Answer:
The ionization equation is
[tex]H_{2}PO_{4}^{-} +H_{2}O[/tex] ⇄[tex]HPO_{4}^{-2} +H_{3}O^{+}[/tex] (1)
Explanation:
The ionization equation is
[tex]H_{2}PO_{4}^{-} +H_{2}O[/tex] ⇄[tex]HPO_{4}^{-2} +H_{3}O^{+}[/tex] (1)
As the Bronsted definition sais, an acid is a substance with the ability to give protons thus, H2PO4 is the acid and HPO42- is the conjugate base.
The Ka expression is the ratio between the concentration of products and reactants of the equilibrium reaction so,
[tex]Ka = \frac{[HPO_{4}^{-2}] [H_{3}O^{+}]}{[H_{2}PO_{4}^{-}] [H_{2}O]} = 6.2x10^{-8}[/tex]
The pKa is
[tex]-Log (Ka) = -Log (6.2x10^{-8}) = 7.2[/tex]
The pKa of H2CO3 is 6,35, thus this a stronger acid than H2PO4. The higher the pKa of an acid greater the capacity to donate protons.
In the body H2CO3 is a more optimal buffer for regulating pH due to the combination of the two acid-base equilibriums and the two pKa.
If the urine is acidified, according to Le Chatlier's Principle the equilibrium (1) moves to the left neutralizing the excess proton concentration.
