The equivalent conductance of an aqueous solution of 1.0283 × 10-3 g equivalent acetic acid per litre is 48.15 π-1 cm 2 equiv -1 at 25° 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.

Question

The equivalent conductance of an aqueous solution of 1.0283 × 10-3 g equivalent acetic acid per litre is 48.15 π-1 cm 2 equiv -1 at 25° 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. (A) 0.1232,1.78 × 10-5 (B) 0.223,10.2 × 10-5 (C) 0.229,1.78 × 10-5 (D) 0.531,2.85 × 10-5

Tardigrade Admin · Accepted Answer

CH 3 COOH + H 2 O leftharpoons CH 3 COO -+ H 3 O + (Degree of ionisation) α=( Lambdac/ Lambda0)=(48.15/390.7) =0.1232 Ka( Ionisation constant )=(c α2/1-α) =(1.02 × 10-3 ×(0.1232)2/(1-0.1232)) =1.78 × 10-5

$0.1232, 1.78 \times 1 0^{- 5}$

$0.223, 10.2 \times 1 0^{- 5}$

$0.229, 1.78 \times 1 0^{- 5}$

$0.531, 2.85 \times 1 0^{- 5}$

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

0.1232,1.78×10−5

0.223,10.2×10−5

0.229,1.78×10−5

0.531,2.85×10−5

CH3​COOH+H2​O⇌CH3​COO−+H3​O+ (Degree of ionisation) α=Λ0​Λc​​=390.748.15​ =0.1232 Ka​( Ionisation constant )=1−αcα2​ =(1−0.1232)1.02×10−3×(0.1232)2​ =1.78×10−5

$0.1232, 1.78 \times 1 0^{- 5}$

$0.223, 10.2 \times 1 0^{- 5}$

$0.229, 1.78 \times 1 0^{- 5}$

$0.531, 2.85 \times 1 0^{- 5}$