Question

how much 5.80 M NaOH must be added to 460.0 mL of a buffer that is 0.0200 M acetic acid and 0.0250 M sodium acetate to raise the pH to 5.75

Answer 1

Henderson-Hasselbalch equation is given as:

pH = pK_a + log ([A^-] / [HA])

The equation can also be written as:

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

K_a for acetic acid (HA) = 1.8 x 10^-5; pK_a = -log K_a = -log (1.8 x 10^-5) = 4.74

initial moles HA = (0.02 mol/L)(0.460) = 0.0092 moles HA
initial moles acetate ion (A^-) = (0.0250)(0.460) = 0.0115 moles A^-

5.75 = 4.74 + log (moles A^- / moles HA)
log (moles A^- / moles HA) = 1.01
(moles A^- / moles HA) = 10^1.01 = 10.23

Let us convert some HA to A- by reacting it with OH-.
HA + OH^- ==> H2O + A^-
moles of HA + moles A^- = 0.0092 + 0.0115 = 0.0207

moles A^- = 0.0207 - moles of HA
moles A^- / moles HA = 10.23

0.0207 - moles of HA)/ (moles of HA) = 10.23
0.0207 - moles of HA = 10.23 x (moles of HA)

11.23 moles of HA = 0.0207

moles of HA = 0.00184

moles A^- = 0.0207 - 0.00184 = 0.0188
So we have to add enough 5.8 M NaOH to reduce moles HA from 0.0092 to 0.00184.

Since HA reacts with OH^- in a 1:1 ratio, then moles OH^- needed = 0.0092 - 0.00184= 0.00736 moles OH^-
volume of NaOH must be added = 0.00736 moles/(5.8 mol/L)

                                                     = 0.0012 L = 1.2 mL