Question

Determine the pH during the titration of 38.2 mL of 0.207 M diethylamine ((C₂H₅)₂NH, K_b = 6.9×10^-4) by 0.207 M HNO₃ at the following points. (Assume the titration is done at 25 °C.)
Note that state symbols are not shown for species in this problem.
(a) Before the addition of any HNO₃ _______

(b) After the addition of 14.6 mL of HNO₃ ______

(c) At the titration midpoint ________

(d) At the equivalence point _______

(e) After adding 59.2 mL of HNO________

Please answer all parts and explain the process

Answer 1

millimoles of (CH3)2NH = 38.2 x 0.207 =7.9074‬

Kb= 6.9 x 10^-4

pKb = -logKb =3.161

a) before the addition of any HNO3

pOH = 1/2 [pKb -logC]

         = 1/2 [3.161 -log(0.207)]

         = 1.92

pH + pOH = 14

pH = 12.08

b) after the addition of 14.6 mL HNO3

millimoles of HNO3 = 14.6 x 0.207 = 3.0222

(CH3)2NH + HNO3 --------------> (CH3)2NH2+

7.9074‬               3.0222                         0

4.8852               0                            3.0222

pOH = 3.161 + log (3.0222 / 4.8852)

          =2.95

pH = 11.05

c) titration mid point :

it is half equivalence point . so

pOH = pKb

pOH =3.16

pH +pOH =14

pH = 10.84

d) equivalence point

concentration of salt =7.9074‬/ 76.4 =0.1035 M

pH = 7 - 1/2 (pKb + log C)

     = 7 - 1/2 ( 3.161 + log 0.1035)

pH = 5.91

e) mmoles of HNO3=0.207*59.2= 12.2544‬

=>[H⁺]= (12.2544-7.9074‬)/97.4= 0.0446

pH =-log [H⁺]= 1.35

Your one thumbs up will help me lot. Thanks!