We have this exercise without solutions

From a 0.2 M $\ce{NaH2PO4}$ solution and a 0.2 M $\ce{Na2HPO4}$ solution, a buffer solution with pH = 6.8 is to be prepared. The total concentration of $\ce{PO_4^3}$- should be 0.1 mol l-¹. Calculate the volumes of the two solutions needed to prepare one litre of buffer solution.

So my understanding is that the $\ce{Na}$ will dissolve in the water. Also, both $\ce{NaH_2PO_4}$ and $\ce{Na_2HPO_4}$ are acidic, but $\ce{NaH_2PO_4}$ will be more acidic in this case because is has one more $\ce{H}$ than the other. So if a buffer consists of those two substances, then $\ce{NaH_2PO_4}$ will be the acid and $\ce{Na_2HPO_4}$ will be the base

First, is this correct ?

What I don't really understand is

The total concentration of $\ce{PO_4^3}$- should be 0.1 mol l-¹.

Does this mean that $c(\ce{NaH_2PO_4}) + c(\ce{Na_2HPO_4})= 0.1 $ ? In other words, we would need to set up the Henderson Hasselbalch equation

$$pH = pK_a + \log{\frac{c(\ce{Na_2HPO_4})}{c(\ce{NaH_2PO_4})}} = pK_a + \log{\frac{c(\ce{Na_2HPO_4})}{c(\ce{Na_2HPO_4) - 0.1}}} $$

and then solve for $c(\ce{Na_2HPO_4})$, and then we would find $$c(\ce{NaH_2PO_4}) = 0.1 - c(\ce{Na_2HPO_4}) $$

We have one litre of buffer, so the number of moles is simply

$$n(\ce{Na_2HPO_4}) = 1 \cdot c(\ce{Na_2HPO_4})$$ respectively

$$n(\ce{NaH_2PO_4}) = 1 \cdot c(\ce{NaH_2PO_4})$$

and then the volume is found by dividing the number of moles by the original concentration of $0.2 M$

Is this how they expect us to proceed ? We have no solutions so I prefer to ask

Are there some other important things I need to be aware of ? The $\ce{Na}$ seems to dissolve in the water, but what happens to the different hydrogen atoms? Do they simply become $\ce{H+}$ and $\ce{HO-}$ ?

  • 1
    $\begingroup$ You need just consider $\ce{H2PO4^-(aq) <=> HPO4^2-(aq) + H+(aq)}$ // Total phosphates means the sum [$\ce{H3PO4}$] +[$\ce{H2PO4-}$] + [$\ce{HPO4^2-}$] + [$\ce{PO4^3-}$] // As both stock solutions are 0.2 M, the sum of their volume is 0.5 L for the final 1 L of 0.1 M solution. // Volume ratio of stock solutions follows Henderson Hasselbalch equation. $\endgroup$
    – Poutnik
    Dec 30, 2022 at 18:21
  • $\begingroup$ @Poutnik But in our case we only have $[\ce{H2PO4-}]$ and $[\ce{HPO4^2-}]$ ? The volume ratio follows from the Henderson Hasselbalch equation: $pH = pK_a + \log(\frac{\ce{H2PO4-}}{\ce{HPO4^2-}} ) \rightarrow 6.8 = 6.8 + log(\frac{\ce{H2PO4-}}{\ce{HPO4^2-}} )\rightarrow$ $ 0 = 0 + log(\frac{\ce{H2PO4-}}{\ce{HPO4^2-}} ) \rightarrow 0 = log(\frac{\ce{H2PO4-}}{\ce{HPO4^2-}} ) = \log{1} \rightarrow \ce{H2PO4-} = \ce{HPO4^2-} = 0.5 $ L $\endgroup$
    – wengen
    Dec 30, 2022 at 21:19
  • 1
    $\begingroup$ You have trace equilibrium amounts of H3PO4 and PO4^3- too, but negligible, several orders lower concentration. You have switched fractions. More of acid form does not mean higher pH. $\endgroup$
    – Poutnik
    Dec 30, 2022 at 21:37
  • $\begingroup$ @Poutnik Yes in fact I switched the fraction, my bad. What do you mean by more acid form does not mean higher pH ? Because it's a buffer, the pH value shouldn't change drastically ? $\endgroup$
    – wengen
    Dec 31, 2022 at 0:13
  • 1
    $\begingroup$ The note was related to the wrong fraction. $\endgroup$
    – Poutnik
    Dec 31, 2022 at 5:55


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