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Determining pH pOH [H+] [H3O+] [OH-]

 

Here are the equations you could use.

pH + pOH=14

pH =-log[H+]   [H+]=10-pH

pOH=-log[OH-]    [OH-]=10-pOH

Kw=1.0 x 10-14 = [H3O+] [OH»]

Here is a table that needs to be complete.

Example

pH

pOH

[H+]

[OH-]

A

6.1

 

 

 

B

 

3.7

 

 

C

 

 

1.5 x 10-6 M

 

D

 

 

 

8.1 x 10-4M

 

Example A

     If pH= 6.1 and pOH=14-pH          pOH=14-6.1=7.9

    [H+]=10-pH   so  [H+]=10-6.1              [H+]=7.9x10-7M

    [OH-]=10-pOH so  [OH-]=10-7.9      [OH-]=1.2x10-8M

Example B

     If pOH= 3.7 and pH=14-pOH          pH=14-3.7=10.3

    [H+]=10-pH   so  [H+]=10-10.3              [H+]=5.0x10-11M

    [OH-]=10-pOH so  [OH-]=10-3.7      [OH-]=2.0x10-4M

Example C

     If [H+]=1.5x10-6M and pH =-log[H+

     pH =-log1.5x10-6M                                 pH=5.82

    If pH= 5.82 and pOH=14-pH          pOH=14-5.82=8.18

    [OH-]=10-pOH so  [OH-]=10-8.18      [OH-]=6.6x10-9M

Example D

     If  [OH-]=8.1x10-4M and pOH =-log[OH-]

     pOH =-log8.1x10-4M                                 pOH =3.09

     If pOH= 3.09 and pH=14-pOH          pH=14-3.09=10.91

     [H+]=10-pH   so  [H+]=10-10.91             [H+]=1.2x10-11M

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