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),

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

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

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

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