Can you please show the work, I really want to learn too. LAB SECTION: NAME EXPE
ID: 941304 • Letter: C
Question
Can you please show the work, I really want to learn too. LAB SECTION: NAME EXPERIMENT 18: Determination of Bleach-A Redox Titration DATA SHEET1 Part A: Standardiration of aIxx M Na S,o, NAME: LAB SECTION EXPERIMENT 18: Determination of Bleach-A Redox Titration DATA SHEET 2 Part B: Preparation and Titration of Diluted Isleach Average Molarity Na,S,o Trial Trial 2 Trial 3 Concntration of K1O, from Experiment #13) Volume of KIO, Pipened mmol or KIO, 00 mmol ofL, Formed Trial 1 Trial 2 Trial 3 Mass of Bleach Final Burette Reading Initial Buretle Reading Volume of NAS,Odaq) Required mmol of S,O Required mmol of l, Present (Equation 2) mmol eof CO Present (Equation 1) Mass of NaCIO Present (mg) Mass of NaC0 Present (g) % (mm) of NaClO in Bleach Avg.%(mm) of NaCO in Bleach (from Equation 3) 43 e. Bnt 211442 Final Burette Reading Initial Burete Reading Volume ofNasoaaq) Required mmol of S,0, Present , 051mL lo 613.18 051 6 12.18 . (from Equation 2) Molarity S,O, Avg. Molanity Na,S,0, % Beand of Bleach 165Explanation / Answer
Part A:
mmol of KIO3(IO3-) = Molarity of KIO3 x volume of KIO3 in ml
= 0.00935 M x 10 ml
= 0.0935 mmol
Volume of Na2S2O3 required -
Trial 1 = 6.47 ml
Trial 2 = 6.14 ml
Trial 3 = 6.30 ml
According to Equation-3 one mole of IO3- generates 3 mole of I3-
mmol of I3- formed = 0.0935 x 3
= 0.2805 mmol
According to Eq 2 one mole of I3- reacts with two moles of S2O32-
According to Eq 3 one mole of IO3- generates 3 mole of I3-
one mole of IO3- reacts with 6 mole of S2O32-
Molarity of KIO3 (M1) = 0.00935 M
Volume of KIO3 (V1) = 10 ml
No, of mole of KIO3 (n1) = 1
Molarity of S2O32- (M2) = ?
Volume of S2O32- (V2) = 6.47 ml (trial 1), 6.14 ml (trial 2), 6.30 ml (trial 3)
No. of mole of S2O32- (n2) = 6
M1V1/ n1 = M2V2/ n2
M2 = (M1V1/ n1) (n2/V2)
= ((0.00935 x 10)/1) (6/6.47)
= 0.0867 M (Trial 1)
= 0.09136 M (Trial 2)
= 0.08904 M (Trial 3)
mmol of S2O32- present (Trial 1) = Molarity of S2O32- x Volume of S2O32- in ml
= 0.0867 M x 6.47 ml
= 0.5609 mmol
Trial 2 = 0.5609 mmol
Trial 3 = 0.5609 mmol
Average Molarity of S2O32- = 0.0890 M
Part B:
Volume of Na2S2O3 required mmol of S2O32- mmol of I3- mmol of ClO-
Trial 1 35.05 ml 3.12 mmol 1.56 1.56
Trial 2 34.34 ml 3.06 mmol 1.53 1.53
Trial 3 35.04 ml 3.14 mmol 1.57 1.57
No. of mole of NaClO is equal to the no. of mole of ClO- present
Mass of NaClO present = No. of mmol of NaClO x M.wt of NaClO in mg
Trial 1 = 1.56 x 74.44 mg
= 116.13 mg
= 0.116 g
Trial 2 = 113.89 mg = 0 .113 g
Tral 3 = 116.87 mg = 0.116 g
%(m/m) of NaClO in bleach = (mg of NaClO/ mg of blech) x 100
trial 1 = (116.13/2000) x 100
= 5.80 %
Trial 2 = 5.69%
Trial 3 = 5.84%
Average %(m/m) of NaClO in bleach = 5.78%