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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?
Verified Answer
In decinormal solution, CH₃COOH acid is ionised to the extent of 1.3%....
The dissociation of acetic acid is:

CH3COOH →  CH3COO- + H+

c – cα  cα

Concentration = 0.1 M

Degree of ionization, α = 1.3 %  

c α = (0.1) * (1.3* 10^-2)  

[H +]=1.3 * 10^-3

pH can be determined as:

-log (H^+) = pH = - log (1.3 * 10^-3) = 2.87

Thus the pH of the solution is 2.87
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Most Upvoted Answer
In decinormal solution, CH₃COOH acid is ionised to the extent of 1.3%....
Understanding the Problem
In this scenario, we are dealing with a decinormal solution of acetic acid (CH₃COOH), which has ionized to an extent of 1.3%. We need to calculate the pH of this solution.
Step 1: Calculate the Concentration of Hydrogen Ions
- A decinormal solution means the initial concentration (C) of acetic acid is 0.1 M.
- Ionization extent (α) = 1.3% = 1.3/100 = 0.013.
- The concentration of hydrogen ions [H⁺] produced can be calculated by:
[H⁺] = C × α = 0.1 M × 0.013 = 0.0013 M.
Step 2: Calculate the pH
- pH is calculated using the formula:
pH = -log[H⁺].
- Here, we need to compute log(0.0013). Given that log(1.3) = 0.11, we can use properties of logarithms:
log(0.0013) = log(1.3) - log(1000) = 0.11 - 3 = -2.89.
- Therefore, pH = -(-2.89) = 2.89.
Conclusion
The pH of the decinormal solution of acetic acid, which is ionized to the extent of 1.3%, is 2.89. Thus, the correct answer is option 'B'.
- Final pH Value: 2.89.
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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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