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The equivalent conductance of M/32 solution of a weak mono basic acid is 8 mho cm^2 and at infinite dilution is 400 mhos cm^2. The dissociation constant of this acid is A) 1.25 x 10^-6 B) 6.25 x 10^-4 C) 1.25 x 10^-4 D) 1.25 x 10^-5?
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The equivalent conductance of M/32 solution of a weak mono basic acid ...
Explanation:
The equivalent conductance of M/32 solution of weak mono basic acid is 8 mho cm^2 and at infinite dilution is 400 mho cm^2. The dissociation constant of this acid is to be determined.
Step 1: Calculate the degree of dissociation (α)
The equivalent conductance at infinite dilution (Λo) is given by the sum of the equivalent conductance of the cation (Λ+o) and anion (Λ-o) at infinite dilution.
Λo = Λ+o + Λ-o
The equivalent conductance of the weak acid at M/32 concentration (Λ) is given as:
Λ = Λo - KcCα^2 (where Kc is the equilibrium constant for the dissociation of the acid, C is the concentration of the acid, and α is the degree of dissociation)
Substituting the given values, we get:
8 = 400 - Kc x (1/32) x α^2
α^2 = (400-8)/(Kc x 1/32)
α^2 = 12.5/Kc
Step 2: Calculate the dissociation constant (Kc)
At infinite dilution, the degree of dissociation approaches unity (α → 1). Substituting this in the equation obtained in Step 1, we get:
Λo = Λ+o + Λ-o
400 = 2Λ+o
Λ+o = 200 mho cm^2
The equivalent conductance at infinite dilution is related to the dissociation constant as:
Λo = λ+o + λ-o = (Kc/λ+o + Kc/λ-o)^-1
Substituting the given values, we get:
400 = (Kc/200 + Kc/200)^-1
Kc = 1.25 x 10^-5 mol/L
Therefore, the dissociation constant of the weak mono basic acid is 1.25 x 10^-5 mol/L.
The equivalent conductance of M/32 solution of weak mono basic acid is 8 mho cm^2 and at infinite dilution is 400 mho cm^2. The dissociation constant of this acid is to be determined.
Step 1: Calculate the degree of dissociation (α)
The equivalent conductance at infinite dilution (Λo) is given by the sum of the equivalent conductance of the cation (Λ+o) and anion (Λ-o) at infinite dilution.
Λo = Λ+o + Λ-o
The equivalent conductance of the weak acid at M/32 concentration (Λ) is given as:
Λ = Λo - KcCα^2 (where Kc is the equilibrium constant for the dissociation of the acid, C is the concentration of the acid, and α is the degree of dissociation)
Substituting the given values, we get:
8 = 400 - Kc x (1/32) x α^2
α^2 = (400-8)/(Kc x 1/32)
α^2 = 12.5/Kc
Step 2: Calculate the dissociation constant (Kc)
At infinite dilution, the degree of dissociation approaches unity (α → 1). Substituting this in the equation obtained in Step 1, we get:
Λo = Λ+o + Λ-o
400 = 2Λ+o
Λ+o = 200 mho cm^2
The equivalent conductance at infinite dilution is related to the dissociation constant as:
Λo = λ+o + λ-o = (Kc/λ+o + Kc/λ-o)^-1
Substituting the given values, we get:
400 = (Kc/200 + Kc/200)^-1
Kc = 1.25 x 10^-5 mol/L
Therefore, the dissociation constant of the weak mono basic acid is 1.25 x 10^-5 mol/L.
Community Answer
The equivalent conductance of M/32 solution of a weak mono basic acid ...
D)1.25×10^-5
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The equivalent conductance of M/32 solution of a weak mono basic acid is 8 mho cm^2 and at infinite dilution is 400 mhos cm^2. The dissociation constant of this acid is A) 1.25 x 10^-6 B) 6.25 x 10^-4 C) 1.25 x 10^-4 D) 1.25 x 10^-5?
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The equivalent conductance of M/32 solution of a weak mono basic acid is 8 mho cm^2 and at infinite dilution is 400 mhos cm^2. The dissociation constant of this acid is A) 1.25 x 10^-6 B) 6.25 x 10^-4 C) 1.25 x 10^-4 D) 1.25 x 10^-5? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The equivalent conductance of M/32 solution of a weak mono basic acid is 8 mho cm^2 and at infinite dilution is 400 mhos cm^2. The dissociation constant of this acid is A) 1.25 x 10^-6 B) 6.25 x 10^-4 C) 1.25 x 10^-4 D) 1.25 x 10^-5? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equivalent conductance of M/32 solution of a weak mono basic acid is 8 mho cm^2 and at infinite dilution is 400 mhos cm^2. The dissociation constant of this acid is A) 1.25 x 10^-6 B) 6.25 x 10^-4 C) 1.25 x 10^-4 D) 1.25 x 10^-5?.
The equivalent conductance of M/32 solution of a weak mono basic acid is 8 mho cm^2 and at infinite dilution is 400 mhos cm^2. The dissociation constant of this acid is A) 1.25 x 10^-6 B) 6.25 x 10^-4 C) 1.25 x 10^-4 D) 1.25 x 10^-5? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The equivalent conductance of M/32 solution of a weak mono basic acid is 8 mho cm^2 and at infinite dilution is 400 mhos cm^2. The dissociation constant of this acid is A) 1.25 x 10^-6 B) 6.25 x 10^-4 C) 1.25 x 10^-4 D) 1.25 x 10^-5? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equivalent conductance of M/32 solution of a weak mono basic acid is 8 mho cm^2 and at infinite dilution is 400 mhos cm^2. The dissociation constant of this acid is A) 1.25 x 10^-6 B) 6.25 x 10^-4 C) 1.25 x 10^-4 D) 1.25 x 10^-5?.
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