# Phosphate buffer from phosphoric acid and K2HPO4?

Is it possible to make a solution of ph 7 phosphate buffer solution using phosphoric acid and $\ce{K2HPO4}$ ? How can I calculate the amount of $\ce{K2HPO4}$ needed for 1L of phosphoric acid ?

Edit: I have 50 mL of 0.2M $\ce{H3PO4}$ solution. How can I convert this solution into 50 mL of pH 7 buffer solution by adding (only) $\ce{K2HPO4}$ ? How can I calculate the weight of $\ce{K2HPO4}$ considering all the equilibria present in the $\ce{H3PO4}$ solution and by the application of Henderson-Hasselbalch equation ?

• What concentration do you want? 50 mM or 1.0 M? – LDC3 Jun 11 '15 at 3:52
• @LDC3 Say 1.0 M. – chemkatku Jun 11 '15 at 4:02

I suggest you first consider the following reaction: $$\ce{H3PO4 + 3K2HPO4 -> 2HPO4^{2-} + 2H2PO4- + 6K+}$$

It appears, that transforming all $\ce{H3PO4}$ to equal amounts of $\ce{HPO2-}$ and $\ce{H2PO4-}$ requires 3 mole equivalents of $\ce{K2HPO4}$. If you add 3 mole equivalents of $\ce{K2HPO4}$ you will end up in a situation where the concentration of $\ce{[HPO2^{-}] = [H2PO4^{-}]}$, i.e. at the $\ce{pH} = pK_{a2} = 7.21$. Then, I suppose you use the $\ce{HH}$-equation to figure out the rest.

At pH = pka2 = 7.21 the concentration of [H2PO4(-)] = [HPO4(2-)] = 0.40 M. This is because we have added 3 mole equivalents of K2HPO4 to 50*0.2 = 10 mmole of phosphoric acid, i.e. we have reached a total concentration of phosphoric acid protolytes of (3*50*0.2 + 50*0.2)/50 = 0.80 M.

Now, since we wanted to reach pH = 7.0, we have theoretically added too much of K2HPO4. We suppose the excess amount is equal to x.

At pH = 7.0: [HPO4(2-)] < [H2PO4(-)]. At this pH, only HPO4(2-) and H2PO4(-) are present in significant amounts in the solution. We can then calculate the following: 7.00 = 7.21 + log ([HPO4(2-)] - x/[H2PO4(-)]) = 7.21 + log (0.4 - x)/0.4) => x = 0,1533.

Now, initially we had 50*0.2 mmole of phosphoric acid. To reach pH = 7.0 we should then add 3*50*0.2 - 0.1533*50 mmole = 30 - 7,66(5) = 22,34 mmole of K2HPO4 or 3.8(9) gram.

• Thanks for the reply. Can you please explain how that reaction happens ? I mean what about $\ce{H3PO4 + K2HPO4 -> 2 H2PO4^- + 2K+}$ ? And I want the pH to be 7.0 not 7.21. – chemkatku Jun 12 '15 at 14:51