Question

How much 6.0M NaOH must be added to .500 L of a buffer that is .0200M acetic acid and .025 M sodium acetate to raise the pH to 5.75 at 25C?

Answer 1

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

pH = pKa + log ([A-] / [HA]) where A- is the conjugate base part of the buffer (acetate ion) and HA is the weak acid part of the buffer (acetic acid). The equation can also be written as

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

Ka for HA = 1.8 x 10^-5; pKa = -log Ka = -log (1.8 x 10^-5) = 4.74

initial moles HA = M HA x L HA = (0.0200)(0.500) = 0.0100 moles HA
initial moles A- = M A- x L A- = (0.0250)(0.500) = 0.0125 moles A-

But what ratio of A- / HA do we need to get a pH of 5.75? Use the Henderson-Hasselbalch equation.

5.75 = 4.74 + log (moles A- / moles HA)
1.01 = moles A- / moles HA
moles A- / moles HA = 10^1.01 = 10.2 . . .so I need to convert some HA to A- by reacting it with OH-.
HA + OH- ==> H2O + A-

moles of HA + moles A- = 0.0100 + 0.0125 = 0.0225
moles A- / moles HA = 10.2

We now have two equations in two unknowns. In eqn 1, we can say moles A- = 0.0225 - moles HA
Substitute that for moles A- in eqn 2 gives

(0.0225 - moles HA) / moles HA = 10.2
0.0225 - moles HA = (10.2)(moles HA)
0.0225 = 11.2 moles HA
moles HA = 0.00201

So I have to add enough 6.0M NaOH to reduce moles HA from 0.0100 to 0.00201. Since HA reacts with OH- in a 1:1 ratio, then moles OH- needed = 0.0100 - 0.00201 = 0.0080 moles OH-

moles OH- = M OH- x L OH-
0.0080 = (6.0)(L OH-)
L OH- = 0.0013 L = 1.3 mL