Question

At 25C what are the molar H3O + and OH- concentrations in:

a. 0.0770 M C6H5 (Ka=6.28e-5)?

b. 0.0770 M HN3 (Ka=2.2e-5)?

c. 0.480 M hydroxylamine hydrochloride (Kb=9.09e-9)?

Answer 1

For weak acids HA ---------> H+ + A-

C 0 0 initial

C(1-x) Cx Cx

Ka = Cx.Cx/C(1-x) as x is tto samll Ka = Cx² or x = (Ka/C)^1/2

thus [H+] = (Ka.C)^1/2

[H+] = [ka x C]^1/2

a) C= 0.077M and Ka = 6.28x 10^-5

Thus [H+] = (0.077x 6.28x 10^-5)^1/2

=2.199 x10^-3 M

and [OH-] = Kw / [H+]

= 1.0x10^-14/2.199 x10^-3 M

= 4.547x10^-12M

b) similarly for Hn3

[H+} = (0.077x 2.2x10^-5)^1/2

=1.3015x10^-3 M

and [OH-] = 7.6x10^-12 M

c) kb = 9.09x10^-9 and hence ka = kw/kb

ka = 1.1x 10^-6

since kb is given you can directly calculate [OH-] and then [H+]

Thus [OH-] = (kbxc)^1/2

   = (0.48M x 9.90x10^-9)^1/2

   = 6.6x10^-4 M

and [H=] = 1.5 x10^-11 M