# 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 $$\ce{HO-}$$ ?

• 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. Dec 30, 2022 at 18:21
• @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 Dec 30, 2022 at 21:19
• 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. Dec 30, 2022 at 21:37
• @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 ? Dec 31, 2022 at 0:13
• The note was related to the wrong fraction. Dec 31, 2022 at 5:55