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40 ml of 0.11M HCl is diluted to 100 ml with water and then titrated with 0.1M NaOH. The pH of the resulting solution after addition of 10 ml of titrant is?
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40 ml of 0.11M HCl is diluted to 100 ml with water and then titrated w...
**Problem:** pH of the solution after titration of 40 ml of 0.11 M HCl with 0.1 M NaOH.
**Given:**
- Volume of HCl solution = 40 ml
- Concentration of HCl solution = 0.11 M
- Volume of water added = 100 ml
- Concentration of NaOH titrant = 0.1 M
- Volume of NaOH titrant added = 10 ml
**Solution Approach:**
1. Calculate the moles of HCl in the initial solution.
2. Calculate the moles of NaOH added during titration.
3. Determine which reactant is in excess.
4. Calculate the moles of the excess reactant.
5. Calculate the moles of the remaining acid or base.
6. Determine the concentration of the remaining acid or base.
7. Calculate the pH of the resulting solution.
**Calculation Steps:**
1. Calculate the moles of HCl in the initial solution:
- Moles of HCl = (concentration) x (volume)
- Moles of HCl = (0.11 M) x (0.04 L)
- Moles of HCl = 0.0044 mol
2. Calculate the moles of NaOH added during titration:
- Moles of NaOH = (concentration) x (volume)
- Moles of NaOH = (0.1 M) x (0.01 L)
- Moles of NaOH = 0.001 mol
3. Determine which reactant is in excess:
- Since the moles of HCl (0.0044 mol) are greater than the moles of NaOH (0.001 mol), HCl is in excess.
4. Calculate the moles of the excess reactant (HCl):
- Moles of excess HCl = Moles of HCl - Moles of NaOH
- Moles of excess HCl = 0.0044 mol - 0.001 mol
- Moles of excess HCl = 0.0034 mol
5. Calculate the moles of the remaining acid (HCl):
- Moles of remaining HCl = Moles of excess HCl - Moles of NaOH
- Moles of remaining HCl = 0.0034 mol - 0.001 mol
- Moles of remaining HCl = 0.0024 mol
6. Determine the concentration of the remaining acid (HCl):
- Concentration of remaining HCl = Moles of remaining HCl / Volume of resulting solution
- Volume of resulting solution = Volume of initial solution + Volume of water added
- Volume of resulting solution = 0.04 L + 0.1 L
- Volume of resulting solution = 0.14 L
- Concentration of remaining HCl = 0.0024 mol / 0.14 L
- Concentration of remaining HCl = 0.017 M
7. Calculate the pH of the resulting solution:
- pH = -log10(H+ concentration)
- pH = -log10(0.017 M)
- pH ≈ 1.77
**Answer:**
The pH of the resulting solution after the addition of 10
**Given:**
- Volume of HCl solution = 40 ml
- Concentration of HCl solution = 0.11 M
- Volume of water added = 100 ml
- Concentration of NaOH titrant = 0.1 M
- Volume of NaOH titrant added = 10 ml
**Solution Approach:**
1. Calculate the moles of HCl in the initial solution.
2. Calculate the moles of NaOH added during titration.
3. Determine which reactant is in excess.
4. Calculate the moles of the excess reactant.
5. Calculate the moles of the remaining acid or base.
6. Determine the concentration of the remaining acid or base.
7. Calculate the pH of the resulting solution.
**Calculation Steps:**
1. Calculate the moles of HCl in the initial solution:
- Moles of HCl = (concentration) x (volume)
- Moles of HCl = (0.11 M) x (0.04 L)
- Moles of HCl = 0.0044 mol
2. Calculate the moles of NaOH added during titration:
- Moles of NaOH = (concentration) x (volume)
- Moles of NaOH = (0.1 M) x (0.01 L)
- Moles of NaOH = 0.001 mol
3. Determine which reactant is in excess:
- Since the moles of HCl (0.0044 mol) are greater than the moles of NaOH (0.001 mol), HCl is in excess.
4. Calculate the moles of the excess reactant (HCl):
- Moles of excess HCl = Moles of HCl - Moles of NaOH
- Moles of excess HCl = 0.0044 mol - 0.001 mol
- Moles of excess HCl = 0.0034 mol
5. Calculate the moles of the remaining acid (HCl):
- Moles of remaining HCl = Moles of excess HCl - Moles of NaOH
- Moles of remaining HCl = 0.0034 mol - 0.001 mol
- Moles of remaining HCl = 0.0024 mol
6. Determine the concentration of the remaining acid (HCl):
- Concentration of remaining HCl = Moles of remaining HCl / Volume of resulting solution
- Volume of resulting solution = Volume of initial solution + Volume of water added
- Volume of resulting solution = 0.04 L + 0.1 L
- Volume of resulting solution = 0.14 L
- Concentration of remaining HCl = 0.0024 mol / 0.14 L
- Concentration of remaining HCl = 0.017 M
7. Calculate the pH of the resulting solution:
- pH = -log10(H+ concentration)
- pH = -log10(0.017 M)
- pH ≈ 1.77
**Answer:**
The pH of the resulting solution after the addition of 10
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40 ml of 0.11M HCl is diluted to 100 ml with water and then titrated with 0.1M NaOH. The pH of the resulting solution after addition of 10 ml of titrant is?
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40 ml of 0.11M HCl is diluted to 100 ml with water and then titrated with 0.1M NaOH. The pH of the resulting solution after addition of 10 ml of titrant is? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about 40 ml of 0.11M HCl is diluted to 100 ml with water and then titrated with 0.1M NaOH. The pH of the resulting solution after addition of 10 ml of titrant is? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 40 ml of 0.11M HCl is diluted to 100 ml with water and then titrated with 0.1M NaOH. The pH of the resulting solution after addition of 10 ml of titrant is?.
40 ml of 0.11M HCl is diluted to 100 ml with water and then titrated with 0.1M NaOH. The pH of the resulting solution after addition of 10 ml of titrant is? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about 40 ml of 0.11M HCl is diluted to 100 ml with water and then titrated with 0.1M NaOH. The pH of the resulting solution after addition of 10 ml of titrant is? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 40 ml of 0.11M HCl is diluted to 100 ml with water and then titrated with 0.1M NaOH. The pH of the resulting solution after addition of 10 ml of titrant is?.
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