In: Chemistry
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?
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