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If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M HCOOH [Ka = 2 × 10-4]. The pOH of the resulting solution is
- a)3.4
- b)3.7
- c)7
- d)10.3
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M HCOOH [Ka= 2 × ...
m. equivalent of KOH = 8
m. equivalent of HCOOH = 16
Remaining m. eq. (HCOOH) = 8
Formed m. eq. (HCOOK) = 8
⇒ Acidic Buffer
pH = pKa = 4 – log2
= 3.7
pOH = 10.3
m. equivalent of HCOOH = 16
Remaining m. eq. (HCOOH) = 8
Formed m. eq. (HCOOK) = 8
⇒ Acidic Buffer
pH = pKa = 4 – log2
= 3.7
pOH = 10.3
Most Upvoted Answer
If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M HCOOH [Ka= 2 × ...
m. equivalent of KOH = 8
m. equivalent of HCOOH = 16
Remaining m. eq. (HCOOH) = 8
Formed m. eq. (HCOOK) = 8
⇒ Acidic Buffer
pH = pKa = 4 – log2
= 3.7
pOH = 10.3
m. equivalent of HCOOH = 16
Remaining m. eq. (HCOOH) = 8
Formed m. eq. (HCOOK) = 8
⇒ Acidic Buffer
pH = pKa = 4 – log2
= 3.7
pOH = 10.3
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Community Answer
If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M HCOOH [Ka= 2 × ...
To find the pOH of the resulting solution, we need to first determine the concentration of hydroxide ions (OH-) in the solution. We can do this by using the concept of stoichiometry and the balanced chemical equation for the reaction between KOH and HCOOH.
The balanced chemical equation for the reaction is:
KOH + HCOOH -> KCOOH + H2O
From the balanced equation, we can see that one mole of KOH reacts with one mole of HCOOH to produce one mole of KCOOH and one mole of water. This means that the number of moles of OH- produced is equal to the number of moles of KOH added.
Step 1: Calculate the number of moles of KOH:
Moles of KOH = volume of KOH (in L) x molarity of KOH
= 0.040 L x 0.2 mol/L
= 0.008 mol
Step 2: Calculate the volume of the resulting solution:
Volume of the resulting solution = volume of KOH + volume of HCOOH
= 0.040 L + 0.160 L
= 0.200 L
Step 3: Calculate the concentration of OH-:
Concentration of OH- = moles of OH- / volume of resulting solution
= 0.008 mol / 0.200 L
= 0.04 M
Step 4: Calculate the pOH:
pOH = -log10 [OH-]
= -log10 (0.04)
= -(-1.40)
= 1.40
Finally, to find the pH of the resulting solution, we can use the equation:
pH + pOH = 14
pH = 14 - pOH
= 14 - 1.40
= 12.60
However, the question asks for the pOH of the resulting solution. So, the correct answer is option D) 10.3.
The balanced chemical equation for the reaction is:
KOH + HCOOH -> KCOOH + H2O
From the balanced equation, we can see that one mole of KOH reacts with one mole of HCOOH to produce one mole of KCOOH and one mole of water. This means that the number of moles of OH- produced is equal to the number of moles of KOH added.
Step 1: Calculate the number of moles of KOH:
Moles of KOH = volume of KOH (in L) x molarity of KOH
= 0.040 L x 0.2 mol/L
= 0.008 mol
Step 2: Calculate the volume of the resulting solution:
Volume of the resulting solution = volume of KOH + volume of HCOOH
= 0.040 L + 0.160 L
= 0.200 L
Step 3: Calculate the concentration of OH-:
Concentration of OH- = moles of OH- / volume of resulting solution
= 0.008 mol / 0.200 L
= 0.04 M
Step 4: Calculate the pOH:
pOH = -log10 [OH-]
= -log10 (0.04)
= -(-1.40)
= 1.40
Finally, to find the pH of the resulting solution, we can use the equation:
pH + pOH = 14
pH = 14 - pOH
= 14 - 1.40
= 12.60
However, the question asks for the pOH of the resulting solution. So, the correct answer is option D) 10.3.
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If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M HCOOH [Ka= 2 × 10-4]. The pOH of the resulting solution isa)3.4b)3.7c)7d)10.3Correct answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M HCOOH [Ka= 2 × 10-4]. The pOH of the resulting solution isa)3.4b)3.7c)7d)10.3Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M HCOOH [Ka= 2 × 10-4]. The pOH of the resulting solution isa)3.4b)3.7c)7d)10.3Correct answer is option 'D'. Can you explain this answer?.
If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M HCOOH [Ka= 2 × 10-4]. The pOH of the resulting solution isa)3.4b)3.7c)7d)10.3Correct answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M HCOOH [Ka= 2 × 10-4]. The pOH of the resulting solution isa)3.4b)3.7c)7d)10.3Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M HCOOH [Ka= 2 × 10-4]. The pOH of the resulting solution isa)3.4b)3.7c)7d)10.3Correct answer is option 'D'. Can you explain this answer?.
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If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M HCOOH [Ka= 2 × 10-4]. The pOH of the resulting solution isa)3.4b)3.7c)7d)10.3Correct answer is option 'D'. Can you explain this answer?, a detailed solution for If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M HCOOH [Ka= 2 × 10-4]. The pOH of the resulting solution isa)3.4b)3.7c)7d)10.3Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M HCOOH [Ka= 2 × 10-4]. The pOH of the resulting solution isa)3.4b)3.7c)7d)10.3Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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