Question

Determine the [OH−] , pH, and pOH of a solution with a [H+] of 0.0018 M at 25 °C.

[OH−]=

pH=

pOH=

Determine the [H+] , pH, and pOH of a solution with an [OH−] of 7.3×10−7 M  at 25 °C.

[H+]=

pH=

pOH=

Determine the [H+] , [OH−] , and pOH of a solution with a pH of 3.67 at 25 °C.

[H+]=

[OH−]=

pOH=

Determine the [H+] , [OH−] , and pH of a solution with a pOH of 5.18 at 25 °C.

[H+]=

[OH−]=

pH=

Answer 1

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)/(1.8*10^-3)
[OH-] = 5.556*10^-12 M

use:
pH = -log [H+]
= -log (1.8*10^-3)
= 2.7447

use:
pOH = -log [OH-]
= -log (5.556*10^-12)
= 11.2553

Since pH < pOH, this is acidic in nature

Answers:
[OH-] = 5.6*10^-12
pH = 2.74
pOH = 11.26

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)/7.3*10^-7
[H+] = 1.37*10^-8 M

use:
pH = -log [H+]
= -log (1.37*10^-8)
= 7.8633

use:
pOH = -log [OH-]
= -log (7.3*10^-7)
= 6.1367

Since pH > pOH, this is basic in nature

Answers:
[H+] = 1.4*10^-8
pH = 7.86
pOH = 6.14

3)

POH = 14 - pH
= 14 - 3.67
= 10.33

use:
pH = -log [H+]
3.67 = -log [H+]
[H+] = 2.138*10^-4 M

use:
pOH = -log [OH-]
10.33 = -log [OH-]
[OH-] = 4.677*10^-11 M

Since pH < pOH, this is acidic in nature

Answers:
[H+] = 2.1*10^-4
[OH-] = 4.7*10^-11
pOH = 10.33

4)
use:
PH = 14 - pOH
= 14 - 5.18
= 8.82

use:
pH = -log [H+]
8.82 = -log [H+]
[H+] = 1.514*10^-9 M

use:
pOH = -log [OH-]
5.18 = -log [OH-]
[OH-] = 6.607*10^-6 M

Since pH > pOH, this is basic in nature

Answers:
[H+] = 1.5*10^-9
[OH-] = 6.6*10^-6
pH = 8.82