Phosphate buffer system

Question

Can anyone show me how to find the masses of $\ce{NaH2PO4}$ and $\ce{Na2HPO4}$ needed to prepare $1.0~\mathrm{L}$ of a $0.1~\mathrm{M}$ phosphate buffer solution with $\mathrm{pH} = 7.4$ using the simultaneous equations method?

This is what I have so far:

$$\ce{NaH2PO4 + H2O <=> H3PO4 + NaOH} \\ \ce{Na2HPO4 + 2H2O <=> H3PO4 + NaOH + H3O+} \\ \ce{NaOH -> Na+ + OH-} \\ \ce{H3O+ + OH- <=> 2H2O} \\ \ce{H3PO4 + H2O <=> H3O+ + H2PO4-} \\ \ce{H2PO4^- + H2O <=> H3O+ + HPO4^2-} \\ \ce{HPO4^2- + H2O <=> H3O+ + PO4^3-}$$

The species present are: $\ce{HPO4^2-}$, $\ce{PO4^3-}$, $\ce{H2PO4-}$, $\ce{H3PO4}$, $\ce{Na+}$, $\ce{OH-}$, and $\ce{H3O+}$.

I have two equations:

$$\begin{align} K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}] &= 1.0 \times 10^{-14} \\ [\ce{OH-}] + 2[\ce{HPO4^2-}] + 3[\ce{PO4^3-}] + [\ce{H2PO4-}] &= [\ce{H3O+}] + [\ce{Na+}] \end{align}$$

I need 5 more equations for this system. How do I get them?

Answer 1

Things to consider:

• Sodium is merely a spectator ion. It will not take part in any reactions, so the initial amount of sodium you weigh in will remain constant. We can thus ignore it.
• The concentrations of $\ce{H3PO4}$ and $\ce{PO4^3-}$ will be sufficiently small to entirely ignore them.

This reduces your system to the following relevant chemical equation:

$$\ce{H2PO4- + H2O <=> HPO4^2- + H3O+}$$

With the unknown concentrations of $\ce{H2PO4-}$, $\ce{HPO4^2-}$ and $\ce{H3O+}$. For these three we only need three equations and luckily we have them. Remember that we know:

1. the final phosphate concentration we want;
2. the final $\mathrm{pH}$ we want; and
3. the $\mathrm{p}K_\mathrm{a}$ value of our acidic species.

This gives us three equations that I am going to hide behind a spoiler tag:

$[\ce{H2PO4-}] + [\ce{HPO4^2-}] = 0.1~\mathrm{M}$
$10^{-7.198} = K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{HPO4^2-}]}{[\ce{H2PO4-}]}$
$10^{-7.4} = [\ce{H3O+}]$

We can now put the third equation into the second resulting in exactly the Henderson-Hasselbalch equation, because this exactly is how that equation was derived. The maths of one possible way to approach this problem is shown in the following spoiler tag.

$10^{-7.198} = \frac{10^{-7.4} \times [\ce{HPO4^2-}]}{[\ce{H2PO4-}]}$
$10^{-7.198 + 7.4} = \frac{[{HPO4^2-}]}{[\ce{H2PO4-}]}$
$[\ce{H2PO4-}] \times 10^{0.202} = [\ce{HPO4^2-}]$ ← Here! You have the Henderson-Hasselbalch right here! (In a different form but equivalent.)

$[\ce{HPO4^2-}] + [\ce{H2PO4-}] = 0.1~\mathrm{M}$
$[\ce{HPO4^2-}] = 0.1~\mathrm{M} - [\ce{H2PO4-}]$
$[\ce{H2PO4-}] \times 10^{0.202} = 0.1~\mathrm{M} - [\ce{H2PO4-}]$
$[\ce{H2PO4-}] \left (10^{0.202} + 1 \right ) = 0.1~\mathrm{M}$
$[\ce{H2PO4-}] = \frac{0.1~\mathrm{M}}{10^{0.202} + 1} = 0.0385~\mathrm{M}$

The last steps to calculate the required concentration of $\ce{Na2HPO4}$ including calculating the respective masses of the water-free salts should be trivial and can be found in another castle answer.

I can only speculate why you are not allowed to directly use the Henderson-Hasselbalch equation since mathematically, every optimal approach will simplify itself to exactly that equation. Maybe the point of the excersize is to derive that equation; only your prof will know.