Question

You need to prepare 100.0 mL of a pH = 4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.200 M sodium benzoate. How much of each solution should be mixed to prepare this buffer?

Answer 1

79.18 mL aqueous benzoic acid
20.82 mL aqueous sodium benzoate

Well, we know the final pH, so the first thing we can solve for is the weak-base/weak-acid ratio via the Henderson-Hasselbalch equation (we are in the buffer region, so this equation works!).

pH=pKa+log[A−]/[HA]

where A− is the benzoate and HA is the benzoic acid.

The ratio is then...

4.00=4.20+log[A−]/[HA]

⇒[A−]/[HA]=10^4.00−4.20=0.6310

It makes sense; pH < pKa, so the solution is more acidic than if there were equal quantities of benzoate and benzoic acid. Hence, there is more weak acid than weak base.

This ratio, however, is NOT the starting concentrations given to you. It's the ratio in the buffer, i.e. the ratio after the buffer has been finalized.

There is a dilution!

Since the solution has only one total volume, the total volumes cancel out for the dilution, and we only need to determine the initial volumes necessary to accomplish the mol:mol−−−−−−−− ratio of 0.6310.

⇒0.6310=( 0.240 ×V_A−/V_tot ) / ( 0.100 ×V_HA/V_tot )

= ( 0.240×V_A− ) / ( 0.100 ×V_HA )

Now, we do actually have to assume something. We assume that the volumes are additive, so that we can find, say, V_A− in terms of V_HA. We know that the total volume is 100 mL, so:

V_A−≈100−V_HA in units of mL

Therefore, we now have:

0.6310=0.240 ×(100−V_HA) / ( 0.100 M×V_HA )

For ease of notation, let x=V_HA. Then we have implied units:

0.6310=0.240(100−x) / 0.100x

=24.0−0.240x / 0.100x

0.0631x=24.0−0.240x

(0.0631+0.240)x=24.0

⇒x=V_HA=24 / (00.0631+0.240)mL

= 79.18 mL aqueous benzoic acid

That means

V_A−=20.82 mL aqueous sodium benzoate.