I. 2 Mn²⁺ + 4 OH^- + O₂(g) → 2 MnO₂(s) + 2 H₂O
II. MnO₂(s) + 2 I^- + 4 H⁺ → Mn²⁺ + I₂(aq) + 2 H₂O
III. 2 S₂O₃ ^2- + I₂(aq) → S₄O₆ ^2- + 2 I^-
The amount of oxygen, O₂, dissolved in water can be determined by titration. First, MnSO₄ and NaOH are added to a sample of water to convert all of the dissolved O₂ to MnO₂, as shown in equation I above. Then H₂SO₄ and KI are added and the reaction represented by equation II proceeds.
Finally, the I2 that is formed is titrated with standard sodium thiosulfate, Na₂S₂O₃,according to equation III.
(a) According to the equation above, how many moles of S₂O₃ ^2- are required for analyzing 1.00 mole of O₂ dissolved in water? (1 point)
(b) A student found that a 50.0 ml sample of water required 4.86 ml of 0.0112 Molar
Na₂S₂O₃ to reach the equivalence point. Calculate the number of moles of O2 dissolved in this sample.(2 points)
(c) How would the results in (b) be affected if some I₂ were lost before the S₂O₃^2- was added?Explain. (1 point)
(d) What volume of dry O2 measured at 25 ̊C and 1.00 atmosphere of pressure would have to be dissolved in 1.00 liter of pure water in order to prepare a solution of the same concentration as that obtained in (b)? Hint: PV=nRT (3 points)
(e) Name an appropriate indictor for the reaction shown in equation III and describe the change you would observe at the end point of the titration. (2 points)

QuoteI. 2 Mn2+ + 4 OH- + O2(g) → 2 MnO2(s) + 2 H2O
II. MnO2(s) + 2 I- + 4 H+ → Mn2+ + I2(aq) + 2 H2O
III. 2 S2O3 2- + I2(aq) → S4O6 2- + 2 I-
The amount of oxygen, O2, dissolved in water can be determined by titration. First, MnSO4 and NaOH are added to a sample of water to convert all of the dissolved O2 to MnO2, as shown in equation I above. Then H2SO4 and KI are added and the reaction represented by equation II proceeds.
Finally, the I2 that is formed is titrated with standard sodium thiosulfate, Na2S2O3,according to equation III.
(a) According to the equation above, how many moles of S2O3 2- are required for analyzing 1.00 mole of O2 dissolved in water? (1 point)

We will use molar ratio from all three equations -
1mole of O₂ gives 2 moles of MnO₂ from equation I.
1mole of MnO₂ gives 1 mole of I₂ from equation II.
1mole of I₂ reacts with 2 moles of 2 S₂O₃ ^2-from equation III.
Now to know moles of  S₂O₃ ^2- =
1mole of O₂ x   2 moles MnO₂/ mole of O₂
        X 1mole of I₂ / 1mol MnO₂
                                              X 2 moles S₂O₃ ^2-  / 1mole of I₂
                                     = 4 mol 1mole of S₂O₃ ^2-

Quote(b) A student found that a 50.0 ml sample of water required 4.86 ml of 0.0112 Molar
Na2S2O3 to reach the equivalence point. Calculate the number of moles of O2 dissolved in this sample.(2 points)

Now from the above post -Moles of thiosulfate needed to determine amount of dissolved O2 in water-Redox

Quoteb]1mole of O₂ x   2 moles MnO₂/ mole of O₂
        X 1mole of I₂ / 1mol MnO₂
                                              X 2 moles S₂O₃ ^2-  / 1mole of I₂
                                     = 4 mol 1mole of S₂O₃ ^2- [/b]

From given information in the  question calculate number of moles of  Na2S2O3 from its molarity and volume.
Moles = Molarity * Volume in L
        =  0.0112 M * 0.00486 L =
          5.44*e-5  mol  S₂O₃ ^2-
                       X 1mol of O₂ / 4 mol  S₂O₃ ^2-
          =1.36 e-5 mol of O₂ dissolved in given 50 mL of water.

(

Quotec) How would the results in (b) be affected if some I₂ were lost before the S₂O₃^2- was added? Explain. (1 point)

Amount of I₂ is very important in calculation of the dissolved amount of oxygen in the given amount of water.
Moles of I₂ ------> moles of MnO₂ -------> moles of O₂ dissolved  in given amount of water.
If some amount of I₂ is lost then thiosulfate which is used to determine the amount of I₂ will be less.
This results in less reporting of I₂  -----> less reporting of mole of MnO₂ ----> reporting  less amount of O₂ dissolved  in given amount of water.

Quote(d) What volume of dry O2 measured at 25 ̊C and 1.00 atmosphere of pressure would have to be dissolved in 1.00 liter of pure water in order to prepare a solution of the same concentration as that obtained in (b)? Hint: PV=nRT (3 points)

Moles of O₂ are already calculated in part a of the question -

Quote1.36 e-5 mol of O₂ dissolved in given 50 mL of water.[/b]

Total volume of water given in this question = 1.00L = 1000 mL

Moles of O₂ in 1000.0mL =
                                1.36 e-5 mol of O₂ * 1000.mL H₂O / 50.00 mL H₂O
                                =   2.72e-4 mol
P = 1atm
T = 298K
V= nRT/P
Plug in all values to know the volume in L
V =  2.72e-4 mol  *0.0821* 298/1 =0.00665L = 6.65 mL in 1L of water.

Quote(e) Name an appropriate indictor for the reaction shown in equation III and describe the change you would observe at the end point of the titration. (2 points)

Indicator used is starch which combines with I₂ to give dark blue complex and once all I₂ is converted to I^-, color changes to light yellow .

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Again one exercise to redoxreaction, after given the solution of oxalate permanganate the student should be able to solve it by her self.
I dont see a learn effect to post also here the full solution.

Quote from: Jag on November 07, 2021, 06:08:57 AM
Again one exercise to redoxreaction, after given the solution of oxalate permanganate the student should be able to solve it by her self.
I dont see a learn effect to post also here the full solution.

It is not only for one student for private tutoring or any kind of HW help. It is for all students who are struggling with these type of questions.
If you have any problem then you stop watching this post.

I have  no problem. Its only my opinion. Its not teaching. The same can be obtain in text books, if the student understand or not is not guaranteed.
Ok I shut up my mouth now.

Here  there is no white board or any other interface to share with student.
Best is to give them solution and let them ask their doubts if they have any.
I advise you not to comment on the answers given by others .This is my way of passing knowledge and teaching according to me is with whiteboard /audio which I am using in my virtual classroom.