If your given pH or pOH, you can also find [H+] or [OH-]
use antilog
The pOH is 6,4.
pH + pOH = 14. So pOH = 14 - 6 = 8 pOH = -log[OH-] [OH-] = 10-8 M
pH= -log[H+] pH + pOH = 14 pOH = 14 - pH pOH= -log[OH], so the antilog of -pOH will give you the OH concentration.
The pOH is the negative log of the OH- concentration. Thus, pOH = -log 2.0x10^-2pOH = 1.699 = 1.7
1.7
pH + pOH = 14. So pOH = 14 - 1.12 = 12.88 pOH = -log[OH-] [OH-] = 1.31 x 10-13 M
The pOH is 6,4.
pH + pOH = 14. So pOH = 14 - 6 = 8 pOH = -log[OH-] [OH-] = 10-8 M
poh=14-ph
pH= -log[H+] pH + pOH = 14 pOH = 14 - pH pOH= -log[OH], so the antilog of -pOH will give you the OH concentration.
12.85 is the pOH.
[OH-] = 3.31 log[OH-] = pOH = .51982 14-pOH = pH = 13.48
The pOH is the negative log of the OH- concentration. Thus, pOH = -log 2.0x10^-2pOH = 1.699 = 1.7
The pOH is the negative log of the OH- concentration. Thus, pOH = -log 2.0x10^-2pOH = 1.699 = 1.7
1.7
Ist step, calculate pOH value by using formula pH + pOH = 14 2nd step, pOH = -log[OH], [OH] = - Antilog of pOH
pH + pOH = 14. So pOH = 14 - 10.95 = 3.05 pOH = -log[OH-] [OH-] = 8.91 x 10-4 M