The ionization constant of a weak monobasic acid is 25× 10- 6 while the equivalent conductance of its 0.01 M solution is 19.6 Scm2eq- 1 . The equivalent conductance of the electrolyte at infinite dilution (.in Scm2eq- 1.) will be

Question

The ionization constant of a weak monobasic acid is 25× 10- 6 while the equivalent conductance of its 0.01 M solution is 19.6 Scm2eq- 1 . The equivalent conductance of the electrolyte at infinite dilution (.in Scm2eq- 1.) will be (A) 250 (B) 196 (C) 392 (D) 384

Tardigrade Admin · Accepted Answer

textK texta = textCÎ±2 25 × 10-6=(0.01) α2 α =5× 10- 2 α =( Lambdae q/ Lambdae qo) 5× 10- 2=(19.6/ Lambdae qo) Lambdae qo=(19.6/5 × 1 0- 2)= 392

Q. The ionization constant of a weak monobasic acid is $25 \times 1 0^{- 6}$ while the equivalent conductance of its $0.01 M$ solution is $19.6 S c m^{2} e q^{- 1}$ . The equivalent conductance of the electrolyte at infinite dilution $(in S c m^{2} e q^{- 1})$ will be

250

196

392

384

$K_{a} = Cα^{2}$

$25 \times 1 0^{- 6} = (0.01) α^{2}$

$α = 5 \times 1 0^{- 2}$

$α = \frac{Λ _{e q}}{Λ _{e q}^{o}}$

$5 \times 1 0^{- 2} = \frac{19.6}{Λ _{e q}^{o}}$

$Λ_{e q}^{o} = \frac{19.6}{5 \times 1 0 ^{- 2}} = 392$

Q. The ionization constant of a weak monobasic acid is 25×10−6 while the equivalent conductance of its 0.01M solution is 19.6Scm2eq−1 . The equivalent conductance of the electrolyte at infinite dilution (inScm2eq−1) will be

250

196

392

384

Ka​=Cα2 25×10−6=(0.01)α2 α=5×10−2 α=Λeqo​Λeq​​ 5×10−2=Λeqo​19.6​ Λeqo​=5×10−219.6​=392

Q. The ionization constant of a weak monobasic acid is $25 \times 1 0^{- 6}$ while the equivalent conductance of its $0.01 M$ solution is $19.6 S c m^{2} e q^{- 1}$ . The equivalent conductance of the electrolyte at infinite dilution $(in S c m^{2} e q^{- 1})$ will be

$K_{a} = Cα^{2}$

$25 \times 1 0^{- 6} = (0.01) α^{2}$

$α = 5 \times 1 0^{- 2}$

$α = \frac{Λ _{e q}}{Λ _{e q}^{o}}$

$5 \times 1 0^{- 2} = \frac{19.6}{Λ _{e q}^{o}}$

$Λ_{e q}^{o} = \frac{19.6}{5 \times 1 0 ^{- 2}} = 392$