Question

1. (4 points) Complete the table below at 25 °C. | Ka рка Кь pKb 0.000092 1.13 0.00711 13.28

Answer 1

ANSWER

Ka	pKa	Kb	pKb
9.96	1.1 $\times$ 10 -10	0.000092	4.04
1.3 $\times$ 10 -13	12.87	0.074	1.13
0.00711	2.15	1.4 $\times$ 10 -12	11.85
5.2 $\times$ 10 -14	13.28	0.19	0.72

PART 1

Kb = 0.000092

We have relation, pKb = - log Kb

$\therefore$ pKb = - log 0.000092

pKb = 4.036

ANSWER : pKb = 4.04

We have relation, pKa + pKb = 14

$\therefore$ pKa = 14 - pKb

pKa = 14 - 4.04

pKa = 9.96

ANSWER : pKa = 9.96

We have relation, pKa = - log Ka

$\therefore$ Ka = 10 - pKa = 10 - 9.96 = 1.096 $\times$ 10 -10

ANSWER : Ka = 1.1 $\times$ 10 -10  

PART 2

We have relation, pKa + pKb = 14

$\therefore$ pKa = 14 - pKb

pKa = 14 -1.13

pKa = 12.87

ANSWER : pKa = 12.87

We have relation, pKa = - log Ka

$\therefore$ Ka = 10 - pKa = 10 - 12.87 = 1.348 $\times$ 10 -13

ANSWER : Ka = 1.3 $\times$ 10 -13   

We have relation, pKb = - log Kb

$\therefore$ Kb = 10 - pKb = 10 - 1.13 =0.074

ANSWER :  Kb = 0.074

PART 3

We have relation, pKa = - log Ka

$\therefore$ pKa = - log 0.00711

pKa = 2.15

ANSWER :pKa = 2.15

We have relation, pKa + pKb = 14

$\therefore$ pKb = 14 - pKa

pKb = 14 - 2.15

pKb = 11.85

ANSWER : pKb = 11.85

We have relation, pKb = - log Kb

$\therefore$ Kb = 10 - pKb = 10 - 11.85 =1.4125 $\times$ 10 -12

ANSWER :  Kb = 1.4 $\times$ 10 -12  

PART 4

We have relation, pKa = - log Ka

$\therefore$ Ka = 10 - pKa = 10 - 13.28 = 5.2481 $\times$ 10 -14

ANSWER : Ka = 5.2 $\times$ 10 -14

We have relation, pKa + pKb = 14

$\therefore$ pKb = 14 - pKa

pKb = 14 -13.28

pKb = 0.72

ANSWER : pKb = 0.72

We have relation, pKb = - log Kb

$\therefore$ Kb = 10 - pKb = 10 - 0.72 =0.1905

ANSWER :  Kb = 0.19