Buffer solution of NaH2PO4 and Na2HPO4

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 $\ce{HO-}$ ?


