JEE Advance - Chemistry (2021 - Paper 2 Online - No. 8)
At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 $$\times$$ 102 S cm2 mol$$-$$1. At 298 K, for an aqueous solution of the acid the degree of dissociation is $$\alpha$$ and the molar conductivity is y $$\times$$ 102 S cm2 mol$$-$$1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y $$\times$$ 102 S cm2 mol$$-$$1.
The value of y is __________.
The value of y is __________.
Answer
0.86
Explanation
Degree of dissociation = $$\alpha$$
Limiting molar conductivity, $$\Lambda _m^o$$ = 4 $$\times$$ 102 S cm2 mol$$-$$1
Molar conductivity, $$\Lambda _m$$ = y $$\times$$ 102 S cm2 mol$$-$$1
Molar conductivity of dilution, $$\Lambda _m^o$$ = 3y $$\times$$ 102 S cm2 mol$$-$$1
Concentration before dilution = C
Concentration after dilution = $${C \over {20}}$$
Using relation, $$\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}}$$ .... (i)
Dissociation constant, $${K_a} = {{C{\alpha ^2}} \over {(1 - \alpha )}}$$
Putting Eq. (i),
$${K_a} = {{C\Lambda _m^2} \over {\Lambda _m^{o2}\left( {1 - {{\Lambda _m^{}} \over {\Lambda _m^o}}} \right)}} = {{C\Lambda _m^2} \over {\Lambda _m^o(\Lambda _m^o - \Lambda _m^{})}}$$
Dissociation constant before dillution,
$${K_a} = {{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}}$$ ..... (ii)
Dissociation constant after dilution,
$${K_a} = {{{C \over {20}}{{(3y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$$ ..... (iii)
Comparing Eqs. (ii) and (iii),
$${{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}} = {{C{{(3y \times {{10}^2})}^2}} \over {20(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$$
$${{{y^2} \times {{10}^4}} \over {{{10}^2}(4 - y)}} = {{9{y^2} \times {{10}^4}} \over {{{10}^2} \times 20(4 - 3y)}}$$
$${1 \over {(4 - y)}} = {9 \over {20(4 - 3y)}}$$
$$80 - 60y = 36 - 9y$$
$$80 - 36 = 60y - 9y$$
$$44 = 51y$$
$$ \Rightarrow y = {{44} \over {51}}$$ = 0.86
Limiting molar conductivity, $$\Lambda _m^o$$ = 4 $$\times$$ 102 S cm2 mol$$-$$1
Molar conductivity, $$\Lambda _m$$ = y $$\times$$ 102 S cm2 mol$$-$$1
Molar conductivity of dilution, $$\Lambda _m^o$$ = 3y $$\times$$ 102 S cm2 mol$$-$$1
Concentration before dilution = C
Concentration after dilution = $${C \over {20}}$$
Using relation, $$\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}}$$ .... (i)
Dissociation constant, $${K_a} = {{C{\alpha ^2}} \over {(1 - \alpha )}}$$
Putting Eq. (i),
$${K_a} = {{C\Lambda _m^2} \over {\Lambda _m^{o2}\left( {1 - {{\Lambda _m^{}} \over {\Lambda _m^o}}} \right)}} = {{C\Lambda _m^2} \over {\Lambda _m^o(\Lambda _m^o - \Lambda _m^{})}}$$
Dissociation constant before dillution,
$${K_a} = {{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}}$$ ..... (ii)
Dissociation constant after dilution,
$${K_a} = {{{C \over {20}}{{(3y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$$ ..... (iii)
Comparing Eqs. (ii) and (iii),
$${{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}} = {{C{{(3y \times {{10}^2})}^2}} \over {20(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$$
$${{{y^2} \times {{10}^4}} \over {{{10}^2}(4 - y)}} = {{9{y^2} \times {{10}^4}} \over {{{10}^2} \times 20(4 - 3y)}}$$
$${1 \over {(4 - y)}} = {9 \over {20(4 - 3y)}}$$
$$80 - 60y = 36 - 9y$$
$$80 - 36 = 60y - 9y$$
$$44 = 51y$$
$$ \Rightarrow y = {{44} \over {51}}$$ = 0.86
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