3
$\begingroup$

$\ce{Na_2HPO_4}$ is amphoteric, which means it can act as a base or as a acid depending on which substance they react with.

This is the reaction in which disodium phosphate appears to be an acid:

$$\ce{Na_2HPO_4 + H_2O \rightarrow H3O+ + Na_2PO_4-}$$

This is when it acts as a base:

$$\ce{Na_2HPO_4 + H_2O \rightarrow OH- + Na_2H_2PO_4+}$$

Did I write those two reactions currently? If not, what's wrong with my reasoning?

Improve this question
$\endgroup$
2
  • $\begingroup$ No it's not. I don't think there is such particle as $\ce{Na2PO4-}$ (let alone $\ce{Na2H2PO4+}$). Either write everything in ionic form, or write everything in molecular form. $\endgroup$
    – Ivan Neretin
    Commented Apr 28, 2016 at 16:30
  • $\begingroup$ Oh dear, consequences of not looking closely enough.. Shameful. $\endgroup$
    – orthocresol
    Commented Apr 28, 2016 at 16:43

1 Answer 1

Reset to default
5
$\begingroup$

$\ce{Na2HPO4}$ is dissociated in solution. The amphoteric species is actually the hydrogen phosphate dianion:

Reaction as an acid: $$\ce{HPO4^{2-} + H2O <=> PO4^{3-} + H3O+}$$

Reaction as a base: $$\ce{HPO4^{2-} + H2O <=> H2PO4- + OH-}$$

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.