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प्रश्न
The molar conductivity of 0.025 mol L−1 methanoic acid is 46.1 S cm2 mol−1. Calculate its degree of dissociation and dissociation constant. Given \[\ce{λ^0_{(H^+)}}\] = 349.6 S cm2 mol−1 and \[\ce{λ^0_{(HCOO^-)}}\] = 54.6 S cm2 mol−1.
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उत्तर
Given: \[\ce{\Lambda_{m(HCOOH)}}\] = 46.1 S cm2 mol−1
\[\ce{λ^0_{(H^+)}}\] = 349.6 S cm2 mol−1
\[\ce{λ^0_{(HCOO^-)}}\] = 54.6 S cm2 mol−1
\[\ce{\Lambda^0_{m(HCOOH)} = \lambda^0_{(H^+)} + \lambda^0_{(HCOO^-)}}\]
= 349.6 + 54.6
= 404.2 S cm2 mol−1
Degree of dissociation (α) = \[\ce{\frac{\Lambda_m}{\Lambda{^0_m}}}\]
= `46.1/404.2`
= 0.114
⇒ α = 11.4%
\[\ce{HCOOH <=> HCOO- + H+}\]
| Initial concentration | c mol L−1 | 0 | 0 |
| Concentration at equilibrium | c(1 − α) | cα | cα |
∴ Kα = `(c alpha * c alpha)/(c (1 - alpha))`
= `(c alpha^2)/(1 - alpha)`
= `(0.025 xx (0.114)^2)/(1 - 0.114)`
= 3.67 × 10−4
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