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![Determine the H . OH, and pOH of a solution with a pH of 4.40 at 25 °C. [OH-] = pOH =](http://img.homeworklib.com/questions/43602ae0-7129-11ea-82a8-b14cdd90d06d.png?x-oss-process=image/resize,w_560)
Homework Answers
1)
use:
[OH-] = Kw/[H+]
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
[OH-] = (1.0*10^-14)/[H+]
[OH-] = (1.0*10^-14)/(1.1*10^-2)
[OH-] = 9.091*10^-13 M
use:
pH = -log [H+]
= -log (1.1*10^-2)
= 1.9586
use:
pOH = -log [OH-]
= -log (9.091*10^-13)
= 12.0414
Answers:
[OH-] = 9.1*10^-13
pH = 1.96
pOH = 12.04
2)
use:
[H+] = Kw/[OH-]
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
[H+] = (1.0*10^-14)/[OH-]
[H+] = (1.0*10^-14)/2.7*10^-6
[H+] = 3.704*10^-9 M
use:
pH = -log [H+]
= -log (3.704*10^-9)
= 8.4314
use:
pOH = -log [OH-]
= -log (2.7*10^-6)
= 5.5686
Answers:
[H+] = 3.7*10^-9
pH = 8.43
pOH = 5.57
3)
POH = 14 - pH
= 14 - 4.4
= 9.6
use:
pH = -log [H+]
4.4 = -log [H+]
[H+] = 3.981*10^-5 M
use:
pOH = -log [OH-]
9.6 = -log [OH-]
[OH-] = 2.512*10^-10 M
Answers:
[H+] = 4.0*10^-5
[OH-] = 2.5*10^-10
pOH = 9.60
4)
use:
PH = 14 - pOH
= 14 - 2.63
= 11.37
use:
pH = -log [H+]
11.37 = -log [H+]
[H+] = 4.266*10^-12 M
use:
pOH = -log [OH-]
2.63 = -log [OH-]
[OH-] = 2.344*10^-3 M
Answers:
[H+] = 4.3*10^-12
[OH-] = 2.3*10^-3
pH = 11.37
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