Question
Molar conductivity of $$0.025\,mol\,{L^{ - 1}}$$ methanoic acid is $$46.1\,S\,c{m^2}mo{l^{ - 1}},$$ the degree of dissociation and dissociation constant will be $$\left( {{\text{Given:}}\,\lambda _{{H^ + }}^ \circ = 349.6\,S\,c{m^2}\,mo{l^{ - 1}}\,\,{\text{and}}\,\,\lambda _{HCO{O^ - }}^ \circ = 54.6\,S\,c{m^2}\,mo{l^{ - 1}}} \right)$$
A.
$$11.4\% ,3.67 \times {10^{ - 4}}\,mol\,{L^{ - 1}}$$
B.
$$22.8\% ,1.83 \times {10^{ - 4}}\,mol\,{L^{ - 1}}$$
C.
$$52.2\% ,4.25 \times {10^{ - 4}}\,mol\,{L^{ - 1}}$$
D.
$$1.14\% ,3.67 \times {10^{ - 6}}\,mol\,{L^{ - 1}}$$
Answer :
$$11.4\% ,3.67 \times {10^{ - 4}}\,mol\,{L^{ - 1}}$$
Solution :
$$\eqalign{
& \lambda _{HCOOH}^ \circ = \lambda _{{H^ + }}^ \circ + \lambda _{HCO{O^ - }}^ \circ \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 349.6 + 54.6 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 404.2\,S\,c{m^2}\,mo{l^{ - 1}} \cr
& \alpha = \frac{{{ \wedge _m}}}{{ \wedge _m^ \circ }} \cr
& \,\,\,\, = \frac{{46.1}}{{404.2}} \cr
& \,\,\,\, = 11.4\% \cr
& {K_a} = \frac{{C{\alpha ^2}}}{{1 - \alpha }} \cr
& \,\,\,\,\,\,\,\, = \frac{{0.025 \times {{\left( {0.114} \right)}^2}}}{{1 - 0.114}} \cr
& \,\,\,\,\,\,\,\, = \frac{{0.025 \times 0.114 \times 0.114}}{{0.886}} \cr
& \,\,\,\,\,\,\,\, = 3.67 \times {10^{ - 4}}\,mol\,{L^{ - 1}} \cr} $$