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 $\ce{HO-}$ ?