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- presentLacid × Nacid = no. of equivalents

Lacid × Nacid = Lbase × Nbase

Cont’ed

H2SO4 = 0.025 M = 0.05 NLacid × 0.05 = 0.525 × 0.06Lacid = (0.525 × 0.06) / 0.05

Lacid = 0.63Acid required = 630 ml

The neutralized solution contains only K2SO4 “a salt” of a strong acid and strong base has no

effect on pHpH =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 OH-

presentLacid × 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 + AcetateHA + A-

[HA[ + ]A- = ]0.2 M

[HA? = ][A-? = ]

HA H+ + A-

Ka=

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

[H+[ ]A-][HA]

Let us assume [A-] = ySince [HA] + [A-] = 0.2 M

[HA] = 0.2 – y Ka =

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 y3.4 × 10-6 = 1 × 10-5 y + 1.7 × 10-5 y

3.4 × 10-6 = 2.7 × 10-5 yy = (3.4 × 10-6 / 2.7 × 10-5)

y = 0.126 M = [A-][HA = ]0.2 – 0.126 = 0.074 M

[H+[ ]A-][HA]

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.