[Mimic]

Can you help me with this homework?
Write the mass balance and the proton balance of phosphoric acid in a 0.5 M solution of KH2PO4

This is my attempt:

[itex]\ce{KH_{2}PO_{4}}[/itex] is completely dissociated

[itex]\ce{KH_{2}PO_{4}} + \ce{NaOH}[/itex] [itex]\ce{K^{+}} + \ce{H_{2}PO^{-}_{4}}[/itex]

The acidic phosphate ion gives rise to the equilibrium

[itex]\ce{H_{2}PO^{-}_{4}} + \ce{H_{2}O}[/itex] [itex]\ce{HPO^{2-}_{4}} + \ce{H_{3}O^{+}}[/itex]

[itex]\ce{HPO^{2-}_{4}}+ \ce{H_{2}O}[/itex] [itex]\ce{PO^{3-}_{4}} + \ce{H_{3}O^{+}}[/itex]

[itex]\ce{H_{2}PO^{-}_{4}} + \ce{H_{2}O}[/itex] [itex]\ce{H_{3}PO_{4}} + \ce{OH^{-}}[/itex]

Water autolysis

[itex]\ce{2H_{2}O}[/itex] [itex]\ce{H_{3}O^{+}} + \ce{OH^{-}}[/itex]

Mass balance

[itex]0.5\ce{\;M} = [\ce{H_{2}PO^{-}_{4}}] + [\ce{HPO^{2-}_{4}}] + [\ce{PO^{3-}_{4}}] + [\ce{H_{3}PO_{4}}][/itex]
[itex][\ce{H_{3}PO_{4}}] = 0.5\ce{\;M} - [\ce{H_{2}PO^{-}_{4}}] - [\ce{HPO^{2-}_{4}}] - [\ce{PO^{3-}_{4}}][/itex]

Proton balance

[itex]
[\ce{OH^{-}}]_{\ce{tot}} = [\ce{OH^{-}}]_{\ce{H_{2}PO^{-}_{4}}} + [\ce{OH^{-}}]_{\ce{H_{2}O}}
[/itex]

[itex][\ce{OH^{-}}]_{\ce{H_{2}PO^{-}_{4}}} = [\ce{H_{3}PO_{4}}][/itex]

[itex]
[\ce{OH^{-}}]_{\ce{tot}} = [\ce{H_{3}PO_{4}}] + [\ce{OH^{-}}]_{\ce{H_{2}O}}
[/itex]

The molar concentration of hydroxide ions and H+ ion from water is the same

[itex][\ce{OH^{-}}]_{\ce{H_{2}O}} = [\ce{H_{3}O^{+}}]_{\ce{H_{2}O}}[/itex]

The total molar concentration of H+ ions in solution is

[itex]
[\ce{H_{3}O^{+}}]_{\ce{tot}} = [\ce{H_{3}O^{+}}]_{\ce{H_{2}PO^{-}_{4}}} + [\ce{H_{3}O^{+}}]_{\ce{HPO^{2-}_{4}}} + [\ce{H_{3}O^{+}}]_{\ce{H_{2}O}}
[/itex]

[itex]
[\ce{H_{3}O^{+}}]_{\ce{tot}} = [\ce{HPO^{2-}_{4}}] + [\ce{PO^{3-}_{4}}] + [\ce{H_{3}O^{+}}]_{\ce{H_{2}O}}
[/itex]

[itex]
[\ce{H_{3}O^{+}}]_{\ce{H_{2}O}} = [\ce{H_{3}O^{+}}]_{\ce{tot}} - [\ce{H_{2}PO^{-}_{4}}] - [\ce{HPO^{2-}_{4}}]
[/itex]

Then, replacing

[itex]
[\ce{OH^{-}}]_{\ce{tot}} = [\ce{H_{3}PO_{4}}] + [\ce{H_{3}O^{+}}]_{\ce{tot}} - [\ce{HPO^{2-}_{4}}] - [\ce{PO^{3-}_{4}}]
[/itex]

[itex]
[\ce{H_{3}PO_{4}}] = [\ce{OH^{-}}]_{\ce{tot}} - [\ce{H_{3}O^{+}}]_{\ce{tot}} + [\ce{HPO^{2-}_{4}}] + [\ce{PO^{3-}_{4}}]
[/itex]

There's something wrong?

Thank you in advance

[Borek]

[itex]\ce{KH_{2}PO_{4}} + \ce{NaOH}[/itex] [itex]\ce{K^{+}} + \ce{H_{2}PO^{-}_{4}}[/itex]

NaOH?

[Mimic]

Sorry, my mistake. Of course, NaOH is not present.