Phosphoric acid [tex]H_{3}PO_{4}[/tex] have three protons which is ionized in different [tex]P^{H}[/tex]. Thats why it is called triprotic acid. The protonisation of phosphoric acid in different[tex]P^{H}[/tex] can be shown as-[tex]H_{3}PO_{4}[/tex]→[tex]H^{+}+H_{2}PO_{4}^{-1}[/tex]. This is the first ionization. The second and third ionizations are shown as-[tex]H_{2}PO_{4}^{-1}[/tex]→[tex]H^{+}[/tex]+[tex]HPO_{4}^{2-}[/tex] and [tex]HPO_{4}^{2-}[/tex]→ [tex]H^{+}[/tex] + P[tex]O_{4}^{3-}[/tex]
[tex]p^{H}[/tex]=5 is near to first equivalence point where [tex]p^{H}[/tex]=[tex]\frac{1}{2}[/tex] ([tex]p_{k}^{1}[/tex] + [tex]p_{k}^{2}[/tex]=[tex]\frac{1}{2}[/tex](2.15+6.82)=4.6. At this [tex]p^{H}[/tex], [tex]H_{2}PO_{4}^{-1}[/tex] will be present.
At [tex]p^{H}[/tex]=10, which is near to second equivalence point[tex]p^{H}=\frac{1}{2} (p_{k}^{2} + p_{k}^{3})[/tex]= [tex]\frac{1}{2}[/tex](6.82+12.38)=9.7. At [tex]p^{H}[/tex]=10, [tex]HPO_{4}^{2-}[/tex] will be present.
