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The molar conductivity of 0.009 M aqueous solution of a weak acid (HA) is 0.005 Sm2mol-1 and the limiting molar conductivity of HA is 0.005 Sm2mol-1 at 298 K. Assuming activity coefficients to be unity, the acid dissociation constant  (Ka) of HA at this temperature is
  • a)
    1×10−4
  • b)
    0.1
  • c)
    9×10−4
  • d)
    1.1×10−5
Correct answer is option 'A'. Can you explain this answer?
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The molar conductivity of 0.009 M aqueous solution of a weak acid (HA)...
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The molar conductivity of 0.009 M aqueous solution of a weak acid (HA)...
The molar conductivity of a weak acid solution is given by the equation:

Λm = Λm° - (A / √C)

where Λm is the molar conductivity of the solution, Λm° is the limiting molar conductivity of the acid, A is a constant, and C is the concentration of the acid.

Given that Λm = 0.005 Sm^2mol^-1, Λm° = 0.05 Sm^2mol^-1, and C = 0.009 M, we can rearrange the equation to solve for A:

A = (Λm° - Λm) * √C
= (0.05 - 0.005) * √0.009
= 0.045 * 0.095
= 0.004275

The acid dissociation constant (Ka) is given by the equation:

Ka = (A^2) / (Λm° - Λm)

Substituting the values we have, we get:

Ka = (0.004275^2) / (0.05 - 0.005)
= 0.00001825625 / 0.045
= 0.000405695

Therefore, the acid dissociation constant (Ka) of HA at this temperature is approximately 0.000405695.
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The molar conductivity of 0.009 M aqueous solution of a weak acid (HA) is0.005 Sm2mol-1and the limiting molar conductivity of HA is0.005 Sm2mol-1at 298 K. Assuming activity coefficients to be unity, the acid dissociation constant (Ka)of HA at this temperature isa)1×10−4b)0.1c)9×10−4d)1.1×10−5Correct answer is option 'A'. Can you explain this answer?
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The molar conductivity of 0.009 M aqueous solution of a weak acid (HA) is0.005 Sm2mol-1and the limiting molar conductivity of HA is0.005 Sm2mol-1at 298 K. Assuming activity coefficients to be unity, the acid dissociation constant (Ka)of HA at this temperature isa)1×10−4b)0.1c)9×10−4d)1.1×10−5Correct answer is option 'A'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about The molar conductivity of 0.009 M aqueous solution of a weak acid (HA) is0.005 Sm2mol-1and the limiting molar conductivity of HA is0.005 Sm2mol-1at 298 K. Assuming activity coefficients to be unity, the acid dissociation constant (Ka)of HA at this temperature isa)1×10−4b)0.1c)9×10−4d)1.1×10−5Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The molar conductivity of 0.009 M aqueous solution of a weak acid (HA) is0.005 Sm2mol-1and the limiting molar conductivity of HA is0.005 Sm2mol-1at 298 K. Assuming activity coefficients to be unity, the acid dissociation constant (Ka)of HA at this temperature isa)1×10−4b)0.1c)9×10−4d)1.1×10−5Correct answer is option 'A'. Can you explain this answer?.
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