The dissociation constant of two acids ${\mathrm{HA}}_1 \&{\mathrm{HA}}_2$ are $3.0\times {10}^{-4}$ & $1.8\times {10}^{-5}$ respectively. The relative strength of the acids are in the ratio

${\mathrm{HA}}_1\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{A}}_1^{-}$

C               0           0

$\mathrm{C}-{\mathrm{C\alpha }}_1 {\mathrm{C\alpha }}_1 {\mathrm{C\alpha }}_1$

${\mathrm{Ka}}_1={\mathrm{C\alpha }}^2\Rightarrow {\mathrm{\alpha }}^2=\frac{{\mathrm{K}}_{{\mathrm{a}}_1}}{\mathrm{C}}$

${\mathrm{HA}}_2\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{A}}_2^{-}$

C          0        0

$\mathrm{C}-{\mathrm{C\alpha }}_2 {\mathrm{C\alpha }}_2 {\mathrm{C\alpha }}_2$

${\mathrm{Ka}}_2={\mathrm{C\alpha }}_2^{\mathrm{r}}\Rightarrow {\mathrm{C\alpha }}_2^{\mathrm{r}}=\frac{{\mathrm{Ka}}_2}{\mathrm{C}}$

$\frac{{\mathrm{\alpha }}_2^1}{{\mathrm{\alpha }}_2^2}=\frac{{\displaystyle \frac{k{a}_1}{C}}}{\frac{k{a}_2}{C}}=\frac{K{a}_1}{K{a}_2}=\frac{3\times {10}^{-4}}{1.8\times {10}^{-5}}=\frac{{10}^2}{6}$

${\left(-\sqrt{16.6}\right)}^2=16.6=\sqrt{16.6}=4$

=4:1