Preparing a buffer solution from a weak acid/strong base mixture

Question

You have at your disposal an ample quantity of a solution of 0.0500 M NaOH(aq) and 500 mL of a solution of 0.100 M formic acid (HCOOH(aq)). How much of the NaOH(aq) solution should be added to the HCOOH(aq) solution to produce a buffer with pH = 4.00? At this pH value, what is the [HCOO– ]/[HCOOH] ratio?

Answer 1

To answer this question, we need to know the pKa for formic acid (HCOOH). Tabular value (Wikipedia) provided the value of pKa = 3.745.

To prepare such a buffer from HCOOH and NaOH, we would have the following reaction:

HCOOH(aq) + NaOH(aq) ==> HCOONa + H2O

We want to add enough NaOH to produce some HCOONa, but not so much NaOH to react with all of the formic acid. We want to have a ratio of HCOONa / HCOOH sufficient to make a buffer. A buffer is defined, in this case, as a solution containing both the weak acid (HCOOH) and the conjugate base (HCOO^-).

We use the Henderson Hasselbalch equation to find this ratio:

pH = pKa + log [HCOO^-] / [HCOOH]

4.00 = 3.745 + log [HCOO^-] / [HCOOH]

log [HCOO^-] / [HCOOH] = 0.255

HCOO^-] / [HCOOH] = 1.799 ANSWER

To find the amount of NaOH needed, we proceed as follows:

moles HCOOH initially present = 500 ml x 1 L / 1000 ml x 0.1 mol / L = 0.0500 mols HCOOH

moles HCOO^- needed = x

x/0.0500 - x = 1.799

x = 0.0321 mols HCOO^- = 0.0321 mols NaOH needed ANSWER (the question didn't specify mols or volume, so this is moles of NaOH needed. If you want volume of NaOH needed, see below).

Check this result as follows:

HCOOH + NaOH ==> HCOONa + H2O

0.05...........0.0321....................0..................Initial

-0.0321.....-0.0321...............+0.0321...........Change

0.0179..........0........................0.0321...........Equilibrium

Ratio of HCOO^- / HCOOH = 0.0321 / 0.0179 = 1.79 (check)

To obtain 0.0321 mols of NaOH from 0.0500 M, what volume do we need?

0.0500 mol / L (x L) = 0.0321 mols

x = 0.642 L NaOH needed ANSWER

Check this result as follows:

Total volume = 0.642 L + 0.5 L = 1.142 L

Final concentrations will be:

[HCOO^-] = 0.0321 mol / 1.142 L = 0.0281 M

[HCOOH] = 0.0179 mol / 1.142 L = 0.0157 M

Ratio = 0.0281 M / 0.0157 M = 1.79 (check)