titration method to standardize a solution of potassium manganate(VII) by an iron(II) salt (ammonium iron(II) sulphate).

Extracts from this document...

Introduction

´╗┐EXPERIMENT 3: Standardize a solution of potassium manganite(VII) by an iron(II) salt (ammonium iron(II) sulphate). Objective: 1. To standardize a solution of potassium manganate(VII) by an iron(II) salt (ammonium iron(II) sulphate). Introduction: In this experiment, it is a redox titration method to standardize a solution of potassium manganate(VII) by an iron(II) salt (ammonium iron(II) sulphate). So, the word of redox is related to the oxidation and reduction. Oxidation numbers describe the number of electrons the atom will gain or lose during a reaction. Each atom in an equation can be assigned an oxidation number according to certain rules. Oxidation occurs when the oxidation number of an atom increases while reduction occurs when the oxidation number decreases. Potassium manganate (VII) (KMnO4) solution is standardised by titration against the ammonium iron(II) sulphate, FeSO4 .(NH4)2SO4.6H2O. Potassium manganate is widely used as an oxidizing agent in volumetric analysis. While the ammonium iron(II) sulphate is used as a primary standard to standardize the KMnO4 solution. In this experiment, ammonium iron(II) sulphate crystals are the stable compound that remain as solid in room temperature. Hence, the ammonium(II) ...read more.

Middle

Calculate the number of moles of Fe2+ ions in 10.0 or 25.0 cm3 of the solution pipetted. No. of moles of Fe2+ = No. of moles of FeSO4 .(NH4)2SO4.6H2O 250.0 cm3 of FeSO4 .(NH4)2SO4.6H2O contain 0.025 mol of Fe2+ ions, Thus, no. of moles of Fe2+ in 25.0cm3 = 0.025250 x 25 = 0.0025 mol 3. Calculate the number of moles of MnO4- which reacted during titration. MnO4- + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O 1 mole of MnO4- = 5 moles of Fe2+ Hence, no. of moles of MnO4- = 0.00255 = 0.0005 mol 4. Calculate the concentration of the manganate(VII) ions, MnO4-, in mol dm-3 and in g dm-3. ( a ) in mol dm-3 Concentration of MnO4- = 0.0005 mol0.02663 dm-3 = 0.02 mol dm-3 ( b ) in g dm-3 Mass = no. of moles × molar mass = 0.0005 × 55+4(16) = 0.06 g Concentration of MnO4- = 0.06 0.02663 = 2.25 g dm-3 5. ...read more.

Conclusion

sulphate, the initial and final readings of the burette is recorded and repeated 3 times, includes 1 for rough reading and another 2 for accurate readings. So, the average value of potassium manganate(VII) is calculated to get a more accurate result. In the summary of this experiment, 25.00 cm3 of the acidified ammonium iron(II) sulphate required 26.63 cm3 of the potassium manganate(VII) solution for reaction. There are some precaution steps in this experiment in order to obtain the results more accurately and precisely. During the process of the experiment, the pipette is rinsed with a little sulphuric acid to remove water present inside the pipette. This is the way to prevent the water from diluting the acid that is poured inside the pipette. In the same way, burette is also rinsed with a little of potassium manganate(VII) solution. Besides, a piece of white tile that placed below the conical flask is used to enable us to detect the end point clearly, which the colourless solution changes to a light pink solution. Moreover, the eyes must be placed at the same level as the meniscus of the solution inside the burette in order to obtain an accurate of initial and final readings. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our AS and A Level Physical 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 AS and A Level Physical Chemistry essays

  1. Peer reviewed

    Titration Lab Report

    4 star(s)

    Readings were taken as accurately as possible, using the apparatus at hand. For example, when filling up burettes with water, we made sure that the bottom of the meniscus touched the calibrated line. Also, by looking at this point from several angles we tried to reduce the error of parallax.

  2. Determination of the Percentage of Oxalate in Iron(II) Oxalate by Redox Titration

    Steps 5 to 9 was repeated 3 more times. Data and Calculation Weight of bottle with FeC2O4.2H2O = 10.093g Weight of bottle after FeC2O4.2H2O was poured into beaker = 8.311g Mass of FeC2O4.2H2O used = 1.782g 1 2 3 4 Final reading 29.9cm3 29.0 cm3 29.9 cm3 29.4 cm3 Initial reading 2.6 cm3 1.7 cm3 2.5 cm3 2.0 cm3 Volume of KMnO4 used 27.3 cm3 (rejected)

  1. Acid-Base Titrations.

    As a consequence, Ka = 1.75 x 10-5 = [H3O+][CH3COO-]/[CH3COOH] and [H3O+] = 1.75 x 10-5, pH = 4.76. The acetic acid is three-quarters titrated when 75 mL of the NaOH solution have been added. At this point 3/4 of the original moles of CH3COOH have been titrated to CH3COO-,

  2. A Redox Titration " Determining the percentage of iron in an iron ore

    b. Pipette 25.0 cm3 of manganate(VII) solution into a 250.0 cm3 of volumetric flask. c. Repeat step(b) again. d. Add deionized water upto the graduated mark of the volumetric flask. e. Stopper the flask and invert it several times in order to mix the contents well.

  1. Investigating the Rate of the Reaction between Bromide and Bromate Ions in Acid Solution

    These reactions suggest a rate equation of the form, where A and B are the reactants, rate = [A][B]. Termolecular (involving three reactants) and higher molecularity reactions are unlikely, since these require three or more reacting particles to collide at the same instant.

  2. The purpose of this experiment was to isolate and characterize macromolecules.

    The correct results from the chromatography paper should have shown the separation of hydrolyzed nucleic acid. These results are the same as what the experiment conducted expected. The chromatography paper revealed three distinct separations. For something to be hydrolyzed, the chemical bonds must be broken.

  1. Double Displacement Reactions

    + (NH4)2SO4(aq) --> Na2SO4(aq) + 2NH4C2H3O2(aq) (NH4)2SO4(aq) + Cu(NO3)2(aq) --> NH4NO3(aq) + CuSO4(aq) (NH4)2SO4(aq) + Pb(NO3)2(aq) --> NH4NO3(aq) + PbSO4(s) 3(NH4)2SO4(aq) + 2Al(NO3)3(aq) --> 6NH4NO3(aq) + Al2(SO4)3(aq) Al(NO3)3(aq) + Cu(NO3)2(aq) --> Al(NO3)3(aq) + Cu(NO3)2(aq) Al(NO3)3(aq) + Pb(NO3)2(aq) --> Al(NO3)3(aq) + Pb(NO3)2(aq) Pb(NO3)2(aq) + Cu(NO3)2(aq) --> Pb(NO3)2(aq) + Cu(NO3)2(aq) IX.

  2. The Determination of rate equation

    Carrying out the experiment in two slots would have certainly affected the reaction rate as stated in the planning, if the temperature of the first day was higher than the second day (or vice versa), this difference in temperature would have had an effect on the overall rate of reaction

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