JEE MAIN - Chemistry (2021 - 27th July Morning Shift - No. 15)
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
12
Explanation
$$\Lambda _m^{} = 1000 \times {\kappa \over M}$$
$$ = 1000 \times {{2 \times {{10}^{ - 5}}} \over {0.001}} = 20$$ S cm2 mol$$-$$1
$$ \Rightarrow \alpha = {{\Lambda _m^{}} \over {\Lambda _m^\infty }} = {{20} \over {190}} = \left( {{2 \over {19}}} \right)$$
HA $$\rightleftharpoons$$ H+ + A$$-$$
$$0.001(1 - \alpha )0.001\alpha 0.001\alpha $$
$$ \Rightarrow {k_a} = 0.001\left( {{{{\alpha ^2}} \over {1 - \alpha }}} \right) = {{0.001 \times {{\left( {{2 \over {19}}} \right)}^2}} \over {1 - \left( {{2 \over {19}}} \right)}}$$
$$ = 12.3 \times {10^{ - 6}}$$
$$ = 1000 \times {{2 \times {{10}^{ - 5}}} \over {0.001}} = 20$$ S cm2 mol$$-$$1
$$ \Rightarrow \alpha = {{\Lambda _m^{}} \over {\Lambda _m^\infty }} = {{20} \over {190}} = \left( {{2 \over {19}}} \right)$$
HA $$\rightleftharpoons$$ H+ + A$$-$$
$$0.001(1 - \alpha )0.001\alpha 0.001\alpha $$
$$ \Rightarrow {k_a} = 0.001\left( {{{{\alpha ^2}} \over {1 - \alpha }}} \right) = {{0.001 \times {{\left( {{2 \over {19}}} \right)}^2}} \over {1 - \left( {{2 \over {19}}} \right)}}$$
$$ = 12.3 \times {10^{ - 6}}$$
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