Question

The equivalent conductance of an aqueous solution of $1.0283 \times \,{10^{ - 3}}\,{\text{g}}$ equivalent acetic acid per liter is\[48.15\,{{\text{$\pi$ }}^{ - 1}}\,{\text{c}}{{\text{m}}^2}\,{\text{equi}}{{\text{v}}^{ - 1}}\] at $25{\,^ \circ }{\text{C}}$ . At infinite dilution the value is \[390.7\,{{\text{$\pi$ }}^{ - 1}}\,{\text{c}}{{\text{m}}^2}\,{\text{equi}}{{\text{v}}^{ - 1}}\].Calculate the degree of ionisation and ionization constant of acetic acid is:
A. \[0.1232,\,1.78\, \times {10^{ - 5}}\]
B. \[0.223,\,102\, \times {10^{ - 5}}\]
C. \[0.229,\,1.78\, \times {10^{ - 5}}\]
D. \[0.531,\,2.85 \times {10^{ - 5}}\]

Answer 1

Hint: The degree of ionization is determined by dividing the equivalent conductance at a concentration by the equivalent conductance at infinite dilution. Ionization constant is determined as the product of degree of ionization and concentration.

Formula used: $\alpha = \,\dfrac{{{\lambda _m}}}{{\lambda _m^{ \circ \pm }}}$
${K_a} = \,\dfrac{{C{\alpha ^2}}}{{1 - \alpha }}$

Complete step by step answer:
The formula to calculate the degree of ionization is as follows:
$\alpha = \,\dfrac{{{\lambda _m}}}{{\lambda _m^ \circ }}$
Where,
$\alpha $ is the degree of ionization.
${\lambda _m}$ is the equivalent conductance at a concentration.
$\lambda _m^ \circ $ is the equivalent conductance at infinite dilution.
Substitute \[48.15\,{{\text{$\pi$ }}^{ - 1}}\,{\text{c}}{{\text{m}}^2}\,{\text{equi}}{{\text{v}}^{ - 1}}\]for the equivalent conductance at a concentration and \[390.7\,{{\text{$\pi$ }}^{ - 1}}\,{\text{c}}{{\text{m}}^2}\,{\text{equi}}{{\text{v}}^{ - 1}}\] for the equivalent conductance at infinite dilution.
$\alpha = \,\dfrac{{48.15\,{{\text{$\pi$ }}^{ - 1}}\,{\text{c}}{{\text{m}}^2}\,{\text{equi}}{{\text{v}}^{ - 1}}}}{{390.7\,{{\text{$\pi$ }}^{ - 1}}\,{\text{c}}{{\text{m}}^2}\,{\text{equi}}{{\text{v}}^{ - 1}}}}$
$\alpha = \,0.1232$
So, the degree of ionization is 0.1232.
The formula to determine the ionization constant is as follows:
${K_a} = \,\dfrac{{C{\alpha ^2}}}{{1 - \alpha }}$
Where,
${K_a}$ is the acid ionization constant.
$\,C$ is the concentration.
Substitute $1.0283 \times \,{10^{ - 3}}$ for concentration, 0.1232 for degree of ionization.
${K_a} = \,\dfrac{{1.0283 \times \,{{10}^{ - 3}}\, \times {{\left( {0.1232} \right)}^2}}}{{1 - 0.1232}}$
${K_a} = \,\dfrac{{1.56 \times \,{{10}^{ - 5}}\,}}{{0.8768}}$
${K_a} = \,1.78 \times \,{10^{ - 5}}$
So, ionization constant for acetic acid is $1.78 \times \,{10^{ - 5}}$.

Therefore, option (A)\[0.1232,\,1.78\, \times {10^{ - 5}}\] is correct.

Note: For very weak electrolyte where, $\alpha < < 1$, the value of $1 - \alpha $ can be taken to equal to 1 so, the formula of ionization constant can be reduced as, ${K_a} = \,C{\alpha ^2}$. The similar formula is used for the determination of ionization constant for weak bases; only the ${K_a}$ is replaced with ${K_b}$ where, ${K_b}$ is the base ionization constant.