Determining PH POH [H+] [H3O+] [OH-]

Có thể bạn quan tâm

KentChemistry HOME

Custom Search

Back to Acid Base Links

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 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

Back to Acid Base Links

Chemical Demonstration Videos

Từ khóa » H30 X Oh