Determine [H3O+] and [OH-] concentrations (in M to two decimal places) for each: a) Pure water at 10 degrees C (Kw=0.293 x 10^-14) Answers:5.41 x 10^-8 M and 5.41 x 10^-8. b) An aqueous solution with a pH of 3.23 Answers: 5.89 x 10^-4 M and 1.70 x 10^-11 M Please show all work
a)
Kw = [H3O+][OH-]
Since water is pure, it will be neutral and [H3O+] = [OH-]
So,
Kw = [H3O+][H3O+]
Kw = [H3O+]^2
0.293*10^-14 = [H3O+]^2
[H3O+] = 5.41*10^-8 M
So,[OH-] = 5.41*10^-8 M
5.41*10^-8 M
5.41*10^-8 M
b)
use:
pH = -log [H3O+]
3.23 = -log [H3O+]
[H3O+] = 5.89*10^-4 M
use:
[OH-] = Kw/[H3O+]
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
[OH-] = (1.0*10^-14)/[H3O+]
[OH-] = (1.0*10^-14)/(5.89*10^-4)
[OH-] = 1.70*10^-11 M
5.89*10^-4 M
1.70*10^-11 M
Get Answers For Free
Most questions answered within 1 hours.