Question

25.0mL of a solution of (CH3CH2)2NH, 0.400M with HCl 0.2000M. If kb (CH3CH2)2NH = 8.6x10^-4 ¿What is the pH on 0.00mL, 25.00mL and on 50.00mL of the titulante?

Answer 1

Kb = 8.6 x 10^-4

pKb = -logKb = -log (8.6 x 10^-4)

pKb = 3.07

pKa + pKb = 14

pKa = 10.93

a) before addition of any base   (0.00 mL)

pOH = 1/2 [pKb - log C]

pOH = 1/2 [3.07 -log 0.4]

pOH = 1.73

pH + pOH = 14

pH = 12.27

b) 25 mL HCl added :

B = 25 x 0.4 = 10

H+ = 25 x 0.2 = 5

it is half eqivalnce point . here pH = pKa

pH = 10.93

c) 50 mL HCl added

B = 25 x 0.4 = 10

millimoles of H+ = 50 x 0.2 = 10

it is equivalence point here BH+ left = 10 / 75 = 0.133 M

pH = 7 -1/ 2[pKb + log C]

pH = 7 -1/2 [ 3.07 -log 0.133 ]

pH = 5.03