The conductivity of a weak acid HA of concentration 0.001 mol L^(-1) is 2.0 x 10^(-5) S cm^(-1). If Lambda _m^o(HA) = 190 S cm^(2) mol^(-1), the ionization constant (K_(a)) of HA is equal to ______________ x 10^(-6). (Round off to the Nearest Integer)

Lambda _m^ = 1000 x kappa / M = 1000 x 2 x 10^ - 5 / 0.001 = 20 S cm^(2) mol^(-1) => alpha = Lambda _m^ / Lambda _m^infinity = 20 / 190 = ( 2 / 19 )HA H^(+) + A^(-)0.001(1 - alpha )0.001alpha 0.001alpha => k_a = 0.001( alpha ^2 / 1 - alpha ) = 0.001 x ( 2 / 19 )^2 / 1 - ( 2 / 19 ) = 12.3 x 10^ - 6

The conductivity of a weak acid HA of concentration 000... | null (Chemistry)

\Lambda _m^{} = 1000 \times {\kappa \over M}

= 1000 \times {{2 \times {{10}^{ - 5}}} \over {0.001}} = 20

S cm² mol^{$-$
1}

\Rightarrow \alpha = {{\Lambda _m^{}} \over {\Lambda _m^\infty }} = {{20} \over {190}} = \left( {{2 \over {19}}} \right)

\rightleftharpoons

H⁺ + A^$-$

0.001(1 - \alpha )0.001\alpha 0.001\alpha

\Rightarrow {k_a} = 0.001\left( {{{{\alpha ^2}} \over {1 - \alpha }}} \right) = {{0.001 \times {{\left( {{2 \over {19}}} \right)}^2}} \over {1 - \left( {{2 \over {19}}} \right)}}

= 12.3 \times {10^{ - 6}}

Explanation

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