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Question: The equivalent conductance of an aqueous solution of \(1.0283 \times \,{10^{ - 3}}\,{\text{g}}\) equ...

The equivalent conductance of an aqueous solution of 1.0283× 10−3 g1.0283 \times \,{10^{ - 3}}\,{\text{g}} equivalent acetic acid per liter is48.15 π −1 cm2 equiv−148.15\,{{\text{$\pi$ }}^{ - 1}}\,{\text{c}}{{\text{m}}^2}\,{\text{equi}}{{\text{v}}^{ - 1}} at 25 ∘C25{\,^ \circ }{\text{C}} . At infinite dilution the value is 390.7 π −1 cm2 equiv−1390.7\,{{\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 ×10−50.1232,\,1.78\, \times {10^{ - 5}}
B. 0.223, 102 ×10−50.223,\,102\, \times {10^{ - 5}}
C. 0.229, 1.78 ×10−50.229,\,1.78\, \times {10^{ - 5}}
D. 0.531, 2.85×10−50.531,\,2.85 \times {10^{ - 5}}

Explanation

Solution

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: α= λmλm∘±\alpha = \,\dfrac{{{\lambda _m}}}{{\lambda _m^{ \circ \pm }}}
Ka= Cα21−α{K_a} = \,\dfrac{{C{\alpha ^2}}}{{1 - \alpha }}

Complete step by step answer:
The formula to calculate the degree of ionization is as follows:
α= λmλm∘\alpha = \,\dfrac{{{\lambda _m}}}{{\lambda _m^ \circ }}
Where,
α\alpha is the degree of ionization.
λm{\lambda _m} is the equivalent conductance at a concentration.
λm∘\lambda _m^ \circ is the equivalent conductance at infinite dilution.
Substitute 48.15 π −1 cm2 equiv−148.15\,{{\text{$\pi$ }}^{ - 1}}\,{\text{c}}{{\text{m}}^2}\,{\text{equi}}{{\text{v}}^{ - 1}}for the equivalent conductance at a concentration and 390.7 π −1 cm2 equiv−1390.7\,{{\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}}}}
α= 0.1232\alpha = \,0.1232
So, the degree of ionization is 0.1232.
The formula to determine the ionization constant is as follows:
Ka= Cα21−α{K_a} = \,\dfrac{{C{\alpha ^2}}}{{1 - \alpha }}
Where,
Ka{K_a} is the acid ionization constant.
 C\,C is the concentration.
Substitute 1.0283× 10−31.0283 \times \,{10^{ - 3}} for concentration, 0.1232 for degree of ionization.
Ka= 1.0283× 10−3 ×(0.1232)21−0.1232{K_a} = \,\dfrac{{1.0283 \times \,{{10}^{ - 3}}\, \times {{\left( {0.1232} \right)}^2}}}{{1 - 0.1232}}
Ka= 1.56× 10−5 0.8768{K_a} = \,\dfrac{{1.56 \times \,{{10}^{ - 5}}\,}}{{0.8768}}
Ka= 1.78× 10−5{K_a} = \,1.78 \times \,{10^{ - 5}}
So, ionization constant for acetic acid is 1.78× 10−51.78 \times \,{10^{ - 5}}.

Therefore, option (A)0.1232, 1.78 ×10−50.1232,\,1.78\, \times {10^{ - 5}} is correct.

Note: For very weak electrolyte where, α<<1\alpha < < 1, the value of 1−α1 - \alpha can be taken to equal to 1 so, the formula of ionization constant can be reduced as, Ka= Cα2{K_a} = \,C{\alpha ^2}. The similar formula is used for the determination of ionization constant for weak bases; only the Ka{K_a} is replaced with Kb{K_b} where, Kb{K_b} is the base ionization constant.