The conductivity of a weak acid HA of concentration 0.001 mol L$-$1 is 2.0 $\times$ 10$-$5 S cm$-$1. If $\Lambda _m^o$(HA) = 190 S cm2 mol$-$1, the ionization constant (Ka) of HA is equal to ______________ $\times$ 10$-$6. (Round off to the Nearest Integer)
Answer (integer)
12
Solution
$\Lambda _m^{} = 1000 \times {\kappa \over M}$<br><br>$= 1000 \times {{2 \times {{10}^{ - 5}}} \over {0.001}} = 20$ S cm<sup>2</sup> mol<sup>$-$1</sup><br><br>$$ \Rightarrow \alpha = {{\Lambda _m^{}} \over {\Lambda _m^\infty }} = {{20} \over {190}} = \left( {{2 \over {19}}} \right)$$<br><br>HA $\rightleftharpoons$ H<sup>+</sup> + A<sup>$-$</sup><br><br>$0.001(1 - \alpha )0.001\alpha 0.001\alpha$<br><br>$$ \Rightarrow {k_a} = 0.001\left( {{{{\alpha ^2}} \over {1 - \alpha }}} \right) = {{0.001 \times {{\left( {{2 \over {19}}} \right)}^2}} \over {1 - \left( {{2 \over {19}}} \right)}}$$<br><br>$= 12.3 \times {10^{ - 6}}$
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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