(D) First,calculate the limiting molar conductivity of methanoic acid $(HCOOH)$: $\Lambda_{m}^{\circ}(HCOOH) = \lambda_{H^{+}}^0 + \lambda_{HCOO^{-}}^

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(D) First,calculate the limiting molar conductivity of methanoic acid $(HCOOH)$: $\Lambda_{m}^{\circ}(HCOOH) = \lambda_{H^{+}}^0 + \lambda_{HCOO^{-}}^0 = 349.6 + 54.6 = 404.2 \ S \ cm^2 \ mol^{-1}$ Next,calculate the degree of dissociation $(\alpha)$: $\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^0} = \frac{40.42}{404.2} = 0.1$ Finally,calculate the dissociation constant $(K_{a})$: $K_{a} = \frac{C\alpha^2}{1-\alpha} = \frac{0.027 \times (0.1)^2}{1-0.1} = \frac{0.027 \times 0.01}{0.9} = \frac{0.00027}{0.9} = 3.0 \times 10^{-4}$

Class 12 Chemistry Electrochemistry Mix Examples-Electrochemistry

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The molar conductivity of $0.027 \ M$ methanoic acid is $40.42 \ S \ cm^2 \ mol^{-1}$. The value of dissociation constant of this acid is
(Given $\lambda_{H^{+}}^{\circ} = 349.6 \ S \ cm^2 \ mol^{-1}$ and $\lambda_{HCOO^{-}}^{\circ} = 54.6 \ S \ cm^2 \ mol^{-1}$)

AP EAMCET 2023 All AP EAMCET

A
$1.5 \times 10^{-5}$
B
$6.0 \times 10^{-5}$
C
$4.5 \times 10^{-4}$
D
$3.0 \times 10^{-4}$

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$A$ hydrogen electrode is immersed in a solution with $pH = 0$ $(HCl)$. By how much will the potential (reduction) change if an equivalent amount of $NaOH$ is added to the solution? (Take $pH_2 = 1 \ atm$,$T = 298 \ K$).

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The equilibrium constant of a $2$ electron redox reaction at $298 \, K$ is $3.8 \times 10^{-3}$. The cell potential $E^{\circ}$ (in $V$) and the free energy change $\Delta G^{\circ}$ (in $kJ \, mol^{-1}$) for this equilibrium,respectively are

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The molar conductivity of $0.025 \ mol \ L^{-1}$ methanoic acid is $46.1 \ S \ cm^2 \ mol^{-1}$. Calculate its degree of dissociation and dissociation constant. Given $\lambda^o(H^{+}) = 349.6 \ S \ cm^2 \ mol^{-1}$ and $\lambda^o(HCOO^{-}) = 54.6 \ S \ cm^2 \ mol^{-1}$.

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Which of the following statements is not correct?

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If the molar conductivity $(\Lambda_{m})$ of a $0.050 \ mol \ L^{-1}$ solution of a monobasic weak acid is $90 \ S \ cm^{2} \ mol^{-1}$,its extent (degree) of dissociation will be. [Assume $\Lambda_{+}^{\circ} = 349.6 \ S \ cm^{2} \ mol^{-1}$ and $\Lambda_{-}^{\circ} = 50.4 \ S \ cm^{2} \ mol^{-1}$.]

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The molar conductivity of $0.027 \ M$ methanoic acid is $40.42 \ S \ cm^2 \ mol^{-1}$. The value of dissociation constant of this acid is (Given $\lambda_{H^{+}}^{\circ} = 349.6 \ S \ cm^2 \ mol^{-1}$ and $\lambda_{HCOO^{-}}^{\circ} = 54.6 \ S \ cm^2 \ mol^{-1}$)

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$A$ hydrogen electrode is immersed in a solution with $pH = 0$ $(HCl)$. By how much will the potential (reduction) change if an equivalent amount of $NaOH$ is added to the solution? (Take $pH_2 = 1 \ atm$,$T = 298 \ K$).

The equilibrium constant of a $2$ electron redox reaction at $298 \, K$ is $3.8 \times 10^{-3}$. The cell potential $E^{\circ}$ (in $V$) and the free energy change $\Delta G^{\circ}$ (in $kJ \, mol^{-1}$) for this equilibrium,respectively are

The molar conductivity of $0.025 \ mol \ L^{-1}$ methanoic acid is $46.1 \ S \ cm^2 \ mol^{-1}$. Calculate its degree of dissociation and dissociation constant. Given $\lambda^o(H^{+}) = 349.6 \ S \ cm^2 \ mol^{-1}$ and $\lambda^o(HCOO^{-}) = 54.6 \ S \ cm^2 \ mol^{-1}$.

Which of the following statements is not correct?

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The molar conductivity of $0.027 \ M$ methanoic acid is $40.42 \ S \ cm^2 \ mol^{-1}$. The value of dissociation constant of this acid is
(Given $\lambda_{H^{+}}^{\circ} = 349.6 \ S \ cm^2 \ mol^{-1}$ and $\lambda_{HCOO^{-}}^{\circ} = 54.6 \ S \ cm^2 \ mol^{-1}$)