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A solution of weak acid HA was titrated with base NaOH. The equivalence point was reaced when 36.12 ml of 0.1 M NaOH has been added. Now 18.06 ml of 0.1 M HCl were added to titration solution, the pH was found to be 4.92. What will be the pH of the solution obtained by mixing 10 ml 0.2 M NaOH and 10 ml of 0.2 M HA.
  • a)
    10.96
  • b)
    8.96
  • c)
    12.96
  • d)
    6.96
Correct answer is option 'B'. Can you explain this answer?
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PH Calculation for the Solution Obtained by Mixing NaOH and HA

Given Information:
- Weak acid HA is titrated with base NaOH.
- Equivalence point is reached when 36.12 ml of 0.1 M NaOH is added.
- Later, 18.06 ml of 0.1 M HCl is added and the pH is found to be 4.92.
- We need to find the pH of the solution obtained by mixing 10 ml of 0.2 M NaOH and 10 ml of 0.2 M HA.

Step 1: Determine the moles of NaOH added at the equivalence point.
- The volume of 0.1 M NaOH added at the equivalence point is 36.12 ml.
- Convert the volume to liters: 36.12 ml = 0.03612 L.
- Calculate the moles of NaOH: moles = concentration x volume = 0.1 M x 0.03612 L = 0.003612 mol.

Step 2: Determine the moles of HCl added after the equivalence point.
- The volume of 0.1 M HCl added is 18.06 ml.
- Convert the volume to liters: 18.06 ml = 0.01806 L.
- Calculate the moles of HCl: moles = concentration x volume = 0.1 M x 0.01806 L = 0.001806 mol.

Step 3: Calculate the moles of HA at the equivalence point.
- Since NaOH is a strong base and HA is a weak acid, the reaction between them is as follows:
HA + NaOH → NaA + H2O
- According to the balanced equation, the moles of HA and NaOH are in a 1:1 ratio.
- Therefore, the moles of HA at the equivalence point is also 0.003612 mol.

Step 4: Calculate the moles of HA remaining after the addition of HCl.
- Since HCl is a strong acid, it completely reacts with the remaining HA.
- Therefore, the moles of HA remaining after the addition of HCl is 0.003612 mol - 0.001806 mol = 0.001806 mol.

Step 5: Determine the concentration of HA in the final solution.
- The total volume of the final solution is 10 ml + 10 ml = 20 ml = 0.02 L.
- Calculate the concentration of HA: concentration = moles/volume = 0.001806 mol/0.02 L = 0.0903 M.

Step 6: Determine the concentration of A- in the final solution.
- Since HA is a weak acid, it partially dissociates in water to form its conjugate base A-.
- At the equivalence point, the concentration of A- is equal to the concentration of HA, which is 0.0903 M.

Step 7: Calculate the pOH of the final solution.
- Since NaOH is a strong base, it completely dissociates in water to form OH- ions.
- The moles of OH- added by NaOH is 0.1 M x 0.01 L = 0.001 mol.
- The moles of OH- added by HA is half the moles of HA, which is 0
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A solution of weak acid HA was titrated with base NaOH. The equivalence point was reaced when 36.12 ml of 0.1 M NaOH has been added. Now 18.06 ml of 0.1 M HCl were added to titration solution, the pH was found to be 4.92. What will be the pH of the solution obtained by mixing 10 ml 0.2 M NaOH and 10 ml of 0.2 M HA.a)10.96b)8.96c)12.96d)6.96Correct answer is option 'B'. Can you explain this answer?
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A solution of weak acid HA was titrated with base NaOH. The equivalence point was reaced when 36.12 ml of 0.1 M NaOH has been added. Now 18.06 ml of 0.1 M HCl were added to titration solution, the pH was found to be 4.92. What will be the pH of the solution obtained by mixing 10 ml 0.2 M NaOH and 10 ml of 0.2 M HA.a)10.96b)8.96c)12.96d)6.96Correct 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 A solution of weak acid HA was titrated with base NaOH. The equivalence point was reaced when 36.12 ml of 0.1 M NaOH has been added. Now 18.06 ml of 0.1 M HCl were added to titration solution, the pH was found to be 4.92. What will be the pH of the solution obtained by mixing 10 ml 0.2 M NaOH and 10 ml of 0.2 M HA.a)10.96b)8.96c)12.96d)6.96Correct 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 A solution of weak acid HA was titrated with base NaOH. The equivalence point was reaced when 36.12 ml of 0.1 M NaOH has been added. Now 18.06 ml of 0.1 M HCl were added to titration solution, the pH was found to be 4.92. What will be the pH of the solution obtained by mixing 10 ml 0.2 M NaOH and 10 ml of 0.2 M HA.a)10.96b)8.96c)12.96d)6.96Correct answer is option 'B'. Can you explain this answer?.
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