Question

part A. Formic acid, HFor, has a Ka value equal to about 1.8 x 10-4. A student is asked to prepare a buffer having a pH of 3.55 from a solution of formic acid and a solution of sodium formate having the same molarity. How many milliliters of the NaFor solution should she add to 20 mL of the HFor solution to make the buffer?

part B. how many ml of 0.10 m naoh should the student add to 20 ml 0.10 M hfor if she wished to prepare a buffer with a ph of 3.55 the same is in part a?

Answer 1

Buffer problems are best solved using the Henderson-Hasselbalch equation:

pH = pKa + log (moles A- / moles HA)

A- in your case is the formate ion, For-
HA in your case is formic acid, HFor

pKa = -log Ka = -log (1.8 x 10^-4) = 3.74

3.55 = 3.74 + log (moles For- / moles HFor)
-0.19 = log (moles For- / moles HFor)
10^-0.19 = (moles For- / moles HFor) = 0.646

So moles For- = 0.646 x moles HFor

Since the For- solution and the HFor solution have the same molarity, then

mL For- = 1.4 x mL HFor = 0.646 x 20 mL = 12.92 mL

Part B

Henderson-Hasselbalch:

pH = pKa + log[For-]/[HFor]

pKa = -log(Ka) = 3.74

3.55 - 3.74 = log[For-]/[HFor]

10^-0.19 = [For-]/[HFor] = 0.646; [For-] =0.646 * [HFor]

Because both are of the same molarity concentrations are proportional to the corresponding volumes. So 0.646 * 20 mL = 12.92 mL of [For-] = [NaFor] must be added.

Again we have same concentrations of NaOH and HFor. And the ratio of [For-]/[HFor] at pH 3.55 is still the same as above 0. 646

At equilibrium:
NaFor(eq)/HFor(eq) = 0.646 and HFor(eq) = HFor(o)-NaFor(eq)

NaFor(eq)/(HFor(o)-NaFor(eq)) = 0.646

All of the NaOH reacts to NaFor(eq) so we can write NaOH/(HFor(o)-NaOH) = 0.646

Solving for NaOH: [NaOH] = 0.645/1.645 * 20 mL = 7.84 mL