I was trying to make a buffer with $\ce{K2HPO4}$ to use it in a bacterial culture. If I use $\ce{K2HPO4}$ I think I have to add $\ce{KH2PO4}$. I am trying to think it writing the equations of the process but I have troubles. I think the buffer's reactions will be:

Salt dissociation in water: $$\ce{KH2PO4_{(aq)} <=> K+_{(aq)} + {H_2PO4^-}_{(aq)}} $$ $$\ce{K2HPO4_{(aq)} <=> 2K+_{(aq)} + {HPO4^{-2}}_{(aq)}}$$

Anion hydrolysis: $$\ce{HPO4^2-_{(aq)} + H2O_{(l)} <=> H2PO4^-_{(aq)} + OH-_{(aq)}}$$

$$ \ce{H2PO4-_{(aq)} + H2O_{(l)} <=> HPO4^2-_{(aq)} + H3O+_{(aq)}}$$

If I add acid: $$\ce{K2HPO4_{(aq)} + H3O+_{(aq)} <=> H2O_{(l)} + H2PO4-_{(aq)}} $$

If I add base: $$\ce{H2PO4-_{(aq)} + OH-_{(aq)} <=> H2O_{(l)} + HPO4^2-_{(aq)}}$$

Is it correct? And if so, is it possible to add only $\ce{K2HPO4}$ and water to make the buffer?

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For use in a bacterial culture, you no doubt want a buffer solution with a $\mathrm{pH = 7.20}$, which corresponds to the $pK_{a2}$ for phosphoric acid. At that pH $\ce{[H2PO4-] = [HPO4^2-]}$. The $\ce{K+}$ ion is basically a spectator ion and doesn't influence the pH directly. So typically you'd start with a solution which has equal molarities of $\ce{KH2PO4}$ and $\ce{K2HPO4}$ and then adjust the pH with $\ce{H3PO4}$ and $\ce{KOH}$ using a pH meter to get $7.20$ "exactly."

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    $\begingroup$ Ok. Thanks. I have understood. So... Are the reactions I wrote fine, @MaxW ? (Because I have to write it, too). @MaxW $\endgroup$ – Geraldine Sep 21 '18 at 11:41
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    $\begingroup$ Well, you went a little overboard. All you need is $$\ce{K+H2PO4- <=> K+ + HPO4^2- + H+}$$ $\endgroup$ – MaxW Sep 21 '18 at 13:25

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