1 / 64

Sodium Thiosulfate Titrations

Sodium Thiosulfate Titrations. Assign oxidation numbers. Which is being oxidised? Reduced? Which is the oxidising agent/ reducing agent?. I 2( aq ) + 2S 2 O 3 2- ( aq )  2I - ( aq ) + S 4 O 6 2- ( aq ). An iodine/ sodium thiosulfate titration .

xylia
Télécharger la présentation

Sodium Thiosulfate Titrations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Sodium Thiosulfate Titrations

  2. Assign oxidation numbers • Which is being oxidised? Reduced? • Which is the oxidising agent/ reducing agent? I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq)

  3. An iodine/ sodium thiosulfate titration • Since the ratio of the reaction is known, this reaction can be used to find the concentration of either solutions, once one is known I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq)

  4. A suitable indicator • The red/brown colour of iodine will fade to yellow and then to colourless as it is used up during the reaction • But this colour change is slow and it is difficult to monitor the exact end point from it • Add a starch indicator when the reaction mixture is pale yellow.. • The iodine /starch complex will change suddenly from Blue/Black to colourless at the end point of the titration

  5. Sodium thiosulfate • Is not a primary standard as it is impure, so a standard solution can not be made up directly. • A standard solution of sodium thiosulfate can only be prepared by titration against a solution of known concentration

  6. Man exp:Preparing a standard solution of Sodium thiosulfate • 2MnO4- + 10I- +16H+ 2Mn+2 +5I2 +8H20 • Iodine is also not a primary standard as free Iodine (I2) is insoluble in water and sublimes By doing the following reaction, an iodine solution of known concentration can be made.. Since Cl- will also react with potassium permanganate it is important that none is present in the reaction mixture Excess Potassium Iodide is used so that all of the potassium permanganate present will react to make iodine. It also increases the solubility of the iodine made Potassium permanganate used is a standard solution (exp*)

  7. Potassium manganate(VII) is not a primary standard. Potassium permanganate, KMnO4solution can be standardised by titration against a standard solution of ammonium iron(II) sulfate solution Titrations involving Potassium permanganate, are always carried out under acidic conditions. Acidic conditions are necessary, because in neutral or alkaline conditions Mn+7 is reduced only as far as Mn+4 (a muddy brown precipitate) Acid used • Dilute sulfuric acid. Sulfuric acid is a good source of H+, and the SO4-2 ions are not reactive. • Hydrochloric acid is a source of Cl- ions which would react with the KMnO4, • Nitric acid is a source of NO3- ion is which would react with the KMnO4, Both would cause more iodine to be released than would be calculated from the balanced equation Note - Manganese(IV) oxide, manganese(II) ions, light, heat, acids and bases all catalyse the decomposition of potassium manganate(VII) solutions so they must be standardised regularly

  8. Using the iodine solution to standardise sodium thiosulfate solution When iodine solution goes pale yellow... Add starch indicator Blue/Back - colourless I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) Sodium thiosulfate solution (unknown concentration) Iodine solution of known concentration made by reacting a standard solution of potassium permanganate with acidified potassium iodide in the conical flask

  9. Calculations • Use the known molarity of potassium permanganate standard solution and the given balanced equation to find the molarity of the iodine solution made • Use to find the molarity of the sodium thiosulfate solution • Once moles/ L is known for sodium thiosulfate you can also work out grams/ L

  10. Question 226 f V1X M1= V2x M2 n1n2 (20)X (M1) = (25) x (0.05) 2 1 M1 = (25) x (0.05) x (2) (1) x (20) M1 = 0.125 The concentration of the sodium thiosulfate solution is 0.125M (moles per litre)

  11. (ii) What is the concentration in grams per litre? X RMM • Moles PER LITRE Grams PER LITRE 0.125 x rmm = grams per litre 0.125 x 248g = 31 • There are 31 g of Na2S2O3.5H2O in one litre.

  12. Question 227 d V1X M1= V2x M2 n1n2 (20)X (M1) = (15) x (0.1) 1 2 M1 = (15) x (0.1) x (1) (2) x (20) M1 = 0.0375 The concentration of the Iodine solution is 0.0375M (moles per litre)

  13. Question 228 V1X M1= V2x M2 n1n2 (18.55)X (M1) = (25) x (0.05) 2 1 M1 = (25) x (0.05) x (2) (1) x (18.55) M1 = 0.1348 The concentration of the Sodium thiosulfate solution is 0.1348M (moles per litre)

  14. (ii) What is the concentration in grams per litre? X RMM • Moles PER LITRE Grams PER LITRE 0.1348 x rmm = grams per litre 0.1348 x 248g = 33.4304 • There are 33.4304 g of Na2S2O3.5H2O in one litre.

  15. Finding the concentration of sodium hypochlorite in bleach • Many commercial bleaches are simply solutions of hypochlorite salts such as sodium hypochlorite (NaOCl) • Hypochlorite ion reacts with excess iodide ion in the presence of acid to generate an iodine solution: • ClO- + 2I- + 2H+→ Cl- + I2 + H2O

  16. Using the iodine solution to standardise sodium thiosulfate solution When iodine solution goes pale yellow... Add starch indicator Blue/Back - colourless I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) Sodium thiosulfate solution (standard solution – last exp **** ) Iodine solution of unknown concentration - made by reaction of sodium hypochlorite from bleach and potassium iodide under acidic conditions

  17. Calculations • Use the known molarity of the standard solution and the given balanced equation to find the molarity of the iodine solution from the titration • Use the balanced equation to see what the molarity of the diluted sodium hypochlorite solution is. • Work out the molarity of sodium hypochlorite in the original bleach solution. 4. Once moles/ L is known for sodium hypochlorite you can also work out grams/ L and v/w%

  18. I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) Question 229 f V1X M1= V2x M2 n1n2 (20)X (M1) = (26.3) x (0.1) 1 2 M1 = (26.3) x (0.1) x (1) (2) x (20) M1 = 0.0658 The concentration of the iodine solution is 0.0658M (moles per litre)

  19. ii) Finding the conc of the sodium thiosulfate in the diluted bleach • The eqt for the 1st reaction was ClO- + 2I- + 2H+→ Cl- + I2 + H2O MOLARITY = 0.0658 SO MOLARITY = 0.0658 The concentration of NaOCl in the diluted bleach was 0.0658 moles per litre

  20. ii) Finding the w/v % of NaOCl in original bleach • Molarity of NaOCl in diluted bleach = 0.0658m/L • Molarity of NaOCl in original bleach = 0.0658 x 10 = 0.658m/L • Concentration of NaOCl in g/L in original bleach = 0.658 x 74.5 = 49.02 g/L • Concentation of NaOCl in w/v % ( grams per 100cm3) in original bleach = 49.02/10 = 4.9%

  21. I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) Question 230 V1X M1= V2x M2 n1n2 (25)X (M1) = (19.56) x (0.25) 1 2 M1= (19.56) x (0.25) x (1) (2) x (25) M1 = 0.0978 The concentration of the iodine solution is 0.0978M (moles per litre)

  22. ii) Finding the conc of the sodium thiosulfate in the diluted bleach • The eqt for the 1st reaction was ClO- + 2I- + 2H+→ Cl- + I2 + H2O MOLARITY = 0.0978 SO MOLARITY = 0.0978 The concentration of NaOCl in the diluted bleach was 0.0978 moles per litre

  23. ii) Finding the w/v % of NaOCl in original bleach • Molarity of NaOCl in diluted bleach = 0.0978m/L • Molarity of NaOCl in original bleach = 0.0978 x 10 = 0.489m/L • Concentration of NaOCl in g/L in original bleach = 0.489 x 74.5 = 34.4305g/L • Concentation of NaOCl in w/v % ( grams per 100cm3) in original bleach = 34.4305/10 = 3.6405%

  24. I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) Question 231 V1X M1= V2x M2 n1n2 (25)X (M1) = (16.1) x (0.1) 1 2 M1= (16.1) x (0.1) x (1) (2) x (25) M1 = 0.0322 The concentration of the iodine solution is 0.0322M (moles per litre)

  25. ii) Finding the conc of the sodium thiosulfate in the diluted bleach • The eqt for the 1st reaction was ClO- + 2I- + 2H+→ Cl- + I2 + H2O MOLARITY = 0.0322 SO MOLARITY = 0.0322 The concentration of NaOCl in the diluted bleach was 0.0322 moles per litre

  26. e)f) Finding the w/v % of NaOCl in original bleach e)i)Molarity of NaOCl in diluted bleach = 0.0322m/L ii)Molarity of NaOCl in original bleach = 0.0322 x 20 = 0.644m/L f)i)Concentration of NaOCl in g/L in original bleach = ii) Concentation of NaOCl in w/v % ( grams per 100cm3) in original bleach =

  27. Biochemical Oxygen Demand • Organic mattere.g. sewage, industrial waste, silage, milk. discharged into a water acts as a food source for the bacteria present there. • The bacteria will multiply and use up the available dissolved oxygen This may cause • fish kills. • bacteria will produce hydrogen sulphide and ammonia( anaerobic conditions)

  28. The level of dissolved oxygen in a water sample is an indicator of the quality of the sample. Biochemical oxygen demand test • The amount of dissolved oxygen used up by biochemical action when a sample of water is kept in the dark at 20oC for 5 days. Note – Compare before and after readings! Sample is kept in the dark to prevent new oxygen being produced by photosynthesis

  29. The Winkler method • It is not possible to directly measure the amount of dissolved oxygen in a water sample directly. • The dissolved oxygen does not directly react with another suitable reagent, an indirect procedure was developed by Winkler. • An iodine/thiosulfate titration can be used to measure the dissolved oxygen present in a water sample.

  30. Manganese sulfate in alkaline conditions 1. Mn2+(aq) + 2OH-(aq) Mn(OH)2(s) ( white precipitate) Mix with sample under water - This reacts with the dissolved oxygen to produce a brown precipitate. 2. 4Mn(OH)2(s) + H20 + O2(aq) 4Mn(OH)3(s) Adding concentrated H2SO4 - enables the Mn(IV) compound to release free iodine from KI. 3. 2Mn(OH)3(s)+ 6H+(aq) + 2I-(aq)  2Mn2+(aq) + I2(aq) + 6H2O(l) Titration : The free iodine is then titrated with standard sodium thiosulfate 4. I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq)

  31. Standard solution = Sodium thiosulfate Iodine solution = unknown concentration Colour change = Red/ Brown – Pale yellow but hard to see the end point Add starch indicator near the end point to see a clear colour change. Blue /Black ( iodine present) – Colourless ( iodine absent

  32. Doing the calculations • Find the concentration of the iodine solution using results from the titration: M1 X V1 = M2 X V2 N1 N2 2. Using the known ratios of how the dissolved oxygen reacted to make the iodine solution you can work out the molarity of the oxygen in the solution 3. Use the molarity of the oxygen in the solution to find g/L and ppm

  33. Questions relating to the experiment • Why is the reagent MnSO4 used? • Why is concentrated H2SO4 used? • Why must the bottles be shaken vigorously after adding the Manganese sulfate and alkaline potassium iodide? • Why are the bottles completely filled with water? • If the white precipitate remains on addition of manganese(II) sulfate solution and alkaline potassium iodide solution, what does this indicate about the water sample? • State and explain what the letters B.O.D. mean. • Why are the bottles used during B.O.D. measurements stored in the dark?

  34. I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) Question 233g V1X M1= V2x M2 n1n2 (50)X (M1) = (6) x (0.01) 1 2 M1= (6) x (0.01) x (1) (2) x (50) M1 = 0.0006 The concentration of the iodine solution is 0.006M (moles per litre)

  35. Ratios of reactions making I2 1. Mn2+(aq) + 2OH-(aq) Mn(OH)2(s) ( white precipitate) 2. 4Mn(OH)2(s) + H20 + O2(aq) 4Mn(OH)3(s) . 3. 2Mn(OH)3(s)+ 6H+(aq) + 2I-(aq)  2Mn2+(aq) + I2(aq) + 6H2O(l) Overall ratio is 1O2 2I2 M1 = 0.0006 /2= .0003 M1 = 0.0006

  36. e)f) Finding the concentration of oxygen in ppm Ppm = parts per million ( mg in 1000cm3) f)i)Concentration of O2 in g/L in water = 0.0003 X 32 = 0.0096g/L ii) Concentation of of O2 in mg/L in water (ppm) .0096 x 1000 = 9.6ppm

  37. I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) Question 235 – Sample 1 V1X M1= V2x M2 n1n2 (200)X (M1) = (27) x (0.01) 1 2 M1= (27) x (0.01) x (1) (2) x (200) M1 = 0.000675 The concentration of the iodine solution is 0.000675M (moles per litre)

  38. Ratios of reactions making I2 1. Mn2+(aq) + 2OH-(aq) Mn(OH)2(s) ( white precipitate) 2. 4Mn(OH)2(s) + H20 + O2(aq) 4Mn(OH)3(s) . 3. 2Mn(OH)3(s)+ 6H+(aq) + 2I-(aq)  2Mn2+(aq) + I2(aq) + 6H2O(l) Overall ratio is 1O2 2I2 M1 = 0.000675 /2= .0003375 M1 = 0.000675

  39. e)f) Finding the concentration of oxygen in ppm Ppm = parts per million ( mg in 1000cm3) f)i)Concentration of O2 in g/L in water tested = 0.0003375 X 32 = 0.0108g/L ii) Concentation of of O2 in mg/L in water tested(ppm) .0108 x 1000 = 10.8ppm Water tested was diluted down ten times so original sample had 10.8 x 10 = 108ppm

  40. I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) Question 235 – Sample 2 V1X M1= V2x M2 n1n2 (200)X (M1) = (4.8) x (0.01) 1 2 M1= (4.8) x (0.01) x (1) (2) x (200) M1 = 0.00012 The concentration of the iodine solution is 0.00012M (moles per litre)

  41. Ratios of reactions making I2 1. Mn2+(aq) + 2OH-(aq) Mn(OH)2(s) ( white precipitate) 2. 4Mn(OH)2(s) + H20 + O2(aq) 4Mn(OH)3(s) . 3. 2Mn(OH)3(s)+ 6H+(aq) + 2I-(aq)  2Mn2+(aq) + I2(aq) + 6H2O(l) Overall ratio is 1O2 2I2 M1 = 0.00012 /2= .00006 M1 = 0.00012

  42. e)f) Finding the concentration of oxygen in ppm Ppm = parts per million ( mg in 1000cm3) f)i)Concentration of O2 in g/L in water tested = 0.00006 X 32 = 0.00192g/L ii) Concentation of of O2 in mg/L in water tested(ppm) .00192x 1000 = 1.92ppm Water tested was diluted down ten times so original sample had 1.92 x 10 = 19.2 ppms

  43. BOD of the water • 1st reading – 2nd reading • 108ppm – 19.2ppm = 88.8ppm

  44. By the end of today’s lesson you should be able to • Describe what hardness in water is • Describe how to estimate of the total hardness of a water sample

  45. Hardness in water • Def: Water is said to be hard when it is difficult to form a lather with soap. • It is caused by the presence of Ca+2and Mg+2ions being dissolved in water

  46. Experiment – estimation of the total hardness of a water sample

  47. Doing the calculations • Find the concentration of the Ca+2 ions in the water using results from the titration: M1 X V1 = M2 X V2 N1 N2 2. Find concentration of CaCO3 in the water in g/L and ppm

  48. Questions relating to the experiment 1. Why is it important that the reaction between the edta and the metal ions in solution (i) is rapid and (ii) goes to completion? 2. The water sample could contain metal ions other than Ca2+ and Mg2+. How would the reliability of the result be affected if this were the case? Suggest two other metal ions that could be present in the water. 3. This reagent cannot distinguish between temporary and permanent hardness. List the compounds of calcium and magnesium that cause hardness, and indicate those which cause temporary hardness. 4. Suggest a method of establishing the amount of permanent hardness in a water sample. 5. What is the function of the buffer solution?

  49. Q243

More Related