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(a) The molar conductance of 0.012 mol dm–3 aqueous solution of chloroacetic acid is 100 Ω-1 cm2 mol-1. The ion conductance of chloroacetate and H+ ions are 50 Ω-1 cm2 mol-1 and 300 Ω-1 cm2 mol-1, respectively. Calculate (i) degree of dissociation and pKa of chloroacetic acid, and (ii) H+ ion concentration in the solution.
(b) Sketch the conductivity versus concentration of base curves for the titration of aqueous solutions of acetic acid (i) with NaOH, and (ii) with NH4OH.
(b) Sketch the conductivity versus concentration of base curves for the titration of aqueous solutions of acetic acid (i) with NaOH, and (ii) with NH4OH.
Correct answer is '(a) Concentration of chloroacetic acid = 0.012 mol dm–3'. Can you explain this answer?
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(a) The molar conductance of 0.012 mol dm3 aqueous solution of chloroa...
(a) Concentration of chloroacetic acid = 0.012 mol dm–3
Dissociation of CH2ClCOOH, CH2ClCOOH →H+ + CH2ClCOO-
Lm∞ = conductance of H+ + conductance of CH2ClCOO-
= 50 + 300 = 350 Ω-1 cm2 mol-1
(b) (1) Weak acid with strong base: CH3COOH with NaOH
(i) Weak acid (CH3COOH) has low conductance than its salt (CH3COONa). Due to its lower dissociation w.r.t. salt. So, addition of base increase conductance (region ab in graph)
(ii) The reson for initial decrease in conductance (region (Oa)) is because small addition of base leads to common anion i.e. CH3COO– ion formation which represses the dissociation of CH3COOH which gives highly conducting H+ ions. (iii) After the end point there is higher rate of increases in conductance because Na+ + OH– has higher conductance than Na+ + CH3COO– (region bc in graph).
(2) Weak acid with weak base: CH3COOH + NH4OH →CH3COO- + NH4+ + H2O
(i) Weak acid (CH3COOH) has low conductance than its salt (CH3COONa). Due to its lower dissociation w.r.t. salt. So, addition of base increase conductance (region ab in graph). (ii) The reason for initial decrease in conductance (region(Oa)) is because small addition of base leads to common anion i.e. CH3COO– ion formation which represses the dissociation of CH3COOH which gives highly conducting H+ ions.
(i) Weak acid (CH3COOH) has low conductance than its salt (CH3COONa). Due to its lower dissociation w.r.t. salt. So, addition of base increase conductance (region ab in graph). (ii) The reason for initial decrease in conductance (region(Oa)) is because small addition of base leads to common anion i.e. CH3COO– ion formation which represses the dissociation of CH3COOH which gives highly conducting H+ ions.
(iii) The region b´c´ of graph is due to weak nature of base. So, conductance remains almost constant with the addition of weak base after end point.
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(a) The molar conductance of 0.012 mol dm3 aqueous solution of chloroa...
Answer:
To solve this question, we will use the concept of molar conductance and the relationship between molar conductance and degree of dissociation.
(a) Calculation of degree of dissociation and pKa:
Given:
Molar conductance of chloroacetic acid solution, Λm = 100 -1 cm2 mol-1
Ion conductance of chloroacetate ions, Λc = 50 -1 cm2 mol-1
Ion conductance of H+ ions, ΛH = 300 -1 cm2 mol-1
We know that the molar conductance of an electrolyte solution is given by the sum of the molar conductance of its ions:
Λm = Λc + ΛH
Substituting the given values, we get:
100 -1 cm2 mol-1 = 50 -1 cm2 mol-1 + 300 -1 cm2 mol-1
Simplifying the equation, we find:
Λc = 50 -1 cm2 mol-1
The degree of dissociation (α) is defined as the ratio of the molar conductance of the electrolyte to the sum of the molar conductances of its ions:
α = Λc / Λm
Substituting the given values, we get:
α = 50 -1 cm2 mol-1 / 100 -1 cm2 mol-1
α = 0.5
The pKa of an acid is defined as the negative logarithm of its acid dissociation constant (Ka). Since we have the degree of dissociation, we can calculate pKa using the following equation:
pKa = -log10(α^2 / (1 - α))
Substituting the given value of α, we get:
pKa = -log10((0.5)^2 / (1 - 0.5))
pKa = -log10(0.25)
pKa ≈ 0.6
Therefore, the degree of dissociation of chloroacetic acid is 0.5 and the pKa is approximately 0.6.
(ii) Calculation of H+ ion concentration:
The H+ ion concentration in the solution can be calculated using the formula:
[H+] = α × C
Substituting the given values of α and C (concentration of chloroacetic acid), we get:
[H+] = 0.5 × 0.012 mol dm–3
[H+] = 0.006 mol dm–3
Therefore, the H+ ion concentration in the solution is 0.006 mol dm–3.
(b) Sketching the conductivity versus concentration curves:
To sketch the conductivity versus concentration curves for the titration of acetic acid with NaOH and NH4OH, we need to consider the ionization of acetic acid and the formation of the respective salts.
(i) Titration of acetic acid with NaOH:
At the beginning of the titration, the conductivity of acetic acid is low because it is a weak electrolyte and does not dissociate completely into ions. As NaOH is added, it reacts with acetic acid to form sodium acetate, which is a strong electrolyte and dissociates completely into ions. This leads to an increase in conductivity
To solve this question, we will use the concept of molar conductance and the relationship between molar conductance and degree of dissociation.
(a) Calculation of degree of dissociation and pKa:
Given:
Molar conductance of chloroacetic acid solution, Λm = 100 -1 cm2 mol-1
Ion conductance of chloroacetate ions, Λc = 50 -1 cm2 mol-1
Ion conductance of H+ ions, ΛH = 300 -1 cm2 mol-1
We know that the molar conductance of an electrolyte solution is given by the sum of the molar conductance of its ions:
Λm = Λc + ΛH
Substituting the given values, we get:
100 -1 cm2 mol-1 = 50 -1 cm2 mol-1 + 300 -1 cm2 mol-1
Simplifying the equation, we find:
Λc = 50 -1 cm2 mol-1
The degree of dissociation (α) is defined as the ratio of the molar conductance of the electrolyte to the sum of the molar conductances of its ions:
α = Λc / Λm
Substituting the given values, we get:
α = 50 -1 cm2 mol-1 / 100 -1 cm2 mol-1
α = 0.5
The pKa of an acid is defined as the negative logarithm of its acid dissociation constant (Ka). Since we have the degree of dissociation, we can calculate pKa using the following equation:
pKa = -log10(α^2 / (1 - α))
Substituting the given value of α, we get:
pKa = -log10((0.5)^2 / (1 - 0.5))
pKa = -log10(0.25)
pKa ≈ 0.6
Therefore, the degree of dissociation of chloroacetic acid is 0.5 and the pKa is approximately 0.6.
(ii) Calculation of H+ ion concentration:
The H+ ion concentration in the solution can be calculated using the formula:
[H+] = α × C
Substituting the given values of α and C (concentration of chloroacetic acid), we get:
[H+] = 0.5 × 0.012 mol dm–3
[H+] = 0.006 mol dm–3
Therefore, the H+ ion concentration in the solution is 0.006 mol dm–3.
(b) Sketching the conductivity versus concentration curves:
To sketch the conductivity versus concentration curves for the titration of acetic acid with NaOH and NH4OH, we need to consider the ionization of acetic acid and the formation of the respective salts.
(i) Titration of acetic acid with NaOH:
At the beginning of the titration, the conductivity of acetic acid is low because it is a weak electrolyte and does not dissociate completely into ions. As NaOH is added, it reacts with acetic acid to form sodium acetate, which is a strong electrolyte and dissociates completely into ions. This leads to an increase in conductivity
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(a) The molar conductance of 0.012 mol dm3 aqueous solution of chloroacetic acid is 100 -1 cm2 mol-1. The ion conductance of chloroacetate and H+ ions are 50 -1 cm2 mol-1 and 300 -1 cm2 mol-1, respectively. Calculate (i) degree of dissociation and pKa of chloroacetic acid, and (ii) H+ ion concentration in the solution.(b) Sketch the conductivity versus concentration of base curves for the titration of aqueous solutions of acetic acid (i) with NaOH, and (ii) with NH4OH.Correct answer is '(a) Concentration of chloroacetic acid = 0.012 mol dm–3'. Can you explain this answer?
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(a) The molar conductance of 0.012 mol dm3 aqueous solution of chloroacetic acid is 100 -1 cm2 mol-1. The ion conductance of chloroacetate and H+ ions are 50 -1 cm2 mol-1 and 300 -1 cm2 mol-1, respectively. Calculate (i) degree of dissociation and pKa of chloroacetic acid, and (ii) H+ ion concentration in the solution.(b) Sketch the conductivity versus concentration of base curves for the titration of aqueous solutions of acetic acid (i) with NaOH, and (ii) with NH4OH.Correct answer is '(a) Concentration of chloroacetic acid = 0.012 mol dm–3'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about (a) The molar conductance of 0.012 mol dm3 aqueous solution of chloroacetic acid is 100 -1 cm2 mol-1. The ion conductance of chloroacetate and H+ ions are 50 -1 cm2 mol-1 and 300 -1 cm2 mol-1, respectively. Calculate (i) degree of dissociation and pKa of chloroacetic acid, and (ii) H+ ion concentration in the solution.(b) Sketch the conductivity versus concentration of base curves for the titration of aqueous solutions of acetic acid (i) with NaOH, and (ii) with NH4OH.Correct answer is '(a) Concentration of chloroacetic acid = 0.012 mol dm–3'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for (a) The molar conductance of 0.012 mol dm3 aqueous solution of chloroacetic acid is 100 -1 cm2 mol-1. The ion conductance of chloroacetate and H+ ions are 50 -1 cm2 mol-1 and 300 -1 cm2 mol-1, respectively. Calculate (i) degree of dissociation and pKa of chloroacetic acid, and (ii) H+ ion concentration in the solution.(b) Sketch the conductivity versus concentration of base curves for the titration of aqueous solutions of acetic acid (i) with NaOH, and (ii) with NH4OH.Correct answer is '(a) Concentration of chloroacetic acid = 0.012 mol dm–3'. Can you explain this answer?.
(a) The molar conductance of 0.012 mol dm3 aqueous solution of chloroacetic acid is 100 -1 cm2 mol-1. The ion conductance of chloroacetate and H+ ions are 50 -1 cm2 mol-1 and 300 -1 cm2 mol-1, respectively. Calculate (i) degree of dissociation and pKa of chloroacetic acid, and (ii) H+ ion concentration in the solution.(b) Sketch the conductivity versus concentration of base curves for the titration of aqueous solutions of acetic acid (i) with NaOH, and (ii) with NH4OH.Correct answer is '(a) Concentration of chloroacetic acid = 0.012 mol dm–3'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about (a) The molar conductance of 0.012 mol dm3 aqueous solution of chloroacetic acid is 100 -1 cm2 mol-1. The ion conductance of chloroacetate and H+ ions are 50 -1 cm2 mol-1 and 300 -1 cm2 mol-1, respectively. Calculate (i) degree of dissociation and pKa of chloroacetic acid, and (ii) H+ ion concentration in the solution.(b) Sketch the conductivity versus concentration of base curves for the titration of aqueous solutions of acetic acid (i) with NaOH, and (ii) with NH4OH.Correct answer is '(a) Concentration of chloroacetic acid = 0.012 mol dm–3'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for (a) The molar conductance of 0.012 mol dm3 aqueous solution of chloroacetic acid is 100 -1 cm2 mol-1. The ion conductance of chloroacetate and H+ ions are 50 -1 cm2 mol-1 and 300 -1 cm2 mol-1, respectively. Calculate (i) degree of dissociation and pKa of chloroacetic acid, and (ii) H+ ion concentration in the solution.(b) Sketch the conductivity versus concentration of base curves for the titration of aqueous solutions of acetic acid (i) with NaOH, and (ii) with NH4OH.Correct answer is '(a) Concentration of chloroacetic acid = 0.012 mol dm–3'. Can you explain this answer?.
Solutions for (a) The molar conductance of 0.012 mol dm3 aqueous solution of chloroacetic acid is 100 -1 cm2 mol-1. The ion conductance of chloroacetate and H+ ions are 50 -1 cm2 mol-1 and 300 -1 cm2 mol-1, respectively. Calculate (i) degree of dissociation and pKa of chloroacetic acid, and (ii) H+ ion concentration in the solution.(b) Sketch the conductivity versus concentration of base curves for the titration of aqueous solutions of acetic acid (i) with NaOH, and (ii) with NH4OH.Correct answer is '(a) Concentration of chloroacetic acid = 0.012 mol dm–3'. Can you explain this answer? in English & in Hindi are available as part of our courses for IIT JAM.
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(a) The molar conductance of 0.012 mol dm3 aqueous solution of chloroacetic acid is 100 -1 cm2 mol-1. The ion conductance of chloroacetate and H+ ions are 50 -1 cm2 mol-1 and 300 -1 cm2 mol-1, respectively. Calculate (i) degree of dissociation and pKa of chloroacetic acid, and (ii) H+ ion concentration in the solution.(b) Sketch the conductivity versus concentration of base curves for the titration of aqueous solutions of acetic acid (i) with NaOH, and (ii) with NH4OH.Correct answer is '(a) Concentration of chloroacetic acid = 0.012 mol dm–3'. Can you explain this answer?, a detailed solution for (a) The molar conductance of 0.012 mol dm3 aqueous solution of chloroacetic acid is 100 -1 cm2 mol-1. The ion conductance of chloroacetate and H+ ions are 50 -1 cm2 mol-1 and 300 -1 cm2 mol-1, respectively. Calculate (i) degree of dissociation and pKa of chloroacetic acid, and (ii) H+ ion concentration in the solution.(b) Sketch the conductivity versus concentration of base curves for the titration of aqueous solutions of acetic acid (i) with NaOH, and (ii) with NH4OH.Correct answer is '(a) Concentration of chloroacetic acid = 0.012 mol dm–3'. Can you explain this answer? has been provided alongside types of (a) The molar conductance of 0.012 mol dm3 aqueous solution of chloroacetic acid is 100 -1 cm2 mol-1. The ion conductance of chloroacetate and H+ ions are 50 -1 cm2 mol-1 and 300 -1 cm2 mol-1, respectively. Calculate (i) degree of dissociation and pKa of chloroacetic acid, and (ii) H+ ion concentration in the solution.(b) Sketch the conductivity versus concentration of base curves for the titration of aqueous solutions of acetic acid (i) with NaOH, and (ii) with NH4OH.Correct answer is '(a) Concentration of chloroacetic acid = 0.012 mol dm–3'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice (a) The molar conductance of 0.012 mol dm3 aqueous solution of chloroacetic acid is 100 -1 cm2 mol-1. The ion conductance of chloroacetate and H+ ions are 50 -1 cm2 mol-1 and 300 -1 cm2 mol-1, respectively. Calculate (i) degree of dissociation and pKa of chloroacetic acid, and (ii) H+ ion concentration in the solution.(b) Sketch the conductivity versus concentration of base curves for the titration of aqueous solutions of acetic acid (i) with NaOH, and (ii) with NH4OH.Correct answer is '(a) Concentration of chloroacetic acid = 0.012 mol dm–3'. Can you explain this answer? tests, examples and also practice IIT JAM tests.
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