are the respective ionisation constants for the following reactions.

${\mathrm{H}}_{2}\mathrm{S}⇌{\mathrm{H}}^{+}+{\mathrm{HS}}^{-}\phantom{\rule{0ex}{0ex}}{\mathrm{HS}}^{-}⇌{\mathrm{H}}^{+}+{\mathrm{S}}^{2-}\phantom{\rule{0ex}{0ex}}{\mathrm{H}}_{2}\mathrm{S}⇌2{\mathrm{H}}^{+}+{\mathrm{S}}^{2-}$

The correct relationship between  is

(1) ${\mathrm{K}}_{{\mathrm{a}}_{3}}={\mathrm{K}}_{{\mathrm{a}}_{1}}×{\mathrm{K}}_{{\mathrm{a}}_{2}}$

(2) ${\mathrm{K}}_{{\mathrm{a}}_{3}}={\mathrm{K}}_{{\mathrm{a}}_{1}}+{\mathrm{K}}_{{\mathrm{a}}_{2}}$

(3) ${\mathrm{K}}_{{\mathrm{a}}_{3}}={\mathrm{K}}_{{\mathrm{a}}_{1}}-{\mathrm{K}}_{{\mathrm{a}}_{2}}$

(4) ${\mathrm{K}}_{{\mathrm{a}}_{3}}={\mathrm{K}}_{{\mathrm{a}}_{1}}/{\mathrm{K}}_{{\mathrm{a}}_{2}}$

(1) HINT: Use basic properties of equilibrium constant.

For the reaction,
${\mathrm{H}}_{2}\mathrm{S}⇌{\mathrm{H}}^{+}+{\mathrm{HS}}^{-}\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{{\mathrm{a}}_{1}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{HS}}^{-}\right]}{\left[{\mathrm{H}}_{2}\mathrm{S}\right]}$
For the reaction,
${\mathrm{HS}}^{-}⇌{\mathrm{H}}^{+}+{\mathrm{S}}^{2-}\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{{\mathrm{a}}_{2}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{S}}^{2-}\right]}{\left[{\mathrm{HS}}^{-}\right]}$
When, the above two reactions are added, their equilibrium constants are multiplied, thus
${\mathrm{K}}_{{\mathrm{a}}_{3}}=\frac{{\left[{\mathrm{H}}^{+}\right]}^{2}\left[{\mathrm{S}}^{2-}\right]}{\left[{\mathrm{H}}_{2}\mathrm{S}\right]}={\mathrm{K}}_{{\mathrm{a}}_{1}}×{\mathrm{K}}_{{\mathrm{a}}_{2}}$
Hence, ${\mathrm{K}}_{{\mathrm{a}}_{3}}={\mathrm{K}}_{{\mathrm{a}}_{1}}×{\mathrm{K}}_{{\mathrm{a}}_{2}}$