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Fixed volume of 0.1 M benzoic acid solution is added into 0.2 M sodium benzoate solution and formed a 300 ml.resultant acidic buffer solution. If pH of this buffer solution is 4.5 then find added volume of benzoic acid - Given: pK, benzoic acid = 4.2 A 100 ml B 150 ml D 200 ml None of these?
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Solution:

Given,
Volume of benzoic acid solution = 0.1 M
Volume of sodium benzoate solution = 0.2 M
Volume of resultant buffer solution = 300 ml
pH of buffer solution = 4.5
pK of benzoic acid = 4.2

Step 1: Calculate the moles of benzoic acid and sodium benzoate present in the solution

Moles of benzoic acid = 0.1 x 0.1 = 0.01 moles
Moles of sodium benzoate = 0.2 x 0.3 = 0.06 moles

Step 2: Calculate the ratio of concentrations of benzoic acid and sodium benzoate

Ratio = (moles of benzoic acid) / (moles of sodium benzoate)
= 0.01 / 0.06
= 1 / 6

Step 3: Calculate the pH of the buffer solution using Henderson-Hasselbalch equation

pH = pK + log ( [A-] / [HA] )

where,
pH = 4.5
pK = 4.2
[A-] = concentration of sodium benzoate = 0.2 M x 0.3 L / 0.3 L = 0.2 M
[HA] = concentration of benzoic acid = 0.1 M

4.5 = 4.2 + log ( 0.2 / [HA] )
log (0.2 / [HA]) = 0.3
0.2 / [HA] = 1.995
[HA] = 0.2 / 1.995 = 0.1 M x 0.3 L / 0.1 L = 0.3 M

Step 4: Calculate the total moles of benzoic acid in the buffer solution

Total moles of benzoic acid = 0.3 M x 0.3 L = 0.09 moles

Step 5: Calculate the moles of benzoic acid added

Moles of benzoic acid added = Total moles of benzoic acid - Initial moles of benzoic acid
= 0.09 - 0.01
= 0.08 moles

Step 6: Calculate the volume of benzoic acid added

Volume of benzoic acid added = Moles of benzoic acid added / Concentration of benzoic acid added
= 0.08 / 0.1
= 0.8 L or 800 ml

Therefore, the added volume of benzoic acid is 800 ml.

Answer: D 200 ml
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Fixed volume of 0.1 M benzoic acid solution is added into 0.2 M sodium benzoate solution and formed a 300 ml.resultant acidic buffer solution. If pH of this buffer solution is 4.5 then find added volume of benzoic acid - Given: pK, benzoic acid = 4.2 A 100 ml B 150 ml D 200 ml None of these?
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Fixed volume of 0.1 M benzoic acid solution is added into 0.2 M sodium benzoate solution and formed a 300 ml.resultant acidic buffer solution. If pH of this buffer solution is 4.5 then find added volume of benzoic acid - Given: pK, benzoic acid = 4.2 A 100 ml B 150 ml D 200 ml None of these? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Fixed volume of 0.1 M benzoic acid solution is added into 0.2 M sodium benzoate solution and formed a 300 ml.resultant acidic buffer solution. If pH of this buffer solution is 4.5 then find added volume of benzoic acid - Given: pK, benzoic acid = 4.2 A 100 ml B 150 ml D 200 ml None of these? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Fixed volume of 0.1 M benzoic acid solution is added into 0.2 M sodium benzoate solution and formed a 300 ml.resultant acidic buffer solution. If pH of this buffer solution is 4.5 then find added volume of benzoic acid - Given: pK, benzoic acid = 4.2 A 100 ml B 150 ml D 200 ml None of these?.
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