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  • Give answer! - forandare 126.4, 425.9 and 100.5 S cm2mol-1, respectively. If the conductivity of 0.001 M HA i - Equilibrium - JEE Main

\wedge ^{o}m for NaCl,\: HCl and NaA are 126.4, 425.9 and 100.5 S cm2 mol-1, respectively. If the conductivity of 0.001 M HA is 5 X 10-5 S cm-1, degree of dissociation of HA is :

  • Option 1)

    0.25

  • Option 2)

    0.75

  • Option 3)

    0.125

  • Option 4)

    0.50

Answers (1)

best_answer

 

Thermodynamics of the reaction -

\bigtriangleup G<0

- wherein

The reaction is spontaneous and proceeds in the forward direction.

 

 

Law of Chemical equilibrium -

At a given temperature, the product of concentration of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentration of the reactants raised to their individual stoichiometric coefficients has a constant value.

- wherein

aA+bB\rightleftharpoons cC+dD


K_{c}=\frac{[C]^{c\:[D]^{d}}}{[A]^{a}\:[B]^{b}}

[A],\:[B],\:[C]\:[D]

are equilibrium concentration

As we have learned in dissassociation

we know

\wedge _{M}^{0}(HA)=\wedge _{M}^{0}(HCL)+V(NaA)-\wedge _{M}^{0}(NaCl)

=425.9+100.5-126.4

=4005cm^{2}mol^{-1}

\wedge _{M}=\frac{1000K}{M}=\frac{1000\times 5\times 10^{-5}}{10^{-3}}=505cm^{2}mol^{-1}

\alpha =\frac{\wedge _{M}}{\wedge _{M}^{0}}=\frac{50}{400}=0.125

 

 

 


Option 1)

0.25

Option 2)

0.75

Option 3)

0.125

Option 4)

0.50

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