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How many millilitres of 0.1 molar H2SO4 must be added to 50 ml of 0.1 molar NaOH to give a solution that has a concentration of 0.05 molar in H2SO4? 1)400ml 2)200ml 3)100ml 100ml is the correct answer. Please explain.?
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How many millilitres of 0.1 molar H2SO4 must be added to 50 ml of 0.1 ...
Calculating the Amount of H2SO4 to Add


To solve this problem, we need to use the formula:


C1V1 = C2V2


where:



  • C1 = initial concentration of H2SO4

  • V1 = volume of H2SO4 to be added

  • C2 = final concentration of H2SO4

  • V2 = total volume of the final solution



Let's fill in the values we know:



  • C1 = 0.1 M (since the initial solution is 0.1 M H2SO4)

  • V1 = ? (this is what we're trying to find)

  • C2 = 0.05 M (since the final solution should be 0.05 M H2SO4)

  • V2 = 50 ml + V1 (since we're adding V1 ml of H2SO4 to 50 ml of NaOH)



Now we can plug these values into the formula and solve for V1:


C1V1 = C2V2

0.1 M x V1 = 0.05 M x (50 ml + V1)

0.1 V1 = 2.5 + 0.05 V1

0.05 V1 = 2.5

V1 = 50 ml


Answer


The amount of 0.1 M H2SO4 we need to add is 50 ml.
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How many millilitres of 0.1 molar H2SO4 must be added to 50 ml of 0.1 ...
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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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How many millilitres of 0.1 molar H2SO4 must be added to 50 ml of 0.1 molar NaOH to give a solution that has a concentration of 0.05 molar in H2SO4? 1)400ml 2)200ml 3)100ml 100ml is the correct answer. Please explain.?
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