Title To fond out how much vitamin C in orange and lemon juice Objective To find out how much vitamin C in orange and lemon juice Theory(1) • Vitamin C is ascorbic acid. This is rapidly and quantitatively oxidized by iodine in an acid medium according to the following equation. Theory(2) • The standard method for determining ascorbic acid prevent in a sample is to titrate against a standard iodine solution. But the low solubility of the iodine makes this procedure less than ideal. • In the experiment, a known excess of iodine is created in site by the reaction between an iodide and an iodate and those not reacting with the ascorbic acid is titrated against standard sodium thiosulphate solution. Theory(3) • During the experiment, there are 2 more equations are involved, • 1. I2+2S2O322I-+S4O62- • 2.6H++5I-+IO3-3I2+3H2O • Washing bottle • Stand and clamp • Squeezer • Dropper • 2*25.00cm3pipette • 1*breaker • 4*conical flash • Test tubes • 250.00cm3volumetric • Test tube holder flask • Orange juice • Lemon juice • 0.5M sulphuric acid • Starch solution • Sodium thiosulphate Procedure • A) Prepare the iodate solution • 1. Accurately weigh out 0.62g of potassium iodate. Dissolve in distilled water and make up to the mark in the 250cm3 volumetric flask. • • B) Prepare the fruit solution • 1. Weighed the mass of lemon/ orange • 2. Extract all the juice out of the lemon. Orange. • 3. Mix the juice with 150cm3 0.5M sulphuric acid in a breaker. • 4. Transfer this solution into a 250cm3 volumetric flask and make up to the mark with distilled water. • C) Mix the iodate solution and fruit solution • 1. Pipette 25.00cm3 lemon/ orange solution into a clean conical flask and add 5cm3 1M KI solution. • 2. Pipette 25.00cm3 iodate solution the flask containing the orange/ lemon solution and KI solution • D) Back titration • 1. Back titrate the excess iodine against the standardized sodium thiosulphate solution. • Calculation No of mole of (KIO3) in 250.00cm 3 =mass/molar =0.62/39.1+126.9+16*3 =2.897*0.002897mol No of mole (KIO3) in 25.00cm 3=0.002897*10 =0.0002897mol No of mole of KI in 5.00cm 3 =molarity*volume = 1*(5/1000) =0.005mol 6H+ + IO3- + 5I- 3I2+3H2O No of mole of I2 form from 25cm3 iodine solution =no of mole of KIO3*3 =0.0002897*3 = 0.0008692mol A)Find in the percentage by mass of vitamin C in lemon No of mole of S2 O3 2-use in back titration =molarity*volume =0.1(17.195/1000) =0.0017195mol I2+2S2 O32-2I-+S 4O6 2- Iodine remain after reaction of 25cm3 iodine and ascorbic acid =no of mole of S2 O32 /2 =0.0017195/2 = 0.00085975mol No of mole of iodine reaction with ascorbic acid in 25cm3 Lemon Solution =0.0008692 - 0.00085975 =0.00000945mol No of mole of ascorbic acid in 25.00cm 3 lemon solution =no of mole of iodine =0.00000945mol No of mole of ascorbic acid in 250.00cm 3 lemon solution =0.00000945*10 =0.0000945mol Mass of ascorbic acid C6H8O6 =no of mole * molar mass =0.0000945*(12*6+1*8+16*6) =0.016632g Percentage by mass of vitamin C in lemon =0.016632/115.62*100% =0.012% B) Find the percentage by mass of vitamin C in orange: No of mole of 2S2 O32- use in back titration =0.1(17.3/1000) =0.00173mol No of mole of I2 remain after reaction of iodine and ascorbic acid =0.00173/2 =0.000865mol No of mole of I2 reaction with ascorbic acid in 25cm3 orange solution =(0.0008692-0.000865)mol =0.0000042mol No of mole of ascorbic acid in 25.00cm3 orange solution =o.ooooo42mol No of mole of ascorbic in 250.00cm 3 orange solution =0.0000042*10 =0.000042mol Mass of ascorbic acid C6H8O6 =0.000042*(12*6+1*8+16*6) +0.007392g Percentage by mass of vitamin C in orange =0.007392/120.9*100% =0.006114% Percentage of vitamin C in an orange is 0.00614% and the percentage of vitamin C in a lemon is 0.014%. Thus, orange has lower percentage of vitamin C than lemon. • During the “iodometric titration”, starch solution is the most commonly used indicator to indicate the end point at which iodine in solution has just reacted completely with thiosulphate solution added. Since starch solution forms a dark blue colour indicates that is no iodine in the reaction mixture. This signals the end point of the titration. • From the calculation in front, we found out that orange juice contain less vitamin C than lemon juice. This means that if we want to get more vitamin C(ascorbic acid), we can prefer to drink lemon juice. • Compare the percentage of vitamin C contained in fruits with those in vitamin C tablet(what I did last time), its shows that vitamin C tablets contain more ascorbic acid than natural fruits. It may right because tablets is artificial, they can add how much ascorbic acid they like in the tablet. • The titration must be carried out immediately after the addition of sulphuric acid, it is because the solution is easily be oxidized by sulphuric acid. If the solution is oxidized then the result is not accurate any more. • Titration involve.KI and KIO3 which produce iodine which is mild oxidized agent are often called “iodemetric titration”. On the other hand, the titration is also “back titration” in order to find the mass of ascorbic acid in vitamin C tablets. • Squeezer may not squeeze enough fruit juice from the fruit, then there may be some vitamin C still left on the fruit but not in the juice. Then the calculation of vitamin C left in juice may not accurate. It is better for us to extract as much as fruit juice form the squeezer as possible in order to accurate the result. • Fruit, KIO, in the powder form and it is needed to mix with water to form mixture in volumetric flask of 250.00cm3. It must be stir the powder thoroughly in the water. Before this mixture mix well with lemon /orange solution and KI solution, we should stopper the volumetric flask of water and KIO3. Also stopper another the volumetric flask of mixture of lemon/ orange solution and iodate solution and KI. Then shake them up and down in order to make sure to make the mixture uniformly. The the result is more accurate.