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n
Experiment 2
Sam le re ort
Topic
: Volumetric analysis – Redox
Purpose
: To determine the ratio of the number of moles of hydroxyammonium ions to the number of moles of iron (III) ions participating in the reaction.
Materials : KA 1 is a solution containing 1.58 g of potassium manganate (VIII) 3 per 500 cm KA 2 is a solution prepared by boiling 4.00 g of hydroxyammonium 3 sulphate, (NH3OH)2 SO4 per dm with excess iron (III) ammonium sulphate and dilute sulphuric acid -3 KA 3 is 1.0 mol dm sulphuric acid 3
Apparatus : One 25 cm pipette and pipette filler Three titration flasks 3 One 50 cm burette One retort stand and clamp 3 One 50 cm measuring cylinder One white tile One wash bottle filled with distilled water Introduction : In the presence of hydrogen ions, the hydroxyammonium ion, + NH3OH will reduce iron (III) ion to iron (II) ion while the NH3OH+ itself will be oxidised to dinitrogen oxide. 3
Procedure : (a) 25.0 cm of KA 2 solution is pipetted into a titration flask. 3 (b) Approximately 25 cm of KA 3 is added into KA 2 and this mixture is titrated with solution KA 1. (c) The titration is repeated as many times to achieve accurate results. (d) The titration readings are recorded in the table below.
Results
:
Titration number
Rough
Accurate 1
2
25.00 0.00
25.00 0.00
25.10 0.00
25.00
25.00
25.10
3
Final reading/cm 3 Initial reading/cm 3
Volume of KA 1/cm
3
3
(i) 25.0 cm of KA 2 required 25.05 cm of KA 1 for a complete reaction.
(ii) Calculate your average titre value showing the suitable titre values that you use. Average titre value =
25.00 + 25.10
2 3 = 25.50 cm
-3
Questions : (a) Calculate the concentration, in mol dm , of manganate (VII) ions in solution KA 1.
The concentration of potassium manganate (VII) in mol dm n = v n = The number of moles of KMnO 4 1.58 v = The volume of KMnO4 mol 158 = 3 0.5 dm -3
= 0.02 mol dm
-3
-3
(b) Calculate the concentration, in mol dm , of iron (II) ions in solution KA 2. The redox equation of the reaction between iron (II) ions and the manganate (VII) ions are as follow. 5Fe
2+
-
+
+ MnO4 + 8H 5Fe
Using the formula
ma va
3+
2+
+ Mn
+ 4H2O
a
=
mb vb
b -3
Where m1 = concentration of KMnO4 in mol dm -3 m2 = concentration of iron(II) ions in mol dm 3 v1 = The volume of KA 1 in cm 3 v2 = The volume of KA 2 in cm -3
(c) Calculate the mass of iron(II) ions in 1 dm of KA 2. 3
The mass of iron(II) ions in 1 dm of KA 2 = the number of moles of iron(II) ions x relative atomic mass = 0.100 mol x 55.8 = 5.58g
(d) Determine the number of moles of iron(III) ions required to oxidise 1 mol of hydroxyammonium ions Fe
2+
Fe
3+
+e
-
1 mol of iron (II) ions gives 1 moles of iron(III) ions ∴
The concentration of iron(II) ions = The concentration of iron(II) ions -3 = 0.100 mol dm The concentration of hydroxyammonium sulphate, (NH3OH)2 SO4 -3 = 4.00 g dm 4 = 2(14.0 + 3.0 + 16.0 ++ 1.0) + 32.1 4(16.0)
4.00 164.1 -3 = 0.02438 mol dm =
+
The concentration of NH3OH ions = 2 x 0.02438 -3 = 0.04876 mol dm The number of moles of iron (III) ions required to oxidise 1 mole of hydroxyammonium ions 0.100 = 0.04876 = 2.05 mol ≈ 2 mol +
(e) Write a balanced redox equation between NH3OH ions and Fe ions. Reduction
2+
: 4Fe
-
+ 4e +
4Fe
3+
2+ +
-
2 2 Oxidation : 2NH3OH + N O 3++ H O + 6H + 4e + 2+ Redox Equation : 2NH3OH + 4Fe N2O +H2O + 6H + 4Fe
(f) Why would the titration not require an external indicator? The titration does not require an external indicator because potassium manganate (VII) can act as indicator. Potassium managanate (VII) turns colourless when manganate (VII) ions are reduced. The end point can be determined when the solution turns to pink colour. Conclusion : The ratio of number of moles of hydroxyammonium ions to the number of moles of iron (III) ions participating in the reaction is 1:2