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 Since the ratio of the reaction is known, this reaction can be used to find the concentration of either solutions, once one is known I 2(aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq)

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

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

2MnO I - +16H + 2Mn +2 +5I 2 +8H 2 0 By doing the following reaction, an iodine solution of known concentration can be made.. 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*) Man exp: Preparing a standard solution of Sodium thiosulfate Iodine is also not a primary standard as free Iodine (I 2 ) is insoluble in water and sublimes Since Cl - will also react with potassium permanganate it is important that none is present in the reaction mixture

Potassium manganate(VII) is not a primary standard. 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 SO 4 -2 ions are not reactive. Hydrochloric acid is a source of Cl - ions which would react with the KMnO 4, Nitric acid is a source of NO 3 - ion is which would react with the KMnO 4, Both would cause more iodine to be released than would be calculated from the balanced equation Potassium permanganate, KMnO 4 solution can be standardised by titration against a standard solution of ammonium iron(II) sulfate solution 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

Using the iodine solution to standardise sodium thiosulfate solution I 2(aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (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 When iodine solution goes pale yellow... Add starch indicator Blue/Back - colourless

Calculations 1.Use the known molarity of potassium permanganate standard solution and the given balanced equation to find the molarity of the iodine solution made 2.Use to find the molarity of the sodium thiosulfate solution 3.Once moles/ L is known for sodium thiosulfate you can also work out grams/ L

Question 226 f V 1 X M 1 = V 2 x M 2 n 1 n 2 (20)X (M 1 ) = (25) x (0.05) 2 1 M 1 = (25) x (0.05) x (2) (1) x (20) M 1 = The concentration of the sodium thiosulfate solution is 0.125M (moles per litre)

(ii) What is the concentration in grams per litre? Moles PER LITRE Grams PER LITRE x rmm = grams per litre x 248g = 31 There are 31 g of Na 2 S 2 O 3.5H 2 O in one litre. X RMM

Question 227 d V 1 X M 1 = V 2 x M 2 n 1 n 2 (20)X (M1) = (15) x (0.1) 1 2 M 1 = (15) x (0.1) x (1) (2) x (20) M 1 = The concentration of the Iodine solution is M (moles per litre)

Question 228 V 1 X M 1 = V 2 x M 2 n 1 n 2 (18.55)X (M1) = (25) x (0.05) 2 1 M 1 = (25) x (0.05) x (2) (1) x (18.55) M 1 = The concentration of the Sodium thiosulfate solution is M (moles per litre)

(ii) What is the concentration in grams per litre? Moles PER LITRE Grams PER LITRE x rmm = grams per litre x 248g = There are g of Na 2 S 2 O 3.5H 2 O in one litre. X RMM

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 - + I 2 + H 2 O

Using the iodine solution to standardise sodium thiosulfate solution I 2(aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (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 When iodine solution goes pale yellow... Add starch indicator Blue/Back - colourless

Calculations 1.Use the known molarity of the standard solution and the given balanced equation to find the molarity of the iodine solution from the titration 2.Use the balanced equation to see what the molarity of the diluted sodium hypochlorite solution is. 3.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%

Question 229 f V 1 X M 1 = V 2 x M 2 n 1 n 2 (20)X (M1) = (26.3) x (0.1) 1 2 M 1 = (26.3) x (0.1) x (1) (2) x (20) M 1 = The concentration of the iodine solution is M (moles per litre) I 2(aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq)

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

Molarity of NaOCl in diluted bleach = m/L Molarity of NaOCl in original bleach = x 10 = 0.658m/L Concentration of NaOCl in g/L in original bleach = x 74.5 = g/L Concentation of NaOCl in w/v % ( grams per 100cm 3 ) in original bleach = 49.02/10 = 4.9% ii) Finding the w/v % of NaOCl in original bleach

Question 230 V 1 X M 1 = V 2 x M 2 n 1 n 2 (25)X (M1) = (19.56) x (0.25) 1 2 M 1 = (19.56) x (0.25) x (1) (2) x (25) M 1 = The concentration of the iodine solution is M (moles per litre) I 2(aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq)

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

Molarity of NaOCl in diluted bleach = m/L Molarity of NaOCl in original bleach = x 10 = 0.489m/L Concentration of NaOCl in g/L in original bleach = x 74.5 = g/L Concentation of NaOCl in w/v % ( grams per 100cm 3 ) in original bleach = /10 = % ii) Finding the w/v % of NaOCl in original bleach

Question 231 V 1 X M 1 = V 2 x M 2 n 1 n 2 (25)X (M1) = (16.1) x (0.1) 1 2 M 1 = (16.1) x (0.1) x (1) (2) x (25) M 1 = The concentration of the iodine solution is M (moles per litre) I 2(aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq)

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

e)i)Molarity of NaOCl in diluted bleach = m/L ii)Molarity of NaOCl in original bleach = 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 100cm 3 ) in original bleach = e)f) Finding the w/v % of NaOCl in original bleach

 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)

 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 20 o C for 5 days. Note – Compare before and after readings! Sample is kept in the dark to prevent new oxygen being produced by photosynthesis

 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.

Manganese sulfate in alkaline conditions 1. Mn 2+ (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) + H O 2 (aq)  4Mn(OH) 3(s) Adding concentrated H 2 SO 4 - enables the Mn(IV) compound to release free iodine from KI. 3. 2Mn(OH) 3(s) + 6H + (aq) + 2I - (aq)  2Mn 2+ (aq) + I 2(aq) + 6H 2 O (l) Titration : The free iodine is then titrated with standard sodium thiosulfate 4. I 2(aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq)

 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

1. 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

1. Why is the reagent MnSO 4 used? 2. Why is concentrated H 2 SO 4 used? 3. Why must the bottles be shaken vigorously after adding the Manganese sulfate and alkaline potassium iodide? 4. Why are the bottles completely filled with water? 5. 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? 6. State and explain what the letters B.O.D. mean. 7. Why are the bottles used during B.O.D. measurements stored in the dark?

Question 233g V 1 X M 1 = V 2 x M 2 n 1 n 2 (50)X (M1) = (6) x (0.01) 1 2 M 1 = (6) x (0.01) x (1) (2) x (50) M 1 = The concentration of the iodine solution is 0.006M (moles per litre) I 2(aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq)

Ratios of reactions making I 2 1. Mn 2+ (aq) + 2OH - (aq)  Mn(OH) 2(s) ( white precipitate) 2. 4Mn(OH) 2(s) + H O 2 (aq)  4Mn(OH) 3(s). 3. 2Mn(OH) 3(s) + 6H + (aq) + 2I - (aq)  2Mn 2+ (aq) + I 2(aq) + 6H 2 O (l) Overall ratio is 1O 2 2I 2 M 1 = M 1 = /2=.0003

Ppm = parts per million ( mg in 1000cm 3 ) f) i)Concentration of O 2 in g/L in water = X 32 = g/L ii) Concentation of of O 2 in mg/L in water (ppm).0096 x 1000 = 9.6ppm e)f) Finding the concentration of oxygen in ppm

Question 235 – Sample 1 V 1 X M 1 = V 2 x M 2 n 1 n 2 (200)X (M1) = (27) x (0.01) 1 2 M 1 = (27) x (0.01) x (1) (2) x (200) M 1 = The concentration of the iodine solution is M (moles per litre) I 2(aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq)

Ratios of reactions making I 2 1. Mn 2+ (aq) + 2OH - (aq)  Mn(OH) 2(s) ( white precipitate) 2. 4Mn(OH) 2(s) + H O 2 (aq)  4Mn(OH) 3(s). 3. 2Mn(OH) 3(s) + 6H + (aq) + 2I - (aq)  2Mn 2+ (aq) + I 2(aq) + 6H 2 O (l) Overall ratio is 1O 2 2I 2 M 1 = M 1 = /2=

Ppm = parts per million ( mg in 1000cm 3 ) f) i)Concentration of O 2 in g/L in water tested = X 32 = g/L ii) Concentation of of O 2 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 e)f) Finding the concentration of oxygen in ppm

Question 235 – Sample 2 V 1 X M 1 = V 2 x M 2 n 1 n 2 (200)X (M1) = (4.8) x (0.01) 1 2 M 1 = (4.8) x (0.01) x (1) (2) x (200) M 1 = The concentration of the iodine solution is M (moles per litre) I 2(aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq)

Ratios of reactions making I 2 1. Mn 2+ (aq) + 2OH - (aq)  Mn(OH) 2(s) ( white precipitate) 2. 4Mn(OH) 2(s) + H O 2 (aq)  4Mn(OH) 3(s). 3. 2Mn(OH) 3(s) + 6H + (aq) + 2I - (aq)  2Mn 2+ (aq) + I 2(aq) + 6H 2 O (l) Overall ratio is 1O 2 2I 2 M 1 = M 1 = /2=.00006

Ppm = parts per million ( mg in 1000cm 3 ) f) i)Concentration of O 2 in g/L in water tested = X 32 = g/L ii) Concentation of of O 2 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 e)f) Finding the concentration of oxygen in ppm

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

Describe what hardness in water is Describe how to estimate of the total hardness of a water sample

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 +2 and Mg +2 ions being dissolved in water

Experiment – estimation of the total hardness of a water sample

1. 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 CaCO 3 in the water in g/L and ppm

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?

Q243

Describe what hardness in water is Explain causes of temporary and permanent hardness Do tests on scale deposits in a kettle Know that removal of hardness is by boiling or ion exchange Describe what deionisation is and how it works

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 +2 and Mg +2 ions being dissolved in water There are two different types of hardness in water: 1. Temporary hardness 2. Permanent hardness

1. Temporary hardness Def: Temporary hardness is hardness in water that can be removed by boiling. Found in parts of the country where limestone (calcium carbonate) is found in the earth.

What causes temporary hardness Water (rain), carbon dioxide( in the air) and Calcium carbonate (limestone) all react together: Calcium hydrogen carbonate Ca(HCO 3 ) 2 is a made and dissolves in the water.

Removing temporary hardness from water 1. It can be removed by boiling! This reaction occurs: Ca(HCO 3 ) 2 + heat CaCO 3 + H 2 O + CO 2

Limescale inside a kettle

Removing temporary hardness from water 2. It can be removed using an ion exchanger - the hardness causing calcium and magnesium ions are swapped with sodium ions, which do not cause hardness.

Permanent hardness Permanent hardness is hardness in water that cannot be removed by boiling. It is caused by the presence of calcium sulfate CaSO 4 and magnesium sulfate MgSO 4 in water.

It can be removed by: using an ion exchanger Removing permanent hardness from water

Deionisation –

Deionisation In a deioniser any dissolved salts present ae replaced by water molecules.

Deionised = no ions other than H+ and OH- Distilled = No dissolved solids of any type Relative purity of deionised and distilled water.

Describe what hardness in water is Explain causes of temporary and permanent hardness Know that removal of hardness is by boiling or ion exchange Describe what deionisation is and how it works