Neutralization
Of strong acids and bases
Example1
• 1- How many ml of 0.025 M H2SO4 are
required to neutralize exactly 525 ml of 0.06
M KOH?
• 2- What is the pH of the neutralized solution?
No. of moles (equivalents) of H+ required = no.
of moles (equivalents) of OH- present
Lacid × Nacid = no. of equivalents
Lacid × Nacid = Lbase × Nbase
Cont’ed
H2SO4 = 0.025 M = 0.05 N
Lacid × 0.05 = 0.525 × 0.06
Lacid = (0.525 × 0.06) / 0.05
Lacid = 0.63
Acid required = 630 ml
The neutralized solution contains only K2SO4 “a
salt” of a strong acid and strong base has no
effect on pH
pH =7
Example 2
• How many ml of 0.05 N HCl are required to
neutralize exactly 8g of NaOH?
At the equivalent point:
The no. of moles H+ added = no. of moles OHpresent
Lacid × Nacid = no. of (moles) equivalents of H+
added
wtNaOH / MwtNaOH = no. of moles of NaOH (OH-)
present
Cont’ed
Lacid × Nacid = wtNaOH / MwtNaOH
Lacid × 0.05 = 8 / 40
Lacid = 8 / 2 = 4 L or 4000 ml
Buffers
Buffers Are Mixtures of Weak Acids
and Their Conjugate Bases
Definition of a buffer system
• Buffers are aqueous systems that tend to
resist changes in pH when small amounts of
acid (H) or base (OH) are added.
• A buffer system consists of a weak acid (the
proton donor) and its conjugate base (the
proton acceptor).
FIGURE 2–19 The acetic acid–acetate pair
as a buffer system
The Handerson Hasselbalch
equation
• The Henderson-Hasselbalch equation also
allows us to:
(1) calculate pKa, given pH and the molar ratio
of proton donor and acceptor
(2) Calculate pH, given pKa and the molar ratio
of proton donor and acceptor
(3) calculate the molar ratio of proton donor and
acceptor, given pH and pKa.
Preparation of buffers
• Example 1: What is the concentration of acetic
acid and acetate in 0.2 M acetate buffer,
which has a pH = 5 and pKa = 4.77
Acetate buffer
Acetic acid + Acetate
HA
+ A[HA] + [A-] = 0.2 M
[HA] = ?
[A-] = ?
HA
Ka =
H+ + A-
[H+] [A-]
[HA]
pKa = - log Ka
4.77 = - log Ka
log Ka = anti log – 4.77
Ka = 1.7 × 10-5
pH = - log [H+]
5 = - log [H+]
[H+] = anti log – 5
[H+] = 1 × 10-5
Let us assume [A-] = y
Since [HA] + [A-] = 0.2 M
[HA] = 0.2 – y
+] [A-]
[H
Ka =
[HA]
1.7 × 10-5 = [(1 × 10-5)(y)] / (0.2 – y)
1.7 × 10-5 (0.2 – y) = 1 × 10-5 y
(3.4 × 10-6) – (1.7 × 10-5 y) = 1 × 10-5 y
3.4 × 10-6 = 1 × 10-5 y + 1.7 × 10-5 y
3.4 × 10-6 = 2.7 × 10-5 y
y = (3.4 × 10-6 / 2.7 × 10-5)
y = 0.126 M = [A-]
[HA] = 0.2 – 0.126 = 0.074 M
Example 2
• Describe the preparation of 3 L of 0.2 M
acetate buffer. Starting from solid sodium
acetate trihydrate (A-), Mwt = 136 and a 1 M
solution of acetic acid (HA) the pKa = 4.77; The
concentration of [A-] = 0.126 M, [HA] = 0.074
M in 0.2 M solution in 1 L.
The no. of moles in buffer = 3 × 0.2 = 0.6 moles
The no. of moles of A- + the no. of moles of HA
= 0.6 moles
SINCE the concentration of [A-] = 0.126 M in 1
L; the Total no. of moles in buffer = 0.126 × 3 =
0.378 moles
SINCE the concentration of [HA] = 0.073 M in 1
L; the Total no. of moles in buffer = 0.073 × 3 =
0.222 moles OR The no. of moles of HA = 0.6 no. of moles of A- = 0.222 moles
SINCE A- is solid the wt needed = M × Mwt =
0.378 × 136 = 51.4 g
The volume if HA needed = no. of moles / M =
0.222 / 1 = 0.222 L = 222 ml
51.4 g of solid sodium acetate trihydrate is
added to 222 ml of acetic acid and the volume
is brought up to 3 L.