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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?
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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
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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.
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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?
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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?.
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