In this reaction H3O+ is the conjugate acid. The original acid in this reaction is H3PO4
In H2O the conjugate base is H2PO4-, being conjugated to the acid H3PO4. As well: H3PO4 is conjugated acid to the base H2PO4-.
h2po3
Since H3PO4 has 3 ionizable hydrogens, it will have three Ka values. Approximate values areKa1 = 7x10^-3; Ka2 = 6x10^-8 and Ka3 = 4.5x10^-13
The conjugate acid of H2PO4- is H3PO4. When H2PO4- donates a proton, it forms the conjugate acid H3PO4.
There are three protolysis steps:H3PO4 ---> H+ + H2PO4-H2PO4- ---> H+ + HPO42-HPO42- ---> H+ + PO43-
In H2O the conjugate base is H2PO4-, being conjugated to the acid H3PO4. As well: H3PO4 is conjugated acid to the base H2PO4-.
h2po3
Since H3PO4 has 3 ionizable hydrogens, it will have three Ka values. Approximate values areKa1 = 7x10^-3; Ka2 = 6x10^-8 and Ka3 = 4.5x10^-13
The conjugate acid of H2PO4- is H3PO4. When H2PO4- donates a proton, it forms the conjugate acid H3PO4.
There are three protolysis steps:H3PO4 ---> H+ + H2PO4-H2PO4- ---> H+ + HPO42-HPO42- ---> H+ + PO43-
The conjugate base of H3PO4 is H2PO4-.
Only H+ (from the strong perchloric acid) and H2PO4- will react, but ClO4- and Na+ will not and stay unchanged in solution.H+ + H2PO4- --> H3PO4
The conjugate base of H3PO4 is H2PO4-. The formula for the conjugate base can be found by removing one proton (H+) from the acid molecule.
H3PO4 > H2PO4-PH 0.1 M SolutionsH3PO4 PH=1.5H2PO4- PH=4.4HPO42- PH=9.3PO43- PH=12
There are two reaction equations for phosphate ion and water. They two equations are: i) HPO4 2- + H2O => H3O+ + PO4 3- ii) HPO4 2- + H2O => OH- + H2PO4 -
The dissociation of 2 protons from trihydrogen phosphate (H3PO4) results in the formation of dihydrogen phosphate (H2PO4-) and a water molecule (H2O). The chemical equation for this process can be written as: H3PO4 ⇌ H2PO4- + H+ + H2O
The second ionization reaction of phosphoric acid (H3PO4) with water is: H2PO4- (aq) + H2O (l) ⇌ H3O+ (aq) + HPO4^2- (aq)