(આપેલ : $K _{ b }\left( NH _4 OH \right)=1 \times 10^{-5}, \log 2=0.30, \log 3=0.48, \log 5=0.69, \log 7=0.84, \log 11=$ $1.04)$
\(\left[ NH _{4}^{+}\right]=\frac{2\,m\,mole }{60\,ml }=\frac{1}{30}\,M\)
\(pH =\frac{ pK _{ w }- pK _{ b }-\log C }{2}=\frac{14-5+1.48}{2}=5.24\)
$(i)\, H_3PO_4+H_2O \rightarrow H_3O^+ + H_2PO_4^-$
$(ii)\, H_2PO 4^- + H_2O \rightarrow HPO_4^{2-} + H_3O^+$
$(iii)\, H_2PO_4^-+ OH^- \rightarrow H_3PO_4 + O^{2-}$
ઉપરના પૈકી શામાં $H_2PO_4^-$ એસિડ તરીકે વર્તે છે ?
$NH_3$ + $H_2O$ $\rightleftharpoons$ $NH_4^ + + O{H^ - }$