The equivalent conductivity of N-10 solution of acetic acid at 25-C is 14-3 ohm-1 cm2 eq-1- Calculate the degree of dissociation of CH3COOH if - CH3COOH is 390-71-

∧_∞ CH_3COOH = 3990.71 ohm^ 1cm^2 eq^ 1 ∧_CH_3COOH = 14.2 ohm^ 1cm^2 eq^ 1 Degree of dissociation,α = ∧_v∧_∞ = 14.3390.71 = 0.0366, i.e.,3.66 ...

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The equivalen...

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The equivalent conductivity of $N / 10$ solution of acetic acid at $25^{\circ} C$ is $14.3 o h m^{- 1} c m^{2} e q^{- 1}$ . Calculate the degree of dissociation of $C H_{3} C O O H$ if $\land_{\infty C H_{3} C O O H}$ is $390.71$ .

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N / 10

25^{o}

14.3

o h m^{- 1}

c m^{2}

e q u i v^{- 1}

(Λ_{\infty C H_{3} C O O H} = 390.71

o h m^{- 1}

c m^{2}

e q u i v^{- 1}

H C l, N a C l

C H_{3} C O O N a

S c m^{2} e q^{- 1}

S c m^{2} e q^{- 1}

C H_{3} C O O H

0.0128

25^{\circ} C

λ_{e q} = 1.4 m h o c m^{2} e q^{- 1}

λ^{\infty} = 391 m h o c m^{2} e q^{- 1}

K_{a}

^{2}

^{- 1}

{CH}_{3} COOH

0.1 N C H_{3} C O O H N

25^{\circ} C

80

400 o h m^{- 1}

C H_{3} C O O H

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