Respuesta :

For the sequential deprotonations of a polyprotic acid, the equilibrium constants will typically decrease with each successive ionization.

For explaining above statement let's take an example of polyprotic acid like phosphoric acid.

In the phospheric acid [tex]H_{3} PO_{4}[/tex], after deprotonation of one hydrogen results as [tex]H_{2} PO_{4}^{-}[/tex] and equilibrium constant as [tex]K_{1}[/tex] . [tex]H_{2} PO_{4}^{-}[/tex]  donate one hydrogen results as [tex]H_{} PO_{4}^{-2}[/tex] and equilibrium constant as [tex]K_{2}[/tex] . Similarly,  [tex]H_{} PO_{4}^{-2}[/tex]  donate one hydrogen results as [tex]PO_{4}^{-3}[/tex] and equilibrium constant as [tex]K_{3}[/tex] . The removing of hygen ion from nuetral molecule is easier than removal of hydrogen ion from ionic species So, the value of [tex]K_{1}[/tex] is greater than  [tex]K_{2}[/tex] and value of  [tex]K_{2}[/tex]  is greater than [tex]K_{3}[/tex].

So, the equilibrium constants of polyprotic acid will typically decrease with each successive ionization.

Learn about deprotonations

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