हिंदी

Calculate the degree of dissociation (α) of acetic acid if its molar conductivity (Λm) is 39.05 S cm2 mol−1. (Given 𝜆⁢∘(H+) = 349.6 S cm2 mol−1 and 𝜆⁢∘(CH3⁢COO−) = 40.95 S cm2 mol−1)

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प्रश्न

Calculate the degree of dissociation (α) of acetic acid if its molar conductivity (Λm) is 39.05 S cm2 mol−1.

(Given \[\ce{\lambda^{\circ}_{(H^+)}}\] = 349.6 S cm2 mol−1 and \[\ce{\lambda^{\circ}_{(CH_3COO^-)}}\] = 40.95 S cm2 mol−1)

संख्यात्मक
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उत्तर

The degree of dissociation of a weak electrolyte is given by the following relation:

α = `Lambda_m/Lambda^circ`    ... (1)

Where,

α = Degree of dissociation of the electrolyte

Λm = Molar conductivity

Λ° = Molar conductivity at infinite dilution

We are given that the ionic molar conductivity at infinite dilution of acetate and hydrogen ions are 349.8 and 40.9 S cm2 mol−1, respectively.

Hence, the limiting molar conductivity of acetic acid would be written as:

\[\ce{\Lambda^{\circ}_{\text{acetic acid}} = \lambda^{\circ}_{(H^+)} + \lambda^{\circ}_{(CH_3COO^-)}}\]

 = 40.9 + 349.8

= 390.7 cm2 mol1

Now, since we are given that the molar conductivity of acetic acid is:

Λm = 39.05 S cm2 mol1

Therefore, substituting the values in equation (1), we get:

Degree of dissociation of acetic acid (α) = `Lambda_m/Lambda^circ`

= \[\ce{\frac{39.05}{390.7}}\]

= 0.0999

≅ 0.1

Hence, the degree of dissociation is 0.1.

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अध्याय 2: Electrochemistry - QUESTIONS FROM ISC EXAMINATION PAPERS [पृष्ठ २१६]

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नूतन Chemistry [English] Class 12 ISC
अध्याय 2 Electrochemistry
QUESTIONS FROM ISC EXAMINATION PAPERS | Q 40. | पृष्ठ २१६

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\[\begin{array}{cc}
\end{array}\]\[\begin{bmatrix}
\ce{\Lambda^{\circ}_{H^+} = 350 S cm^2 mol^{-1}}\\
\ce{\Lambda^{\circ}_{CH_3COO^-} = 50 S cm^2 mol^{-1}}
\end{bmatrix}\]


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