Esercizi svolti stechiometria - parte 2/3

13.1

2?OH^-

[H₃O⁺] = 3,40·10^-2

[H₃O⁺] [OH^-] = 10^-14

[OH^-] = 10^-14

[H₃O⁺] = 9,94·10^-13 M

3,40·10^-2

13.2

2?OH^-

[OH^-] = 2,15·10^-13 M

[H₃O⁺] [OH^-] = 10^-14

[H₃O⁺] = 10^-14

2,15·10^-13

4,65·10^-2 M

13.3

?pH

[H₃O⁺] = 8,40·10^-5 M

pH = -log (8,40·10^-5) = 4,0767

13.4

?pH

[OH^-] = 2,40·10^-4 M

due svolgimenti:

• K_w = [H₃O⁺][OH^-] = 10^-14
• [H₃O⁺] = 10^-14
• 2,40·10^-4 = 4,16·10^-11 M

pH = -log (4,16·10^-11) = 10,380

pOH = -log (2,40·10^-4) = 3,62

pH = 14 - pOH = 10,380

13.5

• pOH

[H₃O⁺] = 2,50 M

pH = -log 2,50 = -0,398

13.6

• pH

[OH^-] = 2,20 M

Due svolgimenti

[H₃O⁺] = ^10-14⁄_{2,20 M} = 4,5 · 10^-15

pH = -log (4,5 · 10^-15) = 14,342

pOH = -log (2,20) = 0,342

pH = 14 - 0,342 = 14,342

13.7

[H₃O⁺]

pH = 2,54

[H₃O⁺] = 10^-2,54 = 3,0 · 10^-3 M

13.8

[OH^-]

pH = 11,04

pOH = 14 - 11,04 = 2,96

[OH^-] = 10^-2,96 = 1,1 · 10^-3 M

Oppure

[H₃O⁺]_10^-11,04 = 3,288 · 10^-12 M

[OH^-] = _10^-14⁄^{3,288 · 10-12} = 3,1 · 10^-3 M

• K_w = [H₃O⁺][OH^-] = 1,0 · 10^-14 (25°C)
• [H₃O⁺] = 10^-pH M [acide] [H₃O⁺] > [OH^-]
• [OH^-] = 10^-pOH M [basiche] [H₃O⁺] < [OH^-]
• pH = -log [H₃O⁺] basiche pH > 7, neutre pH = 7, acide pH < 7
• pOH = -log [OH^-] = 14 - pH
• [OH^-] = _10^-14⁄^[H₃O+]
• [H₃O⁺] = _10^-14⁄^[OH-]

13.5 ACIDI E BASI DEBOLI MONOPROTICI

(13.16)

pH 9% specie 21’ acido formico K_a = 1,8 * 10^-4

HCOOH + H₂O ⇌ H₃O⁺ + HCOO^-

l x x

trascur

l x²

1,8 * 10^-4

3,5 * 10^-2 = x

x = 1,8 * 10^-3 [HCOOH] = 1,9831 M

[H₃O⁺] = [HCOO^-] = 1,87 * 10^-3 M

pH = -log (1,87 * 10^-3) = 1,7283

(13.17)

5,2 * 10^-2 M (HCOOH) pH = 3,907

pK_ow

HCOOH + H₂O ⇌ H₃O⁺ + COO^-

1 x x

trascur

ma [H₃O⁺] 10^-pH

10^-3,907 x 1,94 * 10^-4

5,19 * 10^-2

pK_ow = (1,94 * 10^-4)²

5,19 * 10^-2

= 2,56 * 10^-8

(13.18)

?NH₂ pH = 12,04; K_b = 9,6 * 10^-4

(C₂H₅)₂NH + H₂O ⇌ OH^- + (C₂H₅)₂NH₂⁺

[OH^-] = K_w

[H₃O⁺] = 10

(14-12,04)

= 10^-2,04

[H₃O⁺] = 10^-14,23

= 9,7 * 10^-13

[OH^-] = 10^-14

9,6 * 10^-4 9,7 * 10^-13 = 1,03 * 10^-2

[OH^-] (C₂H₅)₂NH₂⁺] = 1,03 * 10^-2 M

eq. (C₂H₅)₂NH ⇌ 4,03 * 10

(inizia-x)

9,6 * 10^-4

= 4,03 * 10^-2

(x * 1,03 * 10^-2)

x = 1,24 * 10^-3 M concentrazione iniziale (C₂H₅)₂NH

14.2

0,85 M (CH₃COONa) K_a=1,76·10^-5 pH % idrolisi

CH₃COONa → CH₃COO^- + Na⁺

CH₃COO^- + H₂O ← CH₃COOH + OH^-

eq 0,85 - x x x

K_b = 2,10·10^-14 / 1,76·10^-5 = 5,68·10^-10

5,68·10^-10 = x² / 0,85 - x

x = [CH₃COOH] = [OH^-] = 2,20·10^-5

pOH = -log (2,20·10^-5) = 4,658

pH = 14 - 4,658 = 9,342

% idr = [CH₃COOH]_eq / [CH₃COO^-]_inizio = 100 = 2,59·10^-3 %

14.3

? pH NH₄Cl 2,42·10^-1 M K_b (NH₃) = 1,49·10^-5

NH₄Cl → NH₄⁺ + Cl^-

NH₄⁺ + H₂O ← NH₃ + H₃O⁺ idrolisi

eq 2,42·10^-1 - x x x

K_h = 2,10·10^-14 / 1,49·10^-5 = 5,59·10^-10

5,59·10^-10 = x² / 2,42·10^-1 - x

x = 1,16·10^-5 [H₃O⁺] [NH₃]

pH = -log (1,16·10^-5) = 4,936

15.3 Formula generale idrolisi

[H₃O⁺] = √K_wK_a

→ solo

→ K_w e K_a poco che ([A^2-] e [H₂A] trascurabili rispetto [HA^-])

[Sale] ≫ K_w e [Sale] ≫ K₁

15.5

• pH QRM (Na HASc)
• K_f = 9,94·10⁻⁵
• K_e = 1,60·10⁻¹²

Na HASc ⟶ Na⁺ + HASc^-

HASc^- + H₂O ⟶ HSc^2- + H₃O⁺

HASc^- + H₂O ⟶ H₂ASc + OH^-

K_w/