1. Neutralization Of strong acids and bases

2. Example1 1- How many ml of 0.025 M H 2 SO 4 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 L acid × N acid = no. of equivalents L acid × N acid = L base × N base

3. Cont’ed H 2 SO 4 = 0.025 M = 0.05 N L acid × 0.05 = 0.525 × 0.06 L acid = (0.525 × 0.06) / 0.05 L acid = 0.63 Acid required = 630 ml The neutralized solution contains only K 2 SO 4 “a salt” of a strong acid and strong base has no effect on pH pH =7

4. 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 - present L acid × N acid = no. of (moles) equivalents of H + added wt NaOH / Mwt NaOH = no. of moles of NaOH (OH - ) present

5. Cont’ed L acid × N acid = wt NaOH / Mwt NaOH L acid × 0.05 = 8 / 40 L acid = 8 / 2 = 4 L or 4000 ml

6. Buffers Buffers Are Mixtures of Weak Acids and Their Conjugate Bases

7. 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).

8. FIGURE 2–19 The acetic acid–acetate pair as a buffer system

9. The Handerson Hasselbalch equation

12. 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.

14. 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 pK a = 4.77 Acetate buffer Acetic acid + Acetate HA + A - [HA] + [A - ] = 0.2 M

15. [HA] = ? [A - ] = ? HA H + + A - K a = pK a = - log K a 4.77 = - log K a log K a = anti log – 4.77 K a = 1.7 × 10 -5 pH = - log [H + ] 5 = - log [H + ] [H + ] = anti log – 5 [H + ] = 1 × 10 -5 [H + ] [A - ] [HA]

16. Let us assume [A - ] = y Since [HA] + [A - ] = 0.2 M [HA] = 0.2 – y K a = 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 [H + ] [A - ] [HA]

17. 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 pK a = 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

18. 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

19. 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.