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In decinormal solution, CH₃COOH acid is ionised to the extent of 1.3%. If log 1.3=0.11. What is the pH of the solution?
- a)3.89
- b)2.89
- c)4.89
- d)unpredictable
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
In decinormal solution, CH₃COOH acid is ionised to the extent of 1.3%....
Solution:
Given:
Extent of ionization of CH₃COOH = 1.3%
log 1.3 = 0.11
To find:
pH of the solution
Ionization of CH₃COOH:
CH₃COOH ⇌ CH₃COO⁻ + H⁺
Extent of ionization:
Extent of ionization = (Number of ions formed / Number of molecules taken) x 100
Here, extent of ionization = 1.3%
Number of ions formed = 1 (H⁺ ion)
Number of molecules taken = 100
Therefore, 1.3 = (1/100) x 100
1.3 = 1
From the given value of log 1.3 = 0.11, we can find the concentration of H⁺ ions.
Logarithmic relationship:
pH = -log[H⁺]
To find the concentration of H⁺ ions, we need to convert the extent of ionization from percentage to decimal form.
Extent of ionization = 1.3% = 0.013
Now, we can calculate the concentration of H⁺ ions using the formula:
[H⁺] = (Extent of ionization / Volume of solution)
Since the solution is decinormal, the volume of the solution is 1 L.
[H⁺] = (0.013 / 1) = 0.013 M
Finally, we can calculate the pH using the formula:
pH = -log[H⁺]
pH = -log(0.013)
pH ≈ -(-1.89) [Using log(0.01) ≈ -2]
pH ≈ 1.89
Therefore, the pH of the solution is approximately 1.89.
Since none of the given options match with the calculated pH, the correct answer is unpredictable.
Given:
Extent of ionization of CH₃COOH = 1.3%
log 1.3 = 0.11
To find:
pH of the solution
Ionization of CH₃COOH:
CH₃COOH ⇌ CH₃COO⁻ + H⁺
Extent of ionization:
Extent of ionization = (Number of ions formed / Number of molecules taken) x 100
Here, extent of ionization = 1.3%
Number of ions formed = 1 (H⁺ ion)
Number of molecules taken = 100
Therefore, 1.3 = (1/100) x 100
1.3 = 1
From the given value of log 1.3 = 0.11, we can find the concentration of H⁺ ions.
Logarithmic relationship:
pH = -log[H⁺]
To find the concentration of H⁺ ions, we need to convert the extent of ionization from percentage to decimal form.
Extent of ionization = 1.3% = 0.013
Now, we can calculate the concentration of H⁺ ions using the formula:
[H⁺] = (Extent of ionization / Volume of solution)
Since the solution is decinormal, the volume of the solution is 1 L.
[H⁺] = (0.013 / 1) = 0.013 M
Finally, we can calculate the pH using the formula:
pH = -log[H⁺]
pH = -log(0.013)
pH ≈ -(-1.89) [Using log(0.01) ≈ -2]
pH ≈ 1.89
Therefore, the pH of the solution is approximately 1.89.
Since none of the given options match with the calculated pH, the correct answer is unpredictable.
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In decinormal solution, CH₃COOH acid is ionised to the extent of 1.3%. If log 1.3=0.11. What is the pH of the solution?a)3.89b)2.89c)4.89d)unpredictableCorrect answer is option 'B'. Can you explain this answer?
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