Volumetric Analysis Oxidation-Reduction PDF

**[Volumetric Analysis: Oxidation -- Reduction]** Volumetric Analysis may be used to determine the concentrations of oxidising and reducing agents. One of the most important oxidising agents is Potassium permanganate which has a formula of KMnO~4~. **[Potassium Permanganate (Oxidising Agent)]** \...

**[Volumetric Analysis: Oxidation -- Reduction]** Volumetric Analysis may be used to determine the concentrations of oxidising and reducing agents. One of the most important oxidising agents is Potassium permanganate which has a formula of KMnO~4~. **[Potassium Permanganate (Oxidising Agent)]** \- A purple coloured solid which is not a primary standard as it cannot be obtained in a state of very high purity.\ - Must be standardised by titration against a primary standard solution.\ - Acts as an oxidising agent by gaining five electrons in acidic solution.\ M~n~O~4~^-^ + 8H^+^ + 5e^-^ Mn^2+^ + 4H~2~O\ purple colourless \- To undergo the above reaction, it is necessary to add some dilute acid to supply the H+ ions\ - Potassium permanganate acts as its own indicator **[The Reaction of MnO~4~^-^ and Fe^2+^ ions]** It is possible to determine the concentration of KMnO~4~ solution by titrating it against a solution containing Fe^2+^ ions. KMnO~4~ is an oxidising agent when added to a solution containing Fe^2+^ ions, it converts the Fe^2+^ ions to Fe^3+^ ions. Oxidising agent: M~n~O~4~^-^ + 8H^+^ + 5e^-^ Mn^2+^ + 4H~2~O\ Reducing agent: Fe^2+^ Fe^3+^ + e^-^ It takes five Fe^2+^ ions to react with one MnO~4~^-^ ion\ MnO~4~^-^ + 8H^+^ + 5Fe^2+^ Mn^2+^ + 5Fe^3+^ + 4H~2~O \- Ammonium iron (II) sulfate is used to supply Fe^2+^ ions for titration\ - Ammonium iron (II) sulfate can be obtained in a high degree of purity, is not affected by air, and can be used as a primary standard **[Mandatory Experiment: To prepare a standard solution of ammonium iron (II) sulfate and to use this solution to standardise a solution of potassium permanganate by titration.]** 1. Wash the pipette, burette and conical flask with deionised water. Rinse the burette with the potassium manganate(VII) solution and the pipette with the iron(II) solution. 2. Using a pipette filler, fill the pipette with the iron(II) solution and transfer the contents of the pipette to the conical flask. This solution is acidified by addition of about 10 cm^3^ of dilute sulfuric acid. 3. Using a funnel, fill the burette with potassium manganate(VII) solution, making sure that the part below the tap is filled before adjusting to zero. Because of the intense colour of potassium manganate(VII) solutions, readings are taken from the top of the meniscus 4. With the conical flask standing on a white tile, add the solution from the burette to the flask. Swirl the flask continuously and occasionally wash down the walls of the flask with deionised water using a wash bottle. 5. The end-point of the titration is detected by the first persisting pale pink colour. Note the burette reading. 6. Repeat the procedure, adding the potassium manganate(VII) solution dropwise approaching the end-point until two titres agree to within 0.1 cm^3^. 7. Calculate the concentration of the potassium manganate(VII) solution **[Solving Volumetric Problems in Redox Reactions]** [\$\\frac{V\_{\\text{o\\ x\\ }M\_{o}}}{n\_{o}}\$]{.math.inline} = [\$\\frac{V\_{\\text{r\\ x\\ }M\_{\\text{red}}}}{n\_{r}}\$]{.math.inline} V~o~ = Volume of oxidising agent (cm^3^)\ M~0~ = Molarity of oxidising agent\ n~o~ = Number of moles of oxidising agent\ V~r~ = Volume of reducing agent (cm^3^)\ M~red~ = Molarity of reducing agent\ n~r~ = Number of moles of reducing agent A student made up a 0.12 M ammonium iron(II) sulfate solution and used this to standardise a solution of potassium permanganate. 25 cm^3^ of the ammonium iron(II) sulfate solution was titrated against the potassium permanganate solution. The equation for the reaction is:\ MnO~4~^-^ + 8H^+^ + 5Fe^2+^ Mn^2+^ + 5Fe^3+^ + 4H~2~O The average titration figure was 23.75 cm^3^. Calculate the concentration of the potassium permanganate solution in (a) moles per litre and (b) grams per litre. One mole of MnO~4~^-^ reacts with 5 moles of Fe^2+\ ^n~o~ = 1 n~r~ = 5 V~o~ = 23.75 [\$\\frac{V\_{\\text{o\\ x\\ }M\_{o}}}{n\_{o}}\$]{.math.inline} = [\$\\frac{V\_{\\text{r\\ x\\ }M\_{\\text{red}}}}{n\_{r}}\$]{.math.inline}\ M~0~ = ?\ n~o~ = 1\ V~r~ = 25 [\$\\frac{23.75\\ x\\ ?}{1}\$]{.math.inline} = [\$\\frac{25\\ x\\ 0.12}{5}\$]{.math.inline}\ M~red~ = 0.12\ n~r~ = 5 M~o~ = [\$\\frac{25.5\\ x\\ 0.12}{5\\ x\\ 23.75}\$]{.math.inline} = 0.0253 moles per litre\ 0.0253 x 158 grams per litre (Rel. molecular mass KMnO~4~ = 158)\ 3.997 g/L **[Mandatory Experiment: To determine the amount of Iron in an Iron Tablet]** 1. Find the mass of five iron tablets. 2. Crush the weighed tablets in a mortar and pestle. Transfer all the ground material to a beaker where it is dissolved in about 100 cm^3^ of dilute sulfuric acid. 3. All of this solution (including washings) is transferred to a 250 cm^3^ volumetric flask and the solution made up to the mark with deionised water. The volumetric flask should be stoppered and inverted several times. This is the solution containing iron(II) ions. 4. Wash the pipette, burette and conical flask with deionised water. Rinse the burette with the potassium manganate(VII) solution and the pipette with the iron(II) solution. 5. Using a pipette filler, fill the pipette with the iron(II) solution and transfer the contents of the pipette to the conical flask. Acidify this solution by adding about 10 cm^3^ of dilute sulfuric acid. 6. Using a funnel, fill the burette with potassium manganate(VII) solution, making sure that the part below the tap is filled before adjusting to zero. Because of the intense colour of KMnO~4~ solution, readings are taken from the top of the meniscus. 7. With the conical flask standing on a white tile, add the solution from the burette to the flask. Swirl the flask continuously and occasionally wash down the walls of the flask with deionised water using a wash bottle. 8. The end-point of the titration is detected by \'the first persisting pink colour\'. Note the burette reading. 9. Repeat the procedure two or three times, adding the potassium manganate(VII) dropwise approaching the endpoint. These accurate titres should agree to within 0.l cm^3^. 10. Calculate the concentration of the iron(II) solution, and from this calculate the mass of iron in an iron tablet. A student was asked to analyse an iron tablet to determine if the amount of iron sulfate per tablet stated on the packet was correct. He dissolved five tablets whose total mass was 1.2 grams in dilute sulfuric acid and made the solution up to 250 cm^3^ in a volumetric flask. He then titrated 25 cm^3^ portions of the resulting solution against 0.015 M potassium permanganate. The average titration figure was 5.75 cm^3^. Calculate (a) the mass of anhydrous FeSO~4~ in each tablet (b) the mass of iron in each tablet and (c) the percentage of FeSO~4~ in each tablet. The equation for the reaction is:\ MnO~4~^-^ + 8H^+^ + 5Fe^2+^ Mn^2+^ + 5Fe^3+^ + 4H~2~O\ One mole of MnO~4~^-^ reacts with 5 moles of Fe^2+\ ^n~o~ = 1 n~r~ = 5 V~o~ = 5.75 [\$\\frac{V\_{\\text{o\\ x\\ }M\_{o}}}{n\_{o}}\$]{.math.inline} = [\$\\frac{V\_{\\text{r\\ x\\ }M\_{\\text{red}}}}{n\_{r}}\$]{.math.inline}\ M~0~ = 0.015\ n~o~ = 1\ V~r~ = 25 [\$\\frac{5.75\\ \\ x\\ 0.015}{1}\$]{.math.inline} *=* [\$\\frac{25\\ x\\ ?}{5}\$]{.math.inline}\ M~red~ = ?\ n~r~ = 5 M~red~ = [\$\\frac{5.75\\ x\\ 0.015\\ x\\ 5}{25}\$]{.math.inline} = 0.0173 moles per litre\ 0.0173 x 152 g/L (Rel. molecular mass FeSO~4~ = 152)\ 2.63 g/L In the 250 cm^3^ volumetric flask there are [\$\\frac{2.63}{4}\$]{.math.inline} = 0.658 grams of FeSO~4~\ Since we used 5 tablets, 1 tablet contains [\$\\frac{0.658}{5}\$]{.math.inline} grams of FeSO~4~ = 0.132 grams\ The percentage of iron in FeSO~4~ is [\$\\frac{56}{152}\$]{.math.inline} x 100 = 36.84%\ Mass of iron in each tablet is 36.84% of 0.132 g = 0.049 g Mass of one tablet = [\$\\frac{1.2}{5}\$]{.math.inline} = 0.24 g\ % FeSO~4~ in each tablet [\$\\frac{0.132}{0.24}\$]{.math.inline} x 100 = 55% **[Mandatory Experiment: To prepare a solution of sodium thiosulfate and to standardise it by titration against a solution of iodine]** 1. Wash the pipette, burette and conical flask with deionised water. Rinse the pipette with potassium iodate solution and the burette with the sodium thiosulfate solution. 2. Using a pipette filler, fill the pipette with the potassium iodate solution, and transfer the contents of the pipette to the conical flask. 3. Using graduated cylinders, add 20 cm^3^ of dilute sulfuric acid, followed by 10 cm^3^ of 0.5 M potassium iodide solution. 4. Using a funnel, fill the burette with sodium thiosulfate solution, making sure that the part below the tap is filled before adjusting to zero. 5. With the conical flask standing on a white tile, add the solution from the burette to the flask. Swirl the flask continuously and occasionally wash down the walls of the flask with deionised water using a wash bottle. 6. Add a few drops of the starch indicator solution just prior to the end-point, when the colour of the solution fades to **pale** yellow. Upon addition of the indicator a blue-black colour should appear. The thiosulfate solution should now be added dropwise, with thorough swirling. 7. The end-point of the titration is detected by a colour change from blue-black to colourless. Note the burette reading. 8. Repeat the procedure, adding the sodium thiosulfate solution dropwise approaching the end-point until two titres agree to within 0.1 cm^3^. 9. Calculate the concentration of the sodium thiosulfate solution 25 cm^3^ of a 0.045 M iodine solution was titrated against a solution of sodium thiosulfate. The average titration figure was 27.45 cm^3^. Calculate the concentration of the sodium thiosulfate solution in (a) moles per litre and (b) grams per litre of crystalline Na~2~S~2~O~3~.5H~2~O. The reaction between iodine and thiosulfate may be represented as:\ I~2~ + 2S~2~O~3~^2-^ S~4~O~6~^2-^ + 2I^-^ From the balanced reaction we can see that two moles of thiosulfate react with one mole of iodine n~0~ = 1 n~2~ = 2 V~o~ = 25 [\$\\frac{V\_{\\text{o\\ x\\ }M\_{o}}}{n\_{o}}\$]{.math.inline} = [\$\\frac{V\_{\\text{r\\ x\\ }M\_{\\text{red}}}}{n\_{r}}\$]{.math.inline}\ M~0~ = 0.045\ n~o~ = 1\ V~r~ = 27.45 [\$\\frac{25\\ \\ x\\ 0.045}{1}\$]{.math.inline} *=* [\$\\frac{27.45\\ x\\ ?}{2}\$]{.math.inline}\ M~red~ = ?\ n~r~ =2 M~red~ = [\$\\frac{25\\ x\\ 0.045\\ x\\ 2}{27.45}\$]{.math.inline} = 0.082 moles per litre\ 0.082 x248 g/L (rel. molecular mass of Na~2~S~2~O~3~.5H~2~O = 248)\ 20.336 g/L **[Mandatory Experiment: To determine the percentage (w/v) of sodium hypochlorite in household bleach]** 1. The bleach solution must first be diluted to make a solution of suitable concentration for the titration. Using a pipette, add 25 cm^3^ of bleach to a 250 cm^3^ volumetric flask, and make the solution up to the mark with deionised water. The flask should be stoppered and inverted several times. 2. Wash the pipette, burette and conical flask with deionised water. Rinse the pipette with the diluted bleach solution and the burette with the sodium thiosulfate solution. 3. Using a pipette filler, fill the pipette with the diluted bleach solution and transfer the contents of the pipette to the conical flask. 4. Add 1 g potassium iodide and 10 cm^3^ of dilute sulfuric acid to the conical flask -- this liberates iodine. 5. Using a funnel, fill the burette with sodium thiosulfate solution, making sure that the part below the tap is filled before adjusting to zero. 6. With the conical flask standing on a white tile, add the solution from the burette to the flask. Swirl the flask continuously and occasionally wash down the walls of the flask with deionised water using a wash bottle. 7. Add a few drops of the starch indicator solution just prior to the end-point, when the colour of the solution fades to **pale** yellow. A blue-black colour appears. The thiosulfate solution should now be added dropwise, with thorough swirling. 8. The end-point of the titration is detected when the blue-black colour changes to colourless. Note the burette reading. 9. Repeat the procedure, adding the sodium thiosulfate solution dropwise approaching the end-point, until two titres agree to within 0.1 cm^3^. 10. Calculate the concentration of the iodine solution 25 cm^3^ of household bleach is diluted to 250 cm^3^. A 25.0 cm^3^ portion of the solution is added to an excess of potassium iodide solution and titrated against 0.12 M sodium thiosulfate solution. Calculate the concentration of sodium hypochlorite in (i) the diluted bleach and (ii) the original bleach in (a) moles per litre (b) grams per litre and (c) &w/v. CIO^-^ + 2I^-^ + 2H^+^ CI^-^ + I~2~ + H~2~O\ 2S~2~O~3~^2-^ + I~2~ S~4~O~6~^2-^ + 2I^-^ V~o~ = 25 [\$\\frac{V\_{\\text{o\\ x\\ }M\_{o}}}{n\_{o}}\$]{.math.inline} = [\$\\frac{V\_{\\text{r\\ x\\ }M\_{\\text{red}}}}{n\_{r}}\$]{.math.inline}\ M~0~ = ?\ n~o~ = 1\ V~r~ = 32.1 [\$\\frac{25\\ \\ x\\ ?}{1}\$]{.math.inline} *=* [\$\\frac{32.1\\text{\\ x\\ }0.12}{2}\$]{.math.inline}\ M~red~ = 0.12\ n~r~ =2 M~o~ = [\$\\frac{32.1\\ x\\ 0.12}{2\\ x\\ 25}\$]{.math.inline} = 0.077 moles per litre Concentration of NaCIO in diluted bleach = 0.077 moles per litre\ Molecular mass of NaCIO = 74.5 0.077 x 74.5 g NaCIO per litre\ 5.737 g NaCIO per litre\ 0.574 g NaCIO per 100 cm^3^\ 0.574% w/v Concentration of NaCIO in original bleach solution\ Original bleach solution was diluted 10 times\ 0.077 x 10\ = 0.77 moles per litre\ = 0.77 x 74.5 g NaCIO per litre\ = 57.37 g NaCIO per litre\ = 5.74 g NaCIO per 100 cm^3^\ = 5.74 % w/v **[Exam Questions]** [2011 -- HL -- Section -- Question 1] 1\. A chemist determined the concentration of a bleach solution containing NaClO by volumetric analysis. A 25.0 cm^3^ sample of the bleach was first diluted to exactly 500 cm^3^. A pipette was used to measure a 25.0 cm^3^ volume of the diluted bleach and to transfer it into a conical flask. Solutions of potassium iodide, KI, and sulfuric acid were added. The following reaction took place in the conical flask. ClO^--^ + 2I^--^ + 2H^+^ → Cl^--^ + I~2~ + H~2~O\ (a) Describe how the 25.0 cm^3^ sample of the original bleach solution was diluted to exactly 500 cm^3^.\ - Pipette into 500cm^3^ volumetric flask.\ - Add deionised water until near mark.\ - Add dropwise until bottom of meniscus is on the graduation mark and read at eye level.\ - Stopper and invert several times\ (b) What colour developed when the potassium iodide and the sulfuric acid reacted with the diluted bleach in the conical flask? Brown\ Give two reasons why excess potassium iodide was used.\ - To keep the iodine in solution.\ - So that all the bleach has reacted\ The solution in the conical flask was next titrated with a 0.10 M solution of sodium thiosulfate (Na~2~S~2~O~3~). The average volume of sodium thiosulfate required, when the procedure was repeated a number of times, was 16.1 cm^3^. The balanced equation for the titration reaction was: 2S~2~O~3~ ^2--^ + I~2~ → S~4~O~6~ ^2--^ + 2I^--^\ (c) What was the purpose of standing the conical flask on a white tile during the titrations? So the colour change is clearer\ (d) Name the indicator used in the titrations and state the colour change observed at the end point. Starch\ Blue to colourless\ (e) Calculate the concentration of NaClO in moles per litre in\ (i) the diluted bleach, V~o~ = 25 [\$\\frac{V\_{\\text{o\\ x\\ }M\_{o}}}{n\_{o}}\$]{.math.inline} = [\$\\frac{V\_{\\text{r\\ x\\ }M\_{\\text{red}}}}{n\_{r}}\$]{.math.inline}\ M~0~ = ?\ n~o~ = 1\ V~r~ = 16.1 [\$\\frac{25\\ \\ x\\ ?}{1}\$]{.math.inline} *=* [\$\\frac{16.1\\ x\\ 0.1}{2}\$]{.math.inline}\ M~red~ = 0.1\ n~r~ =2 M~o~ = [\$\\frac{16.1\\ x\\ 0.1}{2\\ x\\ 25}\$]{.math.inline} = 0.0322 moles per litre \(ii) the original bleach. 0.0322 × 20 = 0.644 moles/L\ (f) What was the concentration of NaClO in the original bleach\ (i) in grams per litre,\ Rel. molecular mass of NaCIO = 74.5\ 0.644 × 74.5 = 47.978g/L\ (ii) as a % (w/v)?\ 47.978/10 = 4.7978 (w/v)