$=-\log \mathrm{K}_{\mathrm{b}}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Base}]}$
$=-\log 1.8 \times 10^{-5}+\log \frac{0.20}{0.30}$
$=-5-0.25+(-0.176)$
therefore,
$4.75-0.176=4.57$
therefore, $\mathrm{pH}=14-4.57=9.43$
$\mathrm{pOH}=\mathrm{pK} \mathrm{b}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Base}]}$
$=-\log \mathrm{K}_{\mathrm{b}}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Base}]}$
$=-\log 1.8 \times 10^{-5}+\log \frac{0.20}{0.30}$
$=-5-0.25+(-0.176)$
therefore,
$4.75-0.176=4.57$
therefore, $\mathrm{pH}=14-4.57=9.43$
$(i)\, H_3PO_4+H_2O \rightarrow H_3O^+ + H_2PO_4^-$
$(ii)\, H_2PO 4^- + H_2O \rightarrow HPO_4^{2-} + H_3O^+$
$(iii)\, H_2PO_4^-+ OH^- \rightarrow H_3PO_4 + O^{2-}$
ઉપરના પૈકી શામાં $H_2PO_4^-$ એસિડ તરીકે વર્તે છે ?