Question

How much 5.50 M NaOH must be added to 500mL of a buffer that is .0180 M Acetic Acid and .0250 M sodium acetate to raise the pH of 5.75?

Answer 1

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 = (0.0180)(0.500) = 0.009 moles HA
initial moles A- = M A- x L A- = (0.0250)(0.500) = 0.0125 moles A-

Using 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 .
HA + OH- ==> H2O + A-

moles of HA + moles A- = 0.0090 + 0.0125 = 0.0215
moles A- / moles HA = 10.2

Thus tw equations with two unknown values, In eqn 1, we can say moles A- = 0.0225 - moles HA
Substitute that for moles A- in eqn 2 gives

(0.0215 - moles HA) / moles HA = 10.2
0.0215 - moles HA = (10.2)(moles HA)
0.0215 = 11.2 moles HA
moles HA = 0.00192

Hence we have to add enough 6.0M NaOH to reduce moles HA from 0.0100 to 0.00192. Since HA reacts with OH- in a 1:1 ratio, then moles OH- needed = 0.0100 - 0.00192 = 0.00808 moles OH-

moles OH- = M OH- x L OH-
0.0080 = (5.50)(L OH-)
L OH- = 0.00146 L = 1.46 mL