Hexa Aqua Potasium(+) Cation Complex Hydrolise 
by Aris Kaksis Department
of medical Biochemistry, Riga Stradin University RSU

At equilibrium, the following equilibrium equationis established withits conjugate base:

K⁺(H₂O)₆(aq)     +      H₂O(l) K⁺(H₂O)₅(OH^-)(aq)  + H₃O⁺(aq)
acid extremely weak  conjugatestrong base
The acid and its conjugate base have different colours. At low pH values the concentration of H₃O⁺ is high and so
the equilibrium position lies to the left. The equilibrium solution has the colour A. At high pH values,the concentration
of H₃O⁺is low - the equilibrium position thus lies to the right and the equilibrium solution has colour B.

blue (Acid) pHhydrolize=12.75               red (Base)
KOH5HOHacdSM.skc KOH5HOHacdSM.mol KOH5HOHacdSM.tgf

We can apply equilibrium law to Brønsted equilibria- in general for a weak acid potasium complex:
K_eq=( [H₃O⁺]·[K⁺(H₂O)₅(OH^-)]/([K⁺(H₂O)₆]•[H₂O]))_eq
K_eq is known as the Brønsted equilibriaconstant. The pH for hydrolyse is calculated at this point when:
                 [K⁺(H₂O)₅(OH^-)] =   [K⁺(H₂O)₆]
So from equation: K_eq = [H₃O⁺]/[H₂O] =10^-14.5/[H₂O]= 10^-14.5/55.3 = 2•10^-16.24
K_eq = [H₃O⁺]/[H₂O]; K_eq•[H₂O]=[H₃O⁺]=10^-12.75;-log(K_eq•[H₂O])=12.75 = pH = -log([H₃O⁺])
The pH of the solution at its equilibrium point is called the pH andisthe pH at which half of the potasium is in its
acid form and the other half inthe form of its conjugate base.

At a low pH<14.5, a weak acid aqua complex of potasium ion is almostentirely in the [K⁺(H₂O)₆]form, the
concentration of which predominates. As the pH increases pH>14.5- theconcentration of [K⁺(H₂O)₆]decreases and
the equilibrium is pushed to the right to conjugatebase[K⁺(H₂O)₅(OH^-)]form,the concentration of which predominates.

Aqua complex of potassium 

We can apply equilibrium law to complexe stability constant equilibria- in general for a acid form potasium cation:

K⁺(aq)    + OH^- (aq) K⁺(OH^-)(aq)
blue (Acid)                             red (Base)
K_stab=[K⁺(OH^-)]/([Na⁺]•[OH^-])=10^-0.5=([K⁺(OH^-)][H₃O⁺])/([K⁺]•[10^-14])
K_stab is known as the Complexestability constant.The pH for hydrolyse is calculated at this point when:

                   [K⁺(OH^-)] =   [K⁺]
So from equation: K_stab = [H₃O⁺]/10^-14 =10^-0.5
K_stab•10^-14=[H₃O⁺]=10^-14.5; -log(Kstab•10^-14)= 14.5 = pH =-log([H₃O⁺])
The pH of the solution at its equilibrium point is called the pH andisthe pH at which half of the potasium is in its
conjugate base and the other halfin the form of its acid form.

Complex maker of potasium cation
At a low pH<14.5 a weak acid aqua complex of potasium ion is almostentirely in the K⁺form free cation, the
concentration of which predominates. As the pH increases pH>14.5 -the concentration of [K⁺]decreases and the
equilibrium is pushed to the right to conjugatebase [K⁺(OH^-)]form, the concentration of which predominates.

   Literature

Handbook of  Chemical Equilibriain Analytical Chemistry.S.Kortly, L.Sucha. New York, Ellis HorwoodLTD. 1985. 414.p.