All Courses
Explore Courses UPSC Exam UPSC Test Series Free Notes EduRev Infinity Get the App Login Join for Free
Sign in Open App
JEE Exam  >  JEE Questions  >  A 2.5 gm impure sample containing weak monoac... Start Learning for Free
SaveJoin the Discussion on the App
A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. = 45) is dissolved in 100 ml water and titrated with 0.5 M HCl when  of the base was neutralised the pH was found to be 9 and at equivalent point pH of solution is 4.5. Given : All data at 25º C & log 2 = 0.3.
Select correct statement(s).
  • a)
    Kb of base is less than 10-6
  • b)
    Concentration of salt (C) at equivalent point is 0.25 M
  • c)
    Volume of HCl is used at equavalent point is 100 ml
  • d)
    Weight percentage of base in given sample is 80% 
Correct answer is option 'B,C'. Can you explain this answer?
FREE This question is part of
Download PDF Attempt Test
Download PDF Attempt this Test
Verified Answer
A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. = 45)...
BOH + HCl  → BCl + H2O
100 m 0.5 V 20 M
100 M – 0.5 V  0  = 80 M
14–a

⇒ pkb = 5 – log 0.25 ⇒ pkb = 5.6
⇒ Kb = 2.5 × 10–6
greater than 10–6
View all questions of this test
Most Upvoted Answer
A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. = 45)...
Let the initial concentration of base be c. When 1/5 of the base is neutralised amount of base left is equal to 4c/5 and concentration of salt formed is c/5.
Using Henderson's equation
pOH = pKb +log(1/4)

This gives me pKb = 5.6.
At equivalence point
Total amount of base is converted into salt. Thus concentration of salt is c.
Assuming hydrolysis of the salt
pH = 7-1/2pKb-log(c)

This gives me C=0.5. But the correct answer says that it should be 0.25.
< Previous Question Next Question >
Explore Courses for JEE exam

Similar JEE Doubts

Volumetric analysis is based on the principle of equivalence a that minloces lead in the ratio of their equivalents. At the equivalence point of the reaction, involving the reactahts A and B : Number of gram equivalents of A = Number of gram equivalents of B..If VA ml of solution A having normality NA react just completely with VB ml of solution B having normalityNB thenEquation(1), called normality equation is very useful in numerical of volumetric analysis. Equivalent masses of different substances :Basically the equivalent mass of a substance is defined as the pads by seam ctituhich combine with or displace 1.0078 parts (1part) by mass of hydrogen, 8 parts by mansetaisiosaid 35.5 parts by mass of chlorine.Mass of a substance expressed in gram equal to its equivalent man is gam equivalent mass. Equivalent mass of a substance is not constant but depends upon the maw i malice the substance participates.Equivalent mass of an acid in acid-base reaction is its mass in grannehidicodoinisooleaf replaceable Fi ions (= 1.0078 g1g).0n the other hand, equivalent mass of a bagels istolonellfacontairis 1 mole of replaceable 0H ions. 1 g equivalent mass each of an acid and base an ionclinegins saint and 1 mole of water= 18 g). Equivalent mass of an oxidising agent is its mass which gains 1 mole at ellecismitcan be obtained by dividing the molecular mass or formula mass by the total decrease in aididditaiiilberef one or more elements per molecule.0n the other hand, equivalent mass of a reducing agent is the mass of the substance which loses 1 mole of electorns. It can be calculated by dividing the molecular or formula mass of the subtance by the total increase in oxidation number of one or more elements per molecular or formula mass.Q.100 ml of 0.1 M H3 P02 is titrated with 0.1 M NaOH. The volume of NaOH needed will be

Volumetric analysis is based on the principle of equivalence a that minloces lead in the ratio of their equivalents. At the equivalence point of the reaction, involving the reactahts A and B : Number of gram equivalents of A = Number of gram equivalents of B..If VA ml of solution A having normality NA react just completely with VB ml of solution B having normalityNB thenEquation(1), called normality equation is very useful in numerical of volumetric analysis. Equivalent masses of different substances :Basically the equivalent mass of a substance is defined as the pads by seam ctituhich combine with or displace 1.0078 parts (1part) by mass of hydrogen, 8 parts by mansetaisiosaid 35.5 parts by mass of chlorine.Mass of a substance expressed in gram equal to its equivalent man is gam equivalent mass. Equivalent mass of a substance is not constant but depends upon the maw i malice the substance participates.Equivalent mass of an acid in acid-base reaction is its mass in grannehidicodoinisooleaf replaceable Fi ions (= 1.0078 g1g).0n the other hand, equivalent mass of a bagels istolonellfacontairis 1 mole of replaceable 0H ions. 1 g equivalent mass each of an acid and base an ionclinegins saint and 1 mole of water= 18 g). Equivalent mass of an oxidising agent is its mass which gains 1 mole at ellecismitcan be obtained by dividing the molecular mass or formula mass by the total decrease in aididditaiiilberef one or more elements per molecule.0n the other hand, equivalent mass of a reducing agent is the mass of the substance which loses 1 mole of electorns. It can be calculated by dividing the molecular or formula mass of the subtance by the total increase in oxidation number of one or more elements per molecular or formula mass.Q.Which of the following solutions, when mixed with 100 ml of 0.05 M NaOH, will give a neutral solution?

Volumetric analysis is based on the principle of equivalence a that minloces lead in the ratio of their equivalents. At the equivalence point of the reaction, involving the reactahts A and B : Number of gram equivalents of A = Number of gram equivalents of B..If VA ml of solution A having normality NA react just completely with VB ml of solution B having normalityNB thenEquation(1), called normality equation is very useful in numerical of volumetric analysis. Equivalent masses of different substances :Basically the equivalent mass of a substance is defined as the pads by seam ctituhich combine with or displace 1.0078 parts (1part) by mass of hydrogen, 8 parts by mansetaisiosaid 35.5 parts by mass of chlorine.Mass of a substance expressed in gram equal to its equivalent man is gam equivalent mass. Equivalent mass of a substance is not constant but depends upon the maw i malice the substance participates.Equivalent mass of an acid in acid-base reaction is its mass in grannehidicodoinisooleaf replaceable Fi ions (= 1.0078 g1g).0n the other hand, equivalent mass of a bagels istolonellfacontairis 1 mole of replaceable 0H ions. 1 g equivalent mass each of an acid and base an ionclinegins saint and 1 mole of water= 18 g). Equivalent mass of an oxidising agent is its mass which gains 1 mole at ellecismitcan be obtained by dividing the molecular mass or formula mass by the total decrease in aididditaiiilberef one or more elements per molecule.0n the other hand, equivalent mass of a reducing agent is the mass of the substance which loses 1 mole of electorns. It can be calculated by dividing the molecular or formula mass of the subtance by the total increase in oxidation number of one or more elements per molecular or formula mass.Q.How many gram of phosphoric acid (H3PO4) would be required to neutralize 58 g of magnesium hydroxide?

Volumetric analysis is based on the principle of equivalence a that minloces lead in the ratio of their equivalents. At the equivalence point of the reaction, involving the reactahts A and B : Number of gram equivalents of A = Number of gram equivalents of B..If VA ml of solution A having normality NA react just completely with VB ml of solution B having normalityNB thenEquation(1), called normality equation is very useful in numerical of volumetric analysis. Equivalent masses of different substances :Basically the equivalent mass of a substance is defined as the pads by seam ctituhich combine with or displace 1.0078 parts (1part) by mass of hydrogen, 8 parts by mansetaisiosaid 35.5 parts by mass of chlorine.Mass of a substance expressed in gram equal to its equivalent man is gam equivalent mass. Equivalent mass of a substance is not constant but depends upon the maw i malice the substance participates.Equivalent mass of an acid in acid-base reaction is its mass in grannehidicodoinisooleaf replaceable Fi ions (= 1.0078 g1g).0n the other hand, equivalent mass of a bagels istolonellfacontairis 1 mole of replaceable 0H ions. 1 g equivalent mass each of an acid and base an ionclinegins saint and 1 mole of water= 18 g). Equivalent mass of an oxidising agent is its mass which gains 1 mole at ellecismitcan be obtained by dividing the molecular mass or formula mass by the total decrease in aididditaiiilberef one or more elements per molecule.0n the other hand, equivalent mass of a reducing agent is the mass of the substance which loses 1 mole of electorns. It can be calculated by dividing the molecular or formula mass of the subtance by the total increase in oxidation number of one or more elements per molecular or formula mass.Q.9.8 g of an acid on treatment with excess of an alkali forms salt along with 3.6 g of water. What is the equivalent mass of the acid?

Volumetric analysis is based on the principle of equivalence a that minloces lead in the ratio of their equivalents. At the equivalence point of the reaction, involving the reactahts A and B : Number of gram equivalents of A = Number of gram equivalents of B..If VA ml of solution A having normality NA react just completely with VB ml of solution B having normalityNB thenEquation(1), called normality equation is very useful in numerical of volumetric analysis. Equivalent masses of different substances :Basically the equivalent mass of a substance is defined as the pads by seamctituhich combine with or displace 1.0078 parts (1part) by mass of hydrogen, 8 parts by mansetaisiosaid 35.5 parts by mass of chlorine. Mass of a substance expressed in gram equal to its equivalent man is gam equivalent mass. Equivalent mass of a substance is not constant but depends upon the maw i malice the substance participates. Equivalent mass of an acid in acid-base reaction is its mass in grannehidicodoinisooleaf replaceable Fi ions (= 1.0078 g1g).0n the other hand, equivalent mass of a bagels istolonellfacontairis 1 mole of replaceable 0H ions. 1 g equivalent mass each of an acid and base an ionclinegins saint and 1 mole of water ( = 18 g).Equivalent mass of an oxidising agent is its mass which gains 1 mole at ellecismitcan be obtained by dividing the molecular mass or formula mass by the total decrease in aididditaiiilberef one or more elements per molecule. 0n the other hand, equivalent mass of a reducing agent is the mass of the substance which loses 1 mole of electorns. It can be calculated by dividing the molecular or formula mass of the subtance by the total increase in oxidation number of one or more elements per molecular or formula mass.Q.When 5.00 g of a metal is strongly heated, 9.44 g of its oxide is obtained. The equivalent mass of the metal is

Top Courses for JEEView all

Top Courses for JEE

A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. = 45) is dissolved in 100 ml water and titrated with 0.5 M HCl whenof the base was neutralised the pH was found to be 9 and at equivalent point pH of solution is 4.5. Given : All data at 25º C & log 2 = 0.3.Select correct statement(s).a)Kbof base is less than 10-6b)Concentration of salt (C) at equivalent point is 0.25 Mc)Volume of HCl is used at equavalent point is 100 mld)Weight percentage of base in given sample is 80%Correct answer is option 'B,C'. Can you explain this answer?
Question Description
A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. = 45) is dissolved in 100 ml water and titrated with 0.5 M HCl whenof the base was neutralised the pH was found to be 9 and at equivalent point pH of solution is 4.5. Given : All data at 25º C & log 2 = 0.3.Select correct statement(s).a)Kbof base is less than 10-6b)Concentration of salt (C) at equivalent point is 0.25 Mc)Volume of HCl is used at equavalent point is 100 mld)Weight percentage of base in given sample is 80%Correct answer is option 'B,C'. 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 A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. = 45) is dissolved in 100 ml water and titrated with 0.5 M HCl whenof the base was neutralised the pH was found to be 9 and at equivalent point pH of solution is 4.5. Given : All data at 25º C & log 2 = 0.3.Select correct statement(s).a)Kbof base is less than 10-6b)Concentration of salt (C) at equivalent point is 0.25 Mc)Volume of HCl is used at equavalent point is 100 mld)Weight percentage of base in given sample is 80%Correct answer is option 'B,C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. = 45) is dissolved in 100 ml water and titrated with 0.5 M HCl whenof the base was neutralised the pH was found to be 9 and at equivalent point pH of solution is 4.5. Given : All data at 25º C & log 2 = 0.3.Select correct statement(s).a)Kbof base is less than 10-6b)Concentration of salt (C) at equivalent point is 0.25 Mc)Volume of HCl is used at equavalent point is 100 mld)Weight percentage of base in given sample is 80%Correct answer is option 'B,C'. Can you explain this answer?.
Solutions for A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. = 45) is dissolved in 100 ml water and titrated with 0.5 M HCl whenof the base was neutralised the pH was found to be 9 and at equivalent point pH of solution is 4.5. Given : All data at 25º C & log 2 = 0.3.Select correct statement(s).a)Kbof base is less than 10-6b)Concentration of salt (C) at equivalent point is 0.25 Mc)Volume of HCl is used at equavalent point is 100 mld)Weight percentage of base in given sample is 80%Correct answer is option 'B,C'. 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 A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. = 45) is dissolved in 100 ml water and titrated with 0.5 M HCl whenof the base was neutralised the pH was found to be 9 and at equivalent point pH of solution is 4.5. Given : All data at 25º C & log 2 = 0.3.Select correct statement(s).a)Kbof base is less than 10-6b)Concentration of salt (C) at equivalent point is 0.25 Mc)Volume of HCl is used at equavalent point is 100 mld)Weight percentage of base in given sample is 80%Correct answer is option 'B,C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. = 45) is dissolved in 100 ml water and titrated with 0.5 M HCl whenof the base was neutralised the pH was found to be 9 and at equivalent point pH of solution is 4.5. Given : All data at 25º C & log 2 = 0.3.Select correct statement(s).a)Kbof base is less than 10-6b)Concentration of salt (C) at equivalent point is 0.25 Mc)Volume of HCl is used at equavalent point is 100 mld)Weight percentage of base in given sample is 80%Correct answer is option 'B,C'. Can you explain this answer?, a detailed solution for A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. = 45) is dissolved in 100 ml water and titrated with 0.5 M HCl whenof the base was neutralised the pH was found to be 9 and at equivalent point pH of solution is 4.5. Given : All data at 25º C & log 2 = 0.3.Select correct statement(s).a)Kbof base is less than 10-6b)Concentration of salt (C) at equivalent point is 0.25 Mc)Volume of HCl is used at equavalent point is 100 mld)Weight percentage of base in given sample is 80%Correct answer is option 'B,C'. Can you explain this answer? has been provided alongside types of A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. = 45) is dissolved in 100 ml water and titrated with 0.5 M HCl whenof the base was neutralised the pH was found to be 9 and at equivalent point pH of solution is 4.5. Given : All data at 25º C & log 2 = 0.3.Select correct statement(s).a)Kbof base is less than 10-6b)Concentration of salt (C) at equivalent point is 0.25 Mc)Volume of HCl is used at equavalent point is 100 mld)Weight percentage of base in given sample is 80%Correct answer is option 'B,C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice A 2.5 gm impure sample containing weak monoacidic base (Mol. wt. = 45) is dissolved in 100 ml water and titrated with 0.5 M HCl whenof the base was neutralised the pH was found to be 9 and at equivalent point pH of solution is 4.5. Given : All data at 25º C & log 2 = 0.3.Select correct statement(s).a)Kbof base is less than 10-6b)Concentration of salt (C) at equivalent point is 0.25 Mc)Volume of HCl is used at equavalent point is 100 mld)Weight percentage of base in given sample is 80%Correct answer is option 'B,C'. Can you explain this answer? tests, examples and also practice JEE tests.
Explore Courses for JEE exam

Top Courses for JEE

Explore Courses

Suggested Free Tests

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Get App
© EduRev
Education Revolution
Follow Us
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup