Answer – Given,
[C10H15ON] = 0.035 M , pH = 11.33
We know,
pH + pOH = 14
so, pOH = 14- pH
            Â
= 14- 11.33
 Â
           =
2.67
So, [OH-] = 10-pOH
               Â
= 10-2.67
                      Â
  = 0.00214 M
  We know
  C10H15ON + H2O
------> C10H15ONH+ +
OH-
IÂ Â
0.035Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0
C
-x                            Â
 Â
+x            Â
+x
E
0.035-x                         Â
+x             Â
+x
We know, at equilibrium, [OH-] = x = 0.00214 M
Part A) [C10H15ON] =
0.035 -x
                            Â
= 0.035-0.00214
     Â
                        =
0.033 M
Part B)
We know, [C10H15ONH+] =
x = 0.0021 M
Part C) , [OH-] = x = 0.0021
MÂ Â Â Â
Part D)
Kb =
[C10H15ONH+][OH-] /
[C10H15ON]
     = 0.00214 *0.00214 / 0.033
      = 1.4*
10-4