Write equation that showss `H_(2)PO_(4)^(-)` acting both as an acid and as a base.

Explanation

`H_(2)PO_(4)^(-)+H_(2)OtoHPO_(4)^(2-)+H_(3)O^(+)`
Here `H_(2)PO_(4)^(-)` is giving proton to water so it is an acid.
`H_(2)PO_(4)^(-)+HCltoH_(3)PO_(4)+Cl^(-)`
Here `H_(2)PO_(4)^(-)` is taking proton so it is a base.