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If pKb for fluoride ion at 25°C is 10.83, the ionisation constant of hydrofluoric acid in water at this temperature is :
- a)1.74 × 10-5
- b)3.52 × 10-3
- c)6.75 × 10-4
- d)5.38 × 10-2
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
If pKbfor fluoride ion at 25°C is 10.83, the ionisation constant ...
H+ 
H+ + F–
pKw = pKa + pKb
[For conjugate Acid-Base]
⇒ pKa = 14 – 10.87 = 3.17
Ka = 6.76 × 10–4
H+ + F–pKw = pKa + pKb
[For conjugate Acid-Base]
⇒ pKa = 14 – 10.87 = 3.17
Ka = 6.76 × 10–4
Most Upvoted Answer
If pKbfor fluoride ion at 25°C is 10.83, the ionisation constant ...
To find the ionization constant of hydrofluoric acid (HF) in water at 25°C, we can use the relationship between pKb and pKa.
The ionization constant of hydrofluoric acid can be represented as follows:
HF + H2O ⇌ H3O+ + F-
The pKa of HF can be calculated using the pKb of the fluoride ion:
pKa + pKb = 14
Given that pKb for the fluoride ion is 10.83, we can substitute this value into the equation to solve for pKa:
pKa + 10.83 = 14
Subtracting 10.83 from both sides:
pKa = 14 - 10.83
pKa = 3.17
Now we have the pKa value for HF. To find the ionization constant, we need to convert the pKa value to Ka using the formula:
pKa = -log10(Ka)
10^(-pKa) = Ka
Ka = 10^(-3.17)
Using a calculator, we find that Ka is approximately 6.75 × 10^(-4).
Therefore, the ionization constant of hydrofluoric acid in water at 25°C is approximately 6.75 × 10^(-4), which corresponds to option C.
The ionization constant of hydrofluoric acid can be represented as follows:
HF + H2O ⇌ H3O+ + F-
The pKa of HF can be calculated using the pKb of the fluoride ion:
pKa + pKb = 14
Given that pKb for the fluoride ion is 10.83, we can substitute this value into the equation to solve for pKa:
pKa + 10.83 = 14
Subtracting 10.83 from both sides:
pKa = 14 - 10.83
pKa = 3.17
Now we have the pKa value for HF. To find the ionization constant, we need to convert the pKa value to Ka using the formula:
pKa = -log10(Ka)
10^(-pKa) = Ka
Ka = 10^(-3.17)
Using a calculator, we find that Ka is approximately 6.75 × 10^(-4).
Therefore, the ionization constant of hydrofluoric acid in water at 25°C is approximately 6.75 × 10^(-4), which corresponds to option C.
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If pKbfor fluoride ion at 25°C is 10.83, the ionisation constant of hydrofluoric acid in water at this temperature is :a)1.74 × 10-5b)3.52 × 10-3c)6.75 × 10-4d)5.38 × 10-2Correct answer is option 'C'. 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 pKbfor fluoride ion at 25°C is 10.83, the ionisation constant of hydrofluoric acid in water at this temperature is :a)1.74 × 10-5b)3.52 × 10-3c)6.75 × 10-4d)5.38 × 10-2Correct answer is option 'C'. 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 pKbfor fluoride ion at 25°C is 10.83, the ionisation constant of hydrofluoric acid in water at this temperature is :a)1.74 × 10-5b)3.52 × 10-3c)6.75 × 10-4d)5.38 × 10-2Correct answer is option 'C'. Can you explain this answer?.
If pKbfor fluoride ion at 25°C is 10.83, the ionisation constant of hydrofluoric acid in water at this temperature is :a)1.74 × 10-5b)3.52 × 10-3c)6.75 × 10-4d)5.38 × 10-2Correct answer is option 'C'. 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 pKbfor fluoride ion at 25°C is 10.83, the ionisation constant of hydrofluoric acid in water at this temperature is :a)1.74 × 10-5b)3.52 × 10-3c)6.75 × 10-4d)5.38 × 10-2Correct answer is option 'C'. 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 pKbfor fluoride ion at 25°C is 10.83, the ionisation constant of hydrofluoric acid in water at this temperature is :a)1.74 × 10-5b)3.52 × 10-3c)6.75 × 10-4d)5.38 × 10-2Correct answer is option 'C'. Can you explain this answer?.
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