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25ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35ml. The molarity of barium hydroxide solution was [2003]
  • a)
    0.14
  • b)
    0.28
  • c)
    0.35
  • d)
    0.07
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
25ml of a solution of barium hydroxide on titration with a 0.1 molar s...
25 × N = 0.1 × 35 ; N = 0.14
Ba(OH)2 is diacid base hence N = M × 2  or  M =  ⇒  M = 0.07 M
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Most Upvoted Answer
25ml of a solution of barium hydroxide on titration with a 0.1 molar s...
Given:
Volume of barium hydroxide solution (V1) = 25 ml
Volume of hydrochloric acid solution (V2) = 35 ml
Molarity of hydrochloric acid solution (M2) = 0.1 M

To find:
Molarity of barium hydroxide solution (M1)

We can use the equation:
M1V1 = M2V2

Substituting the given values:
M1 * 25 ml = 0.1 M * 35 ml

Simplifying the equation:
M1 = (0.1 M * 35 ml) / 25 ml

M1 = 0.14 M

Therefore, the molarity of the barium hydroxide solution is 0.14 M.

Explanation:
In a titration, the reaction between the acid and base is a neutralization reaction. The balanced chemical equation for the reaction between barium hydroxide (Ba(OH)2) and hydrochloric acid (HCl) is:

Ba(OH)2 + 2HCl → BaCl2 + 2H2O

From the balanced equation, we can see that 1 mole of barium hydroxide reacts with 2 moles of hydrochloric acid. In the given question, we know the volume of hydrochloric acid used in the titration (35 ml), which means we have used 1/2 mole of hydrochloric acid.

Now, using the equation M1V1 = M2V2, we can calculate the molarity of barium hydroxide solution. By substituting the given values, we get the molarity of barium hydroxide as 0.14 M.

Therefore, the correct answer is option D) 0.07 M.
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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.The cell shown below was set upWhat would be the cell e.m.f.? If potassium cyanide solution were added to the left hand half cell (with due care!), what would you expect to happen to the e.m.f. of the cell? E Br2/Br- = 1.07V and use data of previousquestion, if required.

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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25ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35ml. The molarity of barium hydroxide solution was [2003]a)0.14b)0.28c)0.35d)0.07Correct answer is option 'D'. Can you explain this answer?
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25ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35ml. The molarity of barium hydroxide solution was [2003]a)0.14b)0.28c)0.35d)0.07Correct answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 25ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35ml. The molarity of barium hydroxide solution was [2003]a)0.14b)0.28c)0.35d)0.07Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 25ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35ml. The molarity of barium hydroxide solution was [2003]a)0.14b)0.28c)0.35d)0.07Correct answer is option 'D'. Can you explain this answer?.
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