QJEEAdv15Ch418e: JEE Advanced 2015 Chemistry - Class 12 JEEAdv15Ch418e Solution: The molar conductivity of a solution of a weak acid HX (0.01 M) …

Given :

$\Lambda _{m(HX)}^c = {{\Lambda _{m(HY)}^c} \over {10}}$

$\Lambda _{m(HX)}^o = \Lambda _{m(HY)}^o$ ( $\because$ $\lambda _{X - }^o \approx \lambda _{Y - }^o$ )

${K_{a(HX)}} = {\left( {{{C{\alpha ^2}} \over {1 - \alpha }}} \right)_{HX}}$

${K_{a(HX)}} = 0.01{({\alpha _{HX}})^2}$ ( $\because$ \alpha < < < 1 ) ....... (i)

Similarly, ${K_{a(HY)}} = 0.01{({\alpha _{HY}})^2}$ ....... (ii)

On dividing equation (i) by (ii), we get

${{{K_{a(HX)}}} \over {{K_{a(HY)}}}} = {{0.01} \over {0.10}}{\left( {{{{\alpha _{HX}}} \over {{\alpha _{HY}}}}} \right)^2}$ ....... (iii)

$\alpha = {{\Lambda _m^c} \over {\Lambda _m^o}}$

${{{\alpha _{HX}}} \over {{\alpha _{HY}}}} = {{{{\left( {\Lambda _m^c/\Lambda _m^o} \right)}_{HX}}} \over {{{\left( {\Lambda _m^c/\Lambda _m^o} \right)}_{HY}}}} = \left( {{1 \over {10}}\Lambda _{m(HY)}^c} \right) \times {1 \over {\Lambda _{m(HY)}^c}} = {1 \over {10}}$

Substituting above value in equation (iii),

${{{K_{a(HX)}}} \over {{K_{a(HY)}}}} = {{0.01} \over {0.10}}{\left( {{1 \over {10}}} \right)^2} = 1 \times {10^{ - 3}}$

$\log {K_{a(HX)}} - \log {K_{a(HY)}} = \log (1 \times {10^{ - 3}})$

$- \log {K_{a(HX)}} - ( - \log {K_{a(HY)}}) = - \log (1 \times {10^{ - 3}})$

$p{K_{a(HX)}} - p{K_{a(HY)}} = 3$