The equivalent conductance of an aqueous solution of 1-0283- 10-3 g equivalent acetic acid per litre is 48-15-1 cm 2 equiv -1 at 25oC- At infinite dilution value is 390-7-1 cm 2 equiv -1- Calculate the degree of ionisation and ionisation constant of acetic acid is-

The correct option is A 0.1232, 1.78× 10^ 5 The degree of ionisation α = Λλ _0 = 48.15 π^ 1cm^2equiv^ 1390.7π^ 1cm^2equiv^ 1=0.1232 The ionisation c ...

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Standard VII

Physics

Electric Current

The equivalen...

Question

The equivalent conductance of an aqueous solution of $1.0283 \times 10^{- 3}$ g equivalent acetic acid per litre is $48.15 π^{- 1}$ cm $^{2}$ equiv $^{- 1}$ at $25^{o} C$ . At infinite dilution value is $390.7 π^{- 1}$ cm $^{2}$ equiv $^{- 1}$ . Calculate the degree of ionisation and ionisation constant of acetic acid is:

0.1232, 1.78 \times 10^{- 5}

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0.223, 102 \times 10^{- 5}

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0.229, 1.78 \times 10^{- 5}

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0.531, 2.85 \times 10^{- 5}

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Solution

The correct option is A $0.1232, 1.78 \times 10^{- 5}$
The degree of ionisation $α = \frac{Λ}{λ_{0}} = \frac{48.15 π^{- 1} c m^{2} e q u i v^{- 1}}{390.7 π^{- 1} c m^{2} e q u i v^{- 1}} = 0.1232$
The ionisation constant of acetic acid :

$K_{a} = \frac{[C H_{3} C O O^{-}] [H^{+}]}{[C H_{3} C O O H]} = \frac{(C α) \times (C α)}{(C (1 - α))} = C \frac{α^{2}}{1 - α}$

$K_{a} = 1.0283 \times 10^{- 3} \times \frac{(0.1232)^{2}}{1 - 0.1232} = 1.78 \times 10^{- 5}$

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The equivalent conductance of an aqueous solution of 1.0283×10−3 g equivalent acetic acid per litre is 48.15π−1 cm2 equiv−1 at 25oC. At infinite dilution value is 390.7π−1 cm2 equiv−1. Calculate the degree of ionisation and ionisation constant of acetic acid is:

The correct option is A 0.1232,1.78×10−5The degree of ionisation α=Λλ0=48.15π−1cm2equiv−1390.7π−1cm2equiv−1=0.1232The ionisation constant of acetic acid :Ka=[CH3COO−][H+][CH3COOH]=(Cα)×(Cα)(C(1−α))=Cα21−αKa=1.0283×10−3×(0.1232)21−0.1232=1.78×10−5