Please double check one last time have to hand it in tomorrow.Thank-You in Advance.
This was the lab experiment:
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A. Standardization of the Sodium Thiosulfate Solution
Make up a standard solution of potassium iodate by accurately “weighing by difference” about 0.1 g of KIO3 and placing it in your 100 mL volumetric flask. Fill the flask to mark with distilled water. Potassium iodate reacts with excess KI in acid solution according to the following reaction:
IO3- + 5I- + 6H^+ ----->3I2 + 3H2O
RECALL:
2S2O3^2- + I2 --> 2I- + S4O6^ 2-
To perform then standardized, first prepare the buret with the given solution of Na2S2O3. Then dispense 25 mL of KIO3 solution by pipett into a 250 mL Erlenmeyer flask, add 25 mL of 2% (v/v) KI solution and then add 10 mL of 1M HCL, and mix swirling. A deep brown color should appear, indicating the presence of iodine. Titrate immediately with the Na2S2O3 solution in the burett until the brown fades to a pale yellow and then add 5 mL of starch indicator solution and continue the titration until the deep blue color of the starch indicator disappears. Record the volume of the sodium thiosulafte solution used in the titration. Determine the stoichiometric ratio between the iodate and thiosulafte and use this to calculate the accurate concentration if the Na2S2O3 solution.
A.Standardization of Na2S2O3 ( Sodium thiosulfate)
Mass of KIO3 in 100 mL = 0.1255g
Show calculation for molarity for the potassium iodate:
Moles (n)= 0.1255g/214.00= 0.00058645 mols
100mL/1000 = 0.1 L
Molarity= 0.00058645/0.1= 0.00586448
Therefore the Concentration of KIO3 is 5.86*10^-3
Titration:
Average Volume of Na2S2O3 = 27.15 mL
Calculate the concentration of Na2S2O3
_______________M.
KIO3 moles: M x L
= 0.000586448(0.025L)
= .000146612 moles
.000146612 mols KIO3 x (3 mols I2/1 mol KIO3) x (2 mols S2O3^-2/1 mol I2) mols
= .000887972 mols S203^2-
Then M S2O3^-2 = .000887972 mols/0.02715 L.
M = 0.032400
Therefore,the concentration of Na2S2O3 is 0.032406 M.
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B.Titration of a Sample of Household bleaches:
The oxidizing agent in a household bleach is determined by placing 100 mL of distilled water in a 250 mL Erlenmeyer flask and adding 10 mL of the 10% (w/w) KI solution, swirling the contents of the flask to mix the solutions. Add an accurately measured 0.5 mL of your assigned bleach to the flask, and swirl. Then add 10 mL of 2 M H2S04 and proceed with the titration, using the standardized sodium thiosulfate solution. Add the starch indicator when the solution is pale yellow and continue the titration until the deep blue colour of the starch indicator disappear. Record the volume of thiosulfate solution used in the titration.
From the concentration and volume of the added sodium thiosulfate solution used to titrate the different brands of bleach, calculate the numbers of moles of the oxidizing agent presence, assuming it to be sodium hypochlorite, NaOCl. Use this to calculate the number of grams of sodium hypochlorite that was present in the 0.5 mL of bleach. Express the final result as the mass of sodium hypochlorite per 100 mL of bleach.
The sodium thiosulfate reacts with the sodium hypochlorite according to the following stoichiometric redox equations:
NaClO (aq) + 2H^+ + 2I- ---> I2 + Cl- +H2O + Na^+
Recall:
2S203^ 2- + I2 ---> 2I- + S4O6 ^ 2-
B.Titration of household bleach: Javex
Calculated Concentration of Na2S2O3= 0.032400 M.
Reaction # 1:
ClO^- + 2I^- ==> I2 + Cl^-
Reaction # 2:
2S2O3^2-(aq)+I2----> 2I-(aq)+S4O6^2-(aq)
Trial: 1
Volume of bleach ( 0.5mL)
Volume of Na2S2O3 ( 53 mL)
= mass of NaOCl/100 mL bleach
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moles of S203^2- =(MOLARITY)(VOLUME)
=(0.03240 mol/L)(0.053L S2O3^2-)
= 0.001717 moles S2O3^2-
moles of I2= (1/2)(0.001717)
= .0008586 moles of I2
moles of ClO- = (1/1)(0.0008586)
= 0.0008586 moles of ClO-
(Molar mass)(moles of ClO-)= mass
(74.44)(0.0008586)= 0.06391
Change mass/0.5mL to mass/100mL
Therefore: (0.06391)(100/0.5)
= 12.78
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Trail 2:
Volume of bleach ( 0.5mL)
Volume of Na2S2O3 ( 55.4 mL)
= mass of NaOCl/100 mL bleach
=_______________?
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moles of S203^2- =(MOLARITY)(VOLUME)
=(0.03240 mol/L)(0.0554L S2O3^2-)
= 0.001795 moles S2O3^2-
moles of I2= (1/2)(0.001795)
= .0008975 moles of I2
moles of ClO- = (1/1)(.0008975)
= .0008975 moles of ClO-
(Molar mass)(moles of ClO-)= mass
(74.44)(.0008975)= 0.06681
Change mass/0.5mL to mass/100mL
Therefore: (0.06681)(100/0.5)
=13.36
Average mass of NaOCl/100 mL bleach
12.78+ 13.36 = 13.07
Therefore, the average mass of NaOCl/ 100ml = 13.07
A.Standardization of Na2S2O3 ( Sodium thiosulfate)
Mass of KIO3 in 100 mL = 0.1255g
Show calculation for molarity for the potassium iodate:
Moles (n)= 0.1255g/214.00= 0.00058645 mols
100mL/1000 = 0.1 L
Molarity= 0.00058645/0.1= 0.00586448
Therefore the Concentration of KIO3 is 5.86*10^-3
No, the concentration of KIO3 is 5.8645*10^-3 M which I would round to 5.864*10^-3. You carried all the other numbers until here and you should show at least one more place. Your final answer of 0.03240 M is ok; you just need the intermediate steps to follow through the correct number of significant figures.
KIO3 moles: M x L
= 0.000586448(0.025L)I think this should be 0.00586448 but the final answer is ok.
= .000146612 moles
.000146612 mols KIO3 x (3 mols I2/1 mol KIO3) x (2 mols S2O3^-2/1 mol I2) mols
= .000887972 mols S203^2-
Shouldn't this be 0.00087972? I think you have an extra 8 in there. Same in the next step.
Then M S2O3^-2 = .000887972 mols/0.02715 L.
M = 0.032400
You should drop the final 0 of 0.032400 to 0.03240.
Therefore,the concentration of Na2S2O3 is 0.032406 M.It is unclear to me how you went from 0.032400 to 0.032406. I think there was a second titration in here and it may be that you averaged that in but you don't show it anywhere.
oles of S203^2- =(MOLARITY)(VOLUME)
=(0.03240 mol/L)(0.053L S2O3^2-)
= 0.001717 moles S2O3^2-
moles of I2= (1/2)(0.001717)
= .0008586 moles of I2
moles of ClO- = (1/1)(0.0008586)
= 0.0008586 moles of ClO-
(Molar mass)(moles of ClO-)= mass
(74.44)(0.0008586)= 0.06391
Change mass/0.5mL to mass/100mL
Therefore: (0.06391)(100/0.5)
= 12.78
The final answer looks ok except as I mentioned earlier (in your first posts) that 12.8% seems high to me. I think I pointed out that most of the commercial bleaches I see on the market run about 6%. I know this asks for g/100 mL BUT that IS mass/volume percent.
Thank-You so much DrBob.. you really helped a lot..and i had made some typoz up there so that's why some things didn't make sense...and i fixed the problems you told me to before handing it in..Thank- You Once again
can you explain what you're doing here???
KIO3 moles: M x L
= 0.000586448(0.025L)
= .000146612 moles
.000146612 mols KIO3 x (3 mols I2/1 mol KIO3) x (2 mols S2O3^-2/1 mol I2) mols
= .000887972 mols S203^2-
Then M S2O3^-2 = .000887972 mols/0.02715 L.
M = 0.032400
Therefore,the concentration of Na2S2O3 is 0.032406 M.
Well, it looks like you've done all the calculations correctly. I must say, you really know your stuff when it comes to chemistry! But don't forget to proofread your work one last time before handing it in. You wouldn't want any silly mistakes sneaking in and compromising your hard work. Good luck!
In section A, the standardization of the Sodium Thiosulfate solution is carried out. To calculate the concentration of the Potassium Iodate (KIO3) solution, the following steps are followed:
1. Calculate the moles of KIO3:
- The mass of KIO3 used is 0.1255 g.
- The molar mass of KIO3 is 214.00 g/mol.
- Divide the mass of KIO3 by its molar mass to get the moles: 0.1255 g / 214.00 g/mol = 0.00058645 mol.
2. Calculate the volume of the solution:
- The volume of the solution is 100 mL.
- Convert mL to liters by dividing by 1000: 100 mL / 1000 = 0.1 L.
3. Calculate the concentration of KIO3:
- Divide the moles of KIO3 by the volume of the solution: 0.00058645 mol / 0.1 L = 0.00586448 M.
Therefore, the concentration of KIO3 is approximately 0.00586448 M.
In section B, the titration of a sample of household bleaches is conducted using the standardized sodium thiosulfate solution. To calculate the mass of sodium hypochlorite (NaOCl) per 100 mL of bleach, the following steps are followed:
1. Calculate the moles of sodium thiosulfate (Na2S2O3):
- The volume of Na2S2O3 used in the titration is given as 53 mL (for trial 1) and 55.4 mL (for trial 2).
- The concentration of Na2S2O3 is 0.032400 M.
- Multiply the concentration by the volume to get the moles: 0.032400 mol/L * 0.053 L (for trial 1) or 0.0554 L (for trial 2) = 0.001717 moles (for trial 1) or 0.001795 moles (for trial 2).
2. Calculate the moles of iodine (I2):
- Since the stoichiometric ratio between Na2S2O3 and I2 is 1:1, the moles of Na2S2O3 are equal to the moles of I2.
3. Calculate the moles of sodium hypochlorite (NaOCl):
- Since the stoichiometric ratio between NaOCl and I2 is 1:1, the moles of I2 are equal to the moles of NaOCl.
4. Calculate the mass of NaOCl:
- The molar mass of NaOCl is 74.44 g/mol.
- Multiply the moles of NaOCl by its molar mass to get the mass: moles of NaOCl * 74.44 g/mol.
- Calculate the average mass for both trials.
5. Calculate the mass of NaOCl per 100 mL of bleach:
- Divide the mass of NaOCl by the volume of bleach used (0.5 mL) and then multiply by 100 to scale it to 100 mL: mass / 0.5 mL * 100 = mass / 100 mL.
Therefore, the average mass of NaOCl per 100 mL of bleach is approximately 13.07 grams.