Volume expansion from phosphate buffer solution

Question

I want to mix a phosphate buffer at $\mathrm{pH\ 7.2}$ at $20~\mathrm{^\circ C}$. I do this by dissolving $0.0036\ \mathrm{mol}$ of sodium dihydrogen phosphate monohydrate ($\ce{NaH2PO4*H2O}$) and $0.0063\ \mathrm{mol}$ of di-sodium hydrogen phosphate dihydrate ($\ce{Na2HPO4*2H2O}$) in water per liter of buffer. I add $0.15~\mathrm{M}\ \ce{NaCl}$. How much water do I add if I want $1~\mathrm{L}$ of buffer? Do I just use $1~\mathrm{L}$ of water — will the salts not expand the volume? Is it insignificant and at what point do I need take into account the volume expansion of the solution?

Notes:

$$M(\ce{NaH2PO4 . H2O}) = 137.99~\mathrm{g/mol} \Longrightarrow 137.99~\mathrm{g/mol} \times 0.0036~\mathrm{mol} = 0.496764~\mathrm{g}$$

$$M(\ce{Na2HPO4 . 2 H2O}) = 177.99~\mathrm{g/mol} \Longrightarrow 177.99~\mathrm{g/mol} \times 0.0063~\mathrm{mol} = 1.121337~\mathrm{g}$$

$$M(\ce{NaCl}) = 58.44\ \mathrm{g/mol} \Longrightarrow 58.44\ \mathrm{g/mol} \times 0.15~\mathrm{mol/L} = 8.766~\mathrm{g/L}$$

Answer 1

There is an exact way of doing this and there is a practical way of doing this.

The exact way involves a volumetric $1~\mathrm{l}$ flask. You would first fill the salts into the volumetric flash, add some water to dissolve them and then carefully add more water until the flask contains exactly $1~\mathrm{l}$ of solution. That removes the error of the salts increasing the overall volume.

However, it is also part of the exact way to add a pH-monitoring device before you have finised to make sure the pH is exactly at $7.2$ — and to carefully add phosphoric acid or sodium hydrogen phosphate if the pH is off. (A better solution would be to make sure the pH is over the desired value so phosphoric acid can simply be added; as it is a liquid it can be added and dissolved easier.) For most applications outside of thermodynamics, this is an overkill.

The practical way involves one of the following:

• using an Erlenmeyer flask, beaker or similar with a $1~\mathrm{l}$ mark; filling the salts first and then the solution up to the mark

• simply measuring $1~\mathrm{l}$ of water and adding that.

Why are the practical solutions okay? In most cases, it does not matter whether the molar concentration of, say $\ce{NaCl}$ is exactly $0.15~\mathrm{M}$ or whether it is $0.14~\mathrm{M}$. This rather unimportant experimental error is simply accepted. It also does not really matter whether the pH is exactly $7.2$ or whether it is, say $7.5$. Therefore, one often decides to go the not-so-exact but easier and faster way.