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The molar conductivity of a 0.01 M weak acid (HX) at 298 K, measured in a conductivity cell with cell constant of 0.4 cm–1, is 64.4 S cm2 mol–1. The limiting molar conductivities at infinite dilution of H+ and X– at 298 K are 350 and 410 S cm2 mol–1, respectively. Ignoring activity coefficients, the pKa of HX at 298 K is___________. (Up to two decimal places)
Correct answer is '4.01 to 4.21'. Can you explain this answer?
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The molar conductivity of a 0.01 M weak acid (HX) at 298 K, measured i...
$^2$ /ohm, is found to be 3.6 x 10$^{-5}$ S cm$^{-1}$ . Calculate the degree of dissociation of the acid.
We can use the Ostwald's dilution law to find the degree of dissociation of the weak acid:
$\Lambda_m$ = $\Lambda_m^{\circ}$ - K$_a$ / (2$\Lambda_m^{\circ}$) * $\sqrt{C}$
where $\Lambda_m$ is the molar conductivity of the acid, $\Lambda_m^{\circ}$ is the molar conductivity at infinite dilution, K$_a$ is the acid dissociation constant, and C is the concentration of the acid.
Rearranging the equation, we get:
K$_a$ = (2$\Lambda_m^{\circ}$ * ($\Lambda_m^{\circ}$ - $\Lambda_m$)) / ($\Lambda_m$ * $\sqrt{C}$)
At infinite dilution, the acid is completely dissociated, so the molar conductivity at infinite dilution can be approximated as the sum of the molar conductivity of the anion (X$^-$) and cation (H$^+$) formed upon dissociation:
$\Lambda_m^{\circ}$ = $\Lambda_m$(X$^-$) + $\Lambda_m$(H$^+$)
The molar conductivity of H$^+$ is known to be 349.7 S cm$^2$ mol$^{-1}$ at 298 K, while the molar conductivity of X$^-$ is assumed to be equal to that of Cl$^-$, which is 76.3 S cm$^2$ mol$^{-1}$ at 298 K.
Substituting the given values, we get:
K$_a$ = (2*76.3*(76.3-3.6*10$^{-5}$)) / (3.6*10$^{-5}$ * $\sqrt{0.01}$) = 1.65 x 10$^{-5}$
The degree of dissociation ($\alpha$) can be calculated as:
$\alpha$ = $\frac{[\text{H}^+]}{[\text{HX}]}$ = $\sqrt{\frac{K_a}{C}}$ * (1-$\frac{K_a}{C}$)
Substituting the given values, we get:
$\alpha$ = $\sqrt{\frac{1.65\times 10^{-5}}{0.01}}$ * (1-$\frac{1.65\times 10^{-5}}{0.01}$) = 0.017
Therefore, the degree of dissociation of the weak acid is 0.017 or 1.7%.
We can use the Ostwald's dilution law to find the degree of dissociation of the weak acid:
$\Lambda_m$ = $\Lambda_m^{\circ}$ - K$_a$ / (2$\Lambda_m^{\circ}$) * $\sqrt{C}$
where $\Lambda_m$ is the molar conductivity of the acid, $\Lambda_m^{\circ}$ is the molar conductivity at infinite dilution, K$_a$ is the acid dissociation constant, and C is the concentration of the acid.
Rearranging the equation, we get:
K$_a$ = (2$\Lambda_m^{\circ}$ * ($\Lambda_m^{\circ}$ - $\Lambda_m$)) / ($\Lambda_m$ * $\sqrt{C}$)
At infinite dilution, the acid is completely dissociated, so the molar conductivity at infinite dilution can be approximated as the sum of the molar conductivity of the anion (X$^-$) and cation (H$^+$) formed upon dissociation:
$\Lambda_m^{\circ}$ = $\Lambda_m$(X$^-$) + $\Lambda_m$(H$^+$)
The molar conductivity of H$^+$ is known to be 349.7 S cm$^2$ mol$^{-1}$ at 298 K, while the molar conductivity of X$^-$ is assumed to be equal to that of Cl$^-$, which is 76.3 S cm$^2$ mol$^{-1}$ at 298 K.
Substituting the given values, we get:
K$_a$ = (2*76.3*(76.3-3.6*10$^{-5}$)) / (3.6*10$^{-5}$ * $\sqrt{0.01}$) = 1.65 x 10$^{-5}$
The degree of dissociation ($\alpha$) can be calculated as:
$\alpha$ = $\frac{[\text{H}^+]}{[\text{HX}]}$ = $\sqrt{\frac{K_a}{C}}$ * (1-$\frac{K_a}{C}$)
Substituting the given values, we get:
$\alpha$ = $\sqrt{\frac{1.65\times 10^{-5}}{0.01}}$ * (1-$\frac{1.65\times 10^{-5}}{0.01}$) = 0.017
Therefore, the degree of dissociation of the weak acid is 0.017 or 1.7%.
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The molar conductivity of a 0.01 M weak acid (HX) at 298 K, measured in a conductivity cell with cell constant of 0.4 cm–1, is 64.4 S cm2 mol–1. The limiting molar conductivities at infinite dilution of H+ and X– at 298 K are 350 and 410 S cm2 mol–1, respectively. Ignoring activity coefficients, the pKa of HX at 298 K is___________. (Up to two decimal places)Correct answer is '4.01 to 4.21'. Can you explain this answer?
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The molar conductivity of a 0.01 M weak acid (HX) at 298 K, measured in a conductivity cell with cell constant of 0.4 cm–1, is 64.4 S cm2 mol–1. The limiting molar conductivities at infinite dilution of H+ and X– at 298 K are 350 and 410 S cm2 mol–1, respectively. Ignoring activity coefficients, the pKa of HX at 298 K is___________. (Up to two decimal places)Correct answer is '4.01 to 4.21'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The molar conductivity of a 0.01 M weak acid (HX) at 298 K, measured in a conductivity cell with cell constant of 0.4 cm–1, is 64.4 S cm2 mol–1. The limiting molar conductivities at infinite dilution of H+ and X– at 298 K are 350 and 410 S cm2 mol–1, respectively. Ignoring activity coefficients, the pKa of HX at 298 K is___________. (Up to two decimal places)Correct answer is '4.01 to 4.21'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The molar conductivity of a 0.01 M weak acid (HX) at 298 K, measured in a conductivity cell with cell constant of 0.4 cm–1, is 64.4 S cm2 mol–1. The limiting molar conductivities at infinite dilution of H+ and X– at 298 K are 350 and 410 S cm2 mol–1, respectively. Ignoring activity coefficients, the pKa of HX at 298 K is___________. (Up to two decimal places)Correct answer is '4.01 to 4.21'. Can you explain this answer?.
The molar conductivity of a 0.01 M weak acid (HX) at 298 K, measured in a conductivity cell with cell constant of 0.4 cm–1, is 64.4 S cm2 mol–1. The limiting molar conductivities at infinite dilution of H+ and X– at 298 K are 350 and 410 S cm2 mol–1, respectively. Ignoring activity coefficients, the pKa of HX at 298 K is___________. (Up to two decimal places)Correct answer is '4.01 to 4.21'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The molar conductivity of a 0.01 M weak acid (HX) at 298 K, measured in a conductivity cell with cell constant of 0.4 cm–1, is 64.4 S cm2 mol–1. The limiting molar conductivities at infinite dilution of H+ and X– at 298 K are 350 and 410 S cm2 mol–1, respectively. Ignoring activity coefficients, the pKa of HX at 298 K is___________. (Up to two decimal places)Correct answer is '4.01 to 4.21'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The molar conductivity of a 0.01 M weak acid (HX) at 298 K, measured in a conductivity cell with cell constant of 0.4 cm–1, is 64.4 S cm2 mol–1. The limiting molar conductivities at infinite dilution of H+ and X– at 298 K are 350 and 410 S cm2 mol–1, respectively. Ignoring activity coefficients, the pKa of HX at 298 K is___________. (Up to two decimal places)Correct answer is '4.01 to 4.21'. Can you explain this answer?.
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ample number of questions to practice The molar conductivity of a 0.01 M weak acid (HX) at 298 K, measured in a conductivity cell with cell constant of 0.4 cm–1, is 64.4 S cm2 mol–1. The limiting molar conductivities at infinite dilution of H+ and X– at 298 K are 350 and 410 S cm2 mol–1, respectively. Ignoring activity coefficients, the pKa of HX at 298 K is___________. (Up to two decimal places)Correct answer is '4.01 to 4.21'. Can you explain this answer? tests, examples and also practice GATE tests.
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