I have almost covered the entire unit with the help of these videos, however I think you missed the iodine-thiosulfate rxn (titration), it involves iodometry and iodimetry. I was really confused about those. Anyways, thanks a lot:)
When did we find the initial concentration of O2? Im just a bit confused on how you found 8.11 for the initial concentration. Otherwise, thank you so much for this helpful video!
@@lu8509 U basically make the stoichiometric coefficient I2, 2, and by that the s2O3 2- will have 4 as the coefficient. I THINK. So, basically what I am saying is, the 2 of MnO2 in the first eq will be multiplied in the second eq, to get 2MnO2, such that the same amount of MnO2 is maintained. Therefore, the MnO2 : I2 will become 2:2. Then, the ratio of I2 : S2O3 2- will become 2:4. That is how the ratio of O2 : S2O3 2- is 1:4. Hope I am making sense.
The ratio between the O2 and 2MnO2 is 1:2, but the amount of 2MnO2 (from first step) must equal the amount of MnO2 (in the second step). The amount of the MnO2 in the second step is equal to the amount of I2 in the second and third steps. The amount of 2S2O3 2- is twice as much as I2 and the MnO2 (which is twice as much as O2) -- so the ratio between 2S2O3 2- and O2 is 1:4
@@lu8509 you can think of the first ratio as doubling. So one reactant becomes 2 products. Now those 2 products become 2 reactants in the next equation. Since the second equation is 1:1, you don't get a different number of products. Finally, your two products become the 2 reactants in the final equation, and the final equation will double the reactants to 4 products. effectively you get 4 products for 1 reactant
Great video! But I have one question: Why do you calculate the number of moles of thiosulfate if it says *sodium* thiosulfate reacts with the sample in the instructions?
Sodium thiosulfate will dissociate into sodium ions and thiosulfate ions. It is the thiosulfate ions that react with so we calculate the amount in mol of these ions (although it is the same amount as the sodium thiosulfate used in the reaction).
@@MSJChem Thanks! I have a question about a past paper question I was doing. The question asks: Outline with an ionic equation what is observed when magnesium powder is added to a solution of ammonium chloride. In one of your previous videos, you mentioned that metals above H on the activity series will displace H+ ions to form H2. So wouldn't the equation be Mg + 2 NH4Cl --> MgCl2 + 2 H2 + 2 NH2? But the answer says the equation is Mg + 2 NH4Cl → MgCl2 + H2 + 2 NH3. Why did the NH4 split into H2 and NH3?
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thirty more minutes til the IB exam and I got it now ty
I have almost covered the entire unit with the help of these videos, however I think you missed the iodine-thiosulfate rxn (titration), it involves iodometry and iodimetry. I was really confused about those. Anyways, thanks a lot:)
awesome video
Enough for concept clearing
Thanks
Thanks a lot of this video.
How to get the ratio of O2 to S2O3^2-?
I still does not understand that step
It is the ratio of the oxygen in the first step to the thiosulfate ion in the third step.
I have a question.
I don't understand why we are dividing the mass by the volume to calculate the concentration? I thought C = n /V?
The unit is mg/dm3 so the mass (in mg) is divided by the volume (dm3). 1 mg/dm3 is equal to 1 ppm.
When did we find the initial concentration of O2? Im just a bit confused on how you found 8.11 for the initial concentration. Otherwise, thank you so much for this helpful video!
The 1:4 ratio tells you the amount of oxygen.
If we take 20ml of the waste water out of 300ml for out calculation, do we divide 2.43mg of o2 by 20ml?
At the start you said its BOD = final-initial, but you did initial-final, howcome
At 1:25 it says subtract final from
Initial.
@@MSJChem Yes, but at the end of the video you do (BOD = INITIAL - FINAL)
@@anonmela subtract final from initial means the same as (initial - final)
I don’t understand how from the ratios you get the 1:4 ratio, do you have to divide them by each other or something?
The ratio is 1:2 in the first step, 1:1 in the second step and 1:2 in the third step so that is 1:4 ratio of O2 to S2O42-.
I understand how you got the ratio, but not how they become 1:4
@@lu8509 U basically make the stoichiometric coefficient I2, 2, and by that the s2O3 2- will have 4 as the coefficient. I THINK.
So, basically what I am saying is, the 2 of MnO2 in the first eq will be multiplied in the second eq, to get 2MnO2, such that the same amount of MnO2 is maintained. Therefore, the MnO2 : I2 will become 2:2. Then, the ratio of I2 : S2O3 2- will become 2:4. That is how the ratio of O2 : S2O3 2- is 1:4. Hope I am making sense.
The ratio between the O2 and 2MnO2 is 1:2, but the amount of 2MnO2 (from first step) must equal the amount of MnO2 (in the second step). The amount of the MnO2 in the second step is equal to the amount of I2 in the second and third steps. The amount of 2S2O3 2- is twice as much as I2 and the MnO2 (which is twice as much as O2) -- so the ratio between 2S2O3 2- and O2 is 1:4
@@lu8509 you can think of the first ratio as doubling. So one reactant becomes 2 products. Now those 2 products become 2 reactants in the next equation. Since the second equation is 1:1, you don't get a different number of products. Finally, your two products become the 2 reactants in the final equation, and the final equation will double the reactants to 4 products. effectively you get 4 products for 1 reactant
Great video! But I have one question: Why do you calculate the number of moles of thiosulfate if it says *sodium* thiosulfate reacts with the sample in the instructions?
Sodium thiosulfate will dissociate into sodium ions and thiosulfate ions. It is the thiosulfate ions that react with so we calculate the amount in mol of these ions (although it is the same amount as the sodium thiosulfate used in the reaction).
@@MSJChem Ok thank you for the explanation!
Is the titration of iodine against sodium thiosulfate in the syllabus?
It is a common titration so you should be aware of it, yes.
@@MSJChem Alright, thank you so much!
How does the Iodine form in line 2? Is it from the alkaline solution?
Iodide ions are added to the solution.
what are the concentrations of each solution?
0.0200 mol dm-3
Do we need to memorize the Winkler equations for the exam?
No, they will be provided for you.
@@MSJChem Thanks! I have a question about a past paper question I was doing. The question asks: Outline with an ionic equation what is observed when magnesium powder is added to a solution of ammonium chloride. In one of your previous videos, you mentioned that metals above H on the activity series will displace H+ ions to form H2. So wouldn't the equation be Mg + 2 NH4Cl --> MgCl2 + 2 H2 + 2 NH2? But the answer says the equation is Mg + 2 NH4Cl → MgCl2 + H2 + 2 NH3. Why did the NH4 split into H2 and NH3?
From which past paper was this question taken?
@@MSJChem SL 2017 MAY TZ1 question 4d
I found the question - I cannot recall this reaction covered in any of the topics. It’s only come up once so I wouldn’t worry about it too much.
Thank you so much! Your name should be on my diploma tbh