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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 HCL when (1/5)th of the base was neutralised the pH was found to be 9 and at equivalent. PH of solution is 4.5 given all data at and log2= 0.3?
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A 2.5 gm impure sample containing weak monoacidic base (mol wt =45)is ...
Problem Analysis:
- We have an impure sample containing a weak monoacidic base.
- The molecular weight of the base is given as 45 grams/mole.
- The sample is dissolved in 100 ml of water and titrated with 0.5 M HCl.
- When 1/5th of the base is neutralized, the pH is found to be 9.
- At the equivalence point, the pH of the solution is 4.5.
- We are given that log2 = 0.3.
Solution:
1. Calculation of moles of the base:
- The given mass of the impure sample is 2.5 grams.
- The molecular weight of the base is 45 grams/mole.
- Using the formula: Moles = Mass / Molecular Weight, we can calculate the moles of the base.
- Moles of the base = 2.5 g / 45 g/mol = 0.0556 moles.
2. Calculation of moles of HCl:
- At the equivalence point, the moles of the base and moles of HCl will be equal.
- Therefore, the moles of HCl = 0.0556 moles.
3. Calculation of volume of HCl:
- The concentration of HCl is given as 0.5 M.
- Using the formula: Moles = Concentration x Volume, we can calculate the volume of HCl.
- 0.0556 moles = 0.5 M x Volume
- Volume = 0.0556 moles / 0.5 M = 0.1112 L = 111.2 ml.
4. Calculation of moles of HCl used at pH 9:
- At pH 9, the solution is basic.
- The pOH of the solution can be calculated using the formula: pOH = 14 - pH.
- pOH = 14 - 9 = 5.
- The concentration of OH- ions can be calculated using the formula: [OH-] = 10^(-pOH).
- [OH-] = 10^(-5) = 0.00001 M.
- The moles of OH- ions can be calculated using the formula: Moles = Concentration x Volume.
- Moles of OH- = 0.00001 M x 100 ml = 0.001 moles.
- Since 1/5th of the base is neutralized, the moles of the base neutralized = 1/5 x 0.0556 = 0.0111 moles.
- The moles of HCl used at pH 9 = 0.0111 moles - 0.001 moles = 0.0101 moles.
5. Calculation of moles of HCl used at equivalence point:
- At the equivalence point, all the moles of the base are neutralized.
- Therefore, the moles of HCl used at the equivalence point = 0.0556 moles.
6. Calculation of volume of HCl added after pH 9:
- The moles of HCl used after pH 9 = Moles at equivalence point - Moles at pH
- We have an impure sample containing a weak monoacidic base.
- The molecular weight of the base is given as 45 grams/mole.
- The sample is dissolved in 100 ml of water and titrated with 0.5 M HCl.
- When 1/5th of the base is neutralized, the pH is found to be 9.
- At the equivalence point, the pH of the solution is 4.5.
- We are given that log2 = 0.3.
Solution:
1. Calculation of moles of the base:
- The given mass of the impure sample is 2.5 grams.
- The molecular weight of the base is 45 grams/mole.
- Using the formula: Moles = Mass / Molecular Weight, we can calculate the moles of the base.
- Moles of the base = 2.5 g / 45 g/mol = 0.0556 moles.
2. Calculation of moles of HCl:
- At the equivalence point, the moles of the base and moles of HCl will be equal.
- Therefore, the moles of HCl = 0.0556 moles.
3. Calculation of volume of HCl:
- The concentration of HCl is given as 0.5 M.
- Using the formula: Moles = Concentration x Volume, we can calculate the volume of HCl.
- 0.0556 moles = 0.5 M x Volume
- Volume = 0.0556 moles / 0.5 M = 0.1112 L = 111.2 ml.
4. Calculation of moles of HCl used at pH 9:
- At pH 9, the solution is basic.
- The pOH of the solution can be calculated using the formula: pOH = 14 - pH.
- pOH = 14 - 9 = 5.
- The concentration of OH- ions can be calculated using the formula: [OH-] = 10^(-pOH).
- [OH-] = 10^(-5) = 0.00001 M.
- The moles of OH- ions can be calculated using the formula: Moles = Concentration x Volume.
- Moles of OH- = 0.00001 M x 100 ml = 0.001 moles.
- Since 1/5th of the base is neutralized, the moles of the base neutralized = 1/5 x 0.0556 = 0.0111 moles.
- The moles of HCl used at pH 9 = 0.0111 moles - 0.001 moles = 0.0101 moles.
5. Calculation of moles of HCl used at equivalence point:
- At the equivalence point, all the moles of the base are neutralized.
- Therefore, the moles of HCl used at the equivalence point = 0.0556 moles.
6. Calculation of volume of HCl added after pH 9:
- The moles of HCl used after pH 9 = Moles at equivalence point - Moles at pH
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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 HCL when (1/5)th of the base was neutralised the pH was found to be 9 and at equivalent. PH of solution is 4.5 given all data at and log2= 0.3?
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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 HCL when (1/5)th of the base was neutralised the pH was found to be 9 and at equivalent. PH of solution is 4.5 given all data at and log2= 0.3? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 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 HCL when (1/5)th of the base was neutralised the pH was found to be 9 and at equivalent. PH of solution is 4.5 given all data at and log2= 0.3? covers all topics & solutions for Class 12 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 HCL when (1/5)th of the base was neutralised the pH was found to be 9 and at equivalent. PH of solution is 4.5 given all data at and log2= 0.3?.
A 2.5 gm impure sample containing weak monoacidic base (mol wt =45)is dissolved in 100 ml water and titrated with 0.5 HCL when (1/5)th of the base was neutralised the pH was found to be 9 and at equivalent. PH of solution is 4.5 given all data at and log2= 0.3? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 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 HCL when (1/5)th of the base was neutralised the pH was found to be 9 and at equivalent. PH of solution is 4.5 given all data at and log2= 0.3? covers all topics & solutions for Class 12 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 HCL when (1/5)th of the base was neutralised the pH was found to be 9 and at equivalent. PH of solution is 4.5 given all data at and log2= 0.3?.
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