Question
A weak monobasic acid is $$5\% $$ dissociated in $$0.01\,mol\,d{m^{ - 3}}$$ solution. Limiting molar conductivity of acid at infinite dilution is $$4 \times {10^{ - 2}}\,oh{m^{ - 1}}\,{m^2}\,mo{l^{ - 1}}.$$ What will be the conductivity of $$0.05\,mol\,d{m^{ - 3}}$$ solution of the acid?
A.
$$8.94 \times {10^{ - 6}}\,oh{m^{ - 1}}\,c{m^2}\,mo{l^{ - 1}}$$
B.
$$8.92 \times {10^{ - 4}}\,oh{m^{ - 1}}\,c{m^2}\,mo{l^{ - 1}}$$
C.
$$4.46 \times {10^{ - 6}}\,oh{m^{ - 1}}\,c{m^2}\,mo{l^{ - 1}}$$
D.
$$2.23 \times {10^{ - 5}}\,oh{m^{ - 1}}\,c{m^2}\,mo{l^{ - 1}}$$
Answer :
$$8.92 \times {10^{ - 4}}\,oh{m^{ - 1}}\,c{m^2}\,mo{l^{ - 1}}$$
Solution :
$${K_a} = C{\alpha ^2} = 0.01 \times {\left( {0.05} \right)^2}$$ $$ = 2.5 \times {10^{ - 5}}$$
$${K_a} = C{\alpha ^2}$$
$$2.5 \times {10^{ - 5}} = 0.05 \times {\alpha ^2} \Rightarrow \alpha $$ $$ = 0.0223$$
$$\alpha = \frac{{ \wedge _m^c}}{{ \wedge _m^\infty }}$$
$$\eqalign{
& \wedge _m^c = 0.0223 \times 4 \times {10^{ - 2}} \cr
& \,\,\,\,\,\,\,\,\, = 8.92 \times {10^{ - 4}}\,oh{m^{ - 1}}\,c{m^2}\,mo{l^{ - 1}} \cr} $$