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Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is (2007)
  • a)
    3
  • b)
    4
  • c)
    5
  • d)
    6
Correct answer is option 'D'. Can you explain this answer?
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To determine the number of moles of Mohr's salt required per mole of potassium dichromate in the titration, we need to understand the reaction taking place and the stoichiometry of the reaction.

1. Reaction:
The reaction between potassium dichromate (K2Cr2O7) and Mohr's salt (Fe(NH4)2(SO4)2·6H2O) can be represented as follows:

K2Cr2O7 + 6Fe(NH4)2(SO4)2·6H2O + 8H2SO4 → Cr2(SO4)3 + K2SO4 + 6FeSO4 + 14NH4HSO4 + 8H2O

2. Mohr's Salt and Potassium Dichromate:
From the balanced equation, we can see that one mole of potassium dichromate reacts with six moles of Mohr's salt.

3. Diphenylamine Indicator:
Diphenylamine is used as an indicator in this titration. It reacts with oxidizing agents, such as dichromate ions, to form a blue-colored complex. The appearance of this blue color indicates the endpoint of the titration.

4. End Point:
At the end point, all the potassium dichromate has been converted to chromium(III) sulfate. This is indicated by the blue color change of the diphenylamine indicator.

5. Calculation:
To calculate the number of moles of Mohr's salt required per mole of dichromate, we need to consider the stoichiometry of the reaction. Since one mole of dichromate reacts with six moles of Mohr's salt, the correct answer is option D) 6 moles.

In conclusion, during the titration of potassium dichromate with acidified Mohr's salt solution using diphenylamine as an indicator, six moles of Mohr's salt are required per mole of dichromate.
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Titrations are one of the methods we can use to discover the precise concentrations of solution. A typical titration involves adding a solution from a burette to another solution in a flask. The endpoint of the titration is found by watching a colour change taking place. However, a problem arises when a suitable indicator cannot be found, or when the colour changes involved are unclear. In cases redox potential may sometimes come to the rescue.A particularly well known example (Fig.1) is a method of discovering the concentration of iron in a solution by titrating them with a solution of cerium (IV). The redox potential that areof interest here are EFe3+/Fe2+ = + 0.77 V and ECe4+/Ce3+ = + 1.61 V. These tell us that cercium (IV) ions are the oxidizing agents, and iron (II) ions are the reducing agent. They should react according to the equationFe2+ (aq) + Ce4+ (aq) Fe3+ (aq) + Ce3+ (aq)Now imagine that we know the concentration of the cerium (IV) ions solution in the burette. We want to measure the concentration of the iron (II) solution. If we add just one drop of the cerium (IV) solution from the bruette, some of the iron (II) ions will be oxidised. As a consequence the beaker would now contain a large number of unreacted ions, but also some iron (III) ions as well. All of the cerium (III). The solution in the beaker now represents an iron(III)/iron(II) half cell, although not at standard conditions. Thus the e.m.f. of the cell will be near, but not equal, to EFe3+/Fe2+.to ad cerium (IV) solution, the number of iron (II) ions is gradually reduced and eventually only a very few are left (Tabl e).At this stage the next few drops of cerium (IV) solution convert all the remaining iron (II) ions into iron (III), and some of the cerium (IV) ions are left unreacted. Once this happens we no longer ions and a smaller number of cerium (IV) ions. The solution in the beaker now behaves as a cerium (IV)/cerium (III) half-cell (although not a standard one).Just before all the iron (II) ions are converted into iron (III) we have a cell with an e.m.f.of around + 0.77 V. After all the iron (II) ions are oxidised, we have a cell with an e.m.f. of about + 1.61 V. This rapid rise in e.m.f. occurs with the addition of hust one drop of cerium (IV) solution. You should be able to understand why a graph of cell e.m.f. against volume ofcerium (IV) solution added looks like that of Fig. 2. The end point of the titration can be read from the graph and the concentration of the iron (II) solution calculated in the usual wayQ.Imagine you were given a solution of potassium dichromate (VI) in a beaker, and a solution of iron (II) sulphate in a burette. You do not know the concentration of dichromate (VI) ions, but the concentration of the iron (II) solution is known. Your task is to carry out a redox titration using the two solutions in order to determine the concentration of dichromate(VI) ions. Sketch a graph how the e.m.f. changes in the course of above titration. E Cr2O2-7/Cr3+ = 1.33 V, EFe3+/Fe2+ = 0.77 V.

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Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is (2007)a)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer?
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Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is (2007)a)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is (2007)a)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is (2007)a)3b)4c)5d)6Correct answer is option 'D'. Can you explain this answer?.
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