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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 ?

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  • $\begingroup$ What concentration do you want? 50 mM or 1.0 M? $\endgroup$
    – LDC3
    Commented Jun 11, 2015 at 3:52
  • $\begingroup$ @LDC3 Say 1.0 M. $\endgroup$
    – user16096
    Commented Jun 11, 2015 at 4:02

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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.

Please check the calculations.

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    $\begingroup$ The edit of my answer does not look good. It should read HPO4(2-)! $\endgroup$
    – Bive
    Commented Jun 11, 2015 at 23:00
  • $\begingroup$ 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. $\endgroup$
    – user16096
    Commented Jun 12, 2015 at 14:51
  • $\begingroup$ @Bive I think thats the correct equation now isn't it? $\endgroup$
    – bon
    Commented Jun 13, 2015 at 16:20
  • $\begingroup$ You now tell us that the final concentration should be 1,0 M. This cannot be right. Initially, you had 50 ml 0,2 M H3PO4, i.e. 10 mmole. If you add K2HPO4 to reach a final concentration of 1,0 M, the pH of the final solution will have a pH much higher than 7,0. If we approximate the volume of the solution to be constant, you have to add 5 mole equivalents of K2HPO4 to achieve 1, 0 M. Initial: 50 ml*0,2 M = 10 mmole => Final: 50 ml * 1,0 M = 50 mmole? I think you should stick to your original presented problem, which is interesting, since the problem does not state the final concentration. $\endgroup$
    – Bive
    Commented Jun 14, 2015 at 20:46

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