When a weak acid is titrated against a strong base. The pH of solution keeps on changing with amount of base added. In this titration there is a formation of buffer also. If the buffer capacity (here defined as the volume of base of a particular concentration added per unit change in pH), is plotted against volume of base added for titration of 25 ml, 0.1 M HA (weak acid) solution with 0.1 M strong base solution, then the most appropriate curve will be :a)b)c)d)Correct answer is option 'B'. Can you explain this answer?

Explore Courses UPSC Exam UPSC Test Series Free Notes EduRev Infinity

Open App

JEE Exam > JEE Questions > When a weak acid is titrated against a strong...

When a weak acid is titrated against a strong base. The pH of solution keeps on changing with amount of base added. In this titration there is a formation of buffer also. If the buffer capacity

(here defined as the volume of base of a particular concentration added per unit change in pH), is plotted against volume of base added for titration of 25 ml, 0.1 M HA (weak acid) solution with 0.1 M strong base solution, then the most appropriate curve will be :

Correct answer is option 'B'. Can you explain this answer?

Join for Free

Join for Free

Most Upvoted Answer

When a weak acid is titrated against a strong base. The pH of solution...

Initially solution will have low buffer capacity,will have maximum buffer capacity at half neutralisation and will be close to zero at equivalence point.After equivalence point the buffer capacity will keep on increasing with increase in concentration of base

Answered by Infinity Academy · View profile

View all answers Start learning for free

Explore Courses for JEE exam

Similar JEE Doubts

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.

View answer

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.

View answer

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.When an ion is converted into a complex ion, the redox potential changes. You can see this in the case of the e.m.f. of the iron (III)/ iron (II) system (+ 0.77 V) and the hexacyanoferrate (III)/ hexacyanoferrate (II) system (+0.36 V). The cyanide ion is said to stabilize the oxidation state of the iorn. If you were to make up a cell What would be the e.m.f. and what would be the cell reaction?

View answer

The rate of chemical reaction at a particular temperature is proportional to the product of the molar concentration of reactants with each concentration term raised to the power equal to the number of molecules of the respective reactant taking part in the reaction. aA + bB products, rate of reaction a [Aka [Bib = k [Aka [Bib, where k is the rate constant of the reaction.EQUILIBRIUM CONSTANT (k) For a general reaction aA + bB = cC + dD, forward rate rf = kf [A]a [B]b, backward rate rb = kb [Cicpyi ,concentrations of reactants products at equilibrium are related bywhere all theconcentrations are expressed in mole/liter.In the expression of equilibrium constant those components are kept whose concentration changes with time.If equilibrium is estabilished by taking all the components in the reaction then to predict the direction of reactions we calculate the reaction quotient (Q).The values of expression at any time during reaction is called reaction quotient if Q Kc reaction proceed in backward direction until equilibrium in reached if Q Kc reaction will proceed in forward direction until equilibrium is established if Q = KC Reaction is at equilibriumq.Above homogeneous reaction is carried out in a 2 litre container at a particular temperature by taking 1 mole each of A, B, C and D respectively. If Kc for the reaction is then equilibrium concentration of C is

View answer

The rate of chemical reaction at a particular temperature is proportional to the product of the molar concentration of reactants with each concentration term raised to the power equal to the number of molecules of the respective reactant taking part in the reaction. aA + bB products, rate of reaction a [Aka [Bib = k [Aka [Bib, where k is the rate constant of the reaction.EQUILIBRIUM CONSTANT (k) For a general reaction aA + bB = cC + dD, forward rate rf = kf [A]a [B]b, backward rate rb = kb [Cicpyi ,concentrations of reactants products at equilibrium are related bywhere all theconcentrations are expressed in mole/liter.In the expression of equilibrium constant those components are kept whose concentration changes with time.If equilibrium is estabilished by taking all the components in the reaction then to predict the direction of reactions we calculate the reaction quotient (Q).The values of expression at any time during reaction is called reaction quotient if Q Kc reaction proceed in backward direction until equilibrium in reached if Q Kc reaction will proceed in forward direction until equilibrium is established if Q = KC Reaction is at equilibriumQ.Initial concentration of A is twice the initial concentration of `B. At equilibrium concentration of A and a are same then equilibrium constant for the reaction is

View answer

Top Courses for JEEView all

Physics for JEE Main & Advanced

Mock Tests for JEE Main and Advanced 2025

Mathematics (Maths) for JEE Main & Advanced

Chemistry for JEE Main & Advanced

Study Plan for JEE

Top Courses for JEE

View all

When a weak acid is titrated against a strong base. The pH of solution keeps on changing with amount of base added. In this titration there is a formation of buffer also. If the buffer capacity (here defined as the volume of base of a particular concentration added per unit change in pH), is plotted against volume of base added for titration of 25 ml, 0.1 M HA (weak acid) solution with 0.1 M strong base solution, then the most appropriate curve will be :a)b)c)d)Correct answer is option 'B'. Can you explain this answer?

Share with a Friend

Answer this doubt

Question Description
When a weak acid is titrated against a strong base. The pH of solution keeps on changing with amount of base added. In this titration there is a formation of buffer also. If the buffer capacity (here defined as the volume of base of a particular concentration added per unit change in pH), is plotted against volume of base added for titration of 25 ml, 0.1 M HA (weak acid) solution with 0.1 M strong base solution, then the most appropriate curve will be :a)b)c)d)Correct answer is option 'B'. 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 When a weak acid is titrated against a strong base. The pH of solution keeps on changing with amount of base added. In this titration there is a formation of buffer also. If the buffer capacity (here defined as the volume of base of a particular concentration added per unit change in pH), is plotted against volume of base added for titration of 25 ml, 0.1 M HA (weak acid) solution with 0.1 M strong base solution, then the most appropriate curve will be :a)b)c)d)Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When a weak acid is titrated against a strong base. The pH of solution keeps on changing with amount of base added. In this titration there is a formation of buffer also. If the buffer capacity (here defined as the volume of base of a particular concentration added per unit change in pH), is plotted against volume of base added for titration of 25 ml, 0.1 M HA (weak acid) solution with 0.1 M strong base solution, then the most appropriate curve will be :a)b)c)d)Correct answer is option 'B'. Can you explain this answer?.

Solutions for When a weak acid is titrated against a strong base. The pH of solution keeps on changing with amount of base added. In this titration there is a formation of buffer also. If the buffer capacity (here defined as the volume of base of a particular concentration added per unit change in pH), is plotted against volume of base added for titration of 25 ml, 0.1 M HA (weak acid) solution with 0.1 M strong base solution, then the most appropriate curve will be :a)b)c)d)Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of When a weak acid is titrated against a strong base. The pH of solution keeps on changing with amount of base added. In this titration there is a formation of buffer also. If the buffer capacity (here defined as the volume of base of a particular concentration added per unit change in pH), is plotted against volume of base added for titration of 25 ml, 0.1 M HA (weak acid) solution with 0.1 M strong base solution, then the most appropriate curve will be :a)b)c)d)Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of When a weak acid is titrated against a strong base. The pH of solution keeps on changing with amount of base added. In this titration there is a formation of buffer also. If the buffer capacity (here defined as the volume of base of a particular concentration added per unit change in pH), is plotted against volume of base added for titration of 25 ml, 0.1 M HA (weak acid) solution with 0.1 M strong base solution, then the most appropriate curve will be :a)b)c)d)Correct answer is option 'B'. Can you explain this answer?, a detailed solution for When a weak acid is titrated against a strong base. The pH of solution keeps on changing with amount of base added. In this titration there is a formation of buffer also. If the buffer capacity (here defined as the volume of base of a particular concentration added per unit change in pH), is plotted against volume of base added for titration of 25 ml, 0.1 M HA (weak acid) solution with 0.1 M strong base solution, then the most appropriate curve will be :a)b)c)d)Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of When a weak acid is titrated against a strong base. The pH of solution keeps on changing with amount of base added. In this titration there is a formation of buffer also. If the buffer capacity (here defined as the volume of base of a particular concentration added per unit change in pH), is plotted against volume of base added for titration of 25 ml, 0.1 M HA (weak acid) solution with 0.1 M strong base solution, then the most appropriate curve will be :a)b)c)d)Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice When a weak acid is titrated against a strong base. The pH of solution keeps on changing with amount of base added. In this titration there is a formation of buffer also. If the buffer capacity (here defined as the volume of base of a particular concentration added per unit change in pH), is plotted against volume of base added for titration of 25 ml, 0.1 M HA (weak acid) solution with 0.1 M strong base solution, then the most appropriate curve will be :a)b)c)d)Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.

Explore Courses for JEE exam