$Ca{F_2} \rightleftharpoons C{a^{2 + }} + 2{F^ - }$
Ionic product $ = [C{a^{2 - }}]{[{F^ - }]^2}$
when. $[C{a^{2 + }}] = 1 \times {10^{ - 2}}\,M$
${[{F^ - }]^2} = {(1 \times {10^{ - 3}})^2}\,M$
$ = 1 \times {10^{ - 6}}\,M$
$\therefore \,[C{a^{2 + }}]{[{F^ - }]^2} = (1 \times {10^{ - 2}})(1 \times {10^{ - 6}}) = 1 \times {10^{ - 5}}$
In this case,
Ionic product $(1 \times {10^{ - 8}}) > $ solubility product $(1.7 \times {10^{ - 10}})$
$\therefore $ Hence $(a)$ is correct option.
$(i)\, H_3PO_4+H_2O \rightarrow H_3O^+ + H_2PO_4^-$
$(ii)\, H_2PO 4^- + H_2O \rightarrow HPO_4^{2-} + H_3O^+$
$(iii)\, H_2PO_4^-+ OH^- \rightarrow H_3PO_4 + O^{2-}$
ઉપરના પૈકી શામાં $H_2PO_4^-$ એસિડ તરીકે વર્તે છે ?