Question

How much 5.60 M NaoOH must be added to 630.0 mL of a buffer that is 0.0205 M acetic acid and 0.0275 M sodium acetate to raise the pH to 5.75?

Answer 1

The pH of a buffer solution can be calculated using Henderson-Hasselbalch equation given by

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 acetic acid = 1.8 x 10^-5;

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

initial moles CH3COOH = Molarity of CH3COOH x Vol .of CH3COOH = (0.0205)(0.630) = 0.0120 moles CH3COOH
initial moles of CH3COONa = Molarity of CH3COONa x Vol .of CH3COONa = (0.0275)(0.630) = 0.0173 moles CH3COONa

The ratio of A- / HA needed to get a pH of 5.75 is

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 = 10 . . .so we need to convert some HA to A- by reacting it with OH-.
HA + OH- ==> H2O + A-

moles of HA + moles A- = 0.0120 + 0.0173 = 0.0293
moles A- / moles HA = 10

We now have two equations in two unknowns. In eqn 1, we can say moles CH3COONa = 0.0293 - moles CH3COOH
Substitute that for moles A- in eqn 2 gives

(0.0293 - moles HA) / moles HA = 10
0.0293 - moles HA = (10.0)(moles HA)
0.0293 = 11.0 moles HA
moles HA = 0.00266

So we have to add enough 5.60M NaOH to reduce moles HA(CH3COOH) from 0.0120 to 0.00266. Since CH3COOH reacts with OH- in a 1:1 ratio, then moles OH- needed = 0.0120 - 0.00266 = 0.00934 moles OH-

moles OH- = M OH- x L OH-
0.00934 = (5.60)(L OH-)
L OH- = 0.0017 L = 1.7 mL

The volume of 5.60 M NaOH that must be added to raise the pH is 1.7 mL