pH = -log(concentration of H+ ion) which indicates acid
pOH = -log(concentration of OH- ion) which indicates base
In pure water at 25 degree Celsius, concentration of H+ and OH- is equal i.e. 10^-7 mol/dm^3
so, the sum of pH and pOH is -log(10^-7) + -log(10^-7) = 7 + 7 = 14
Ultrapure water is neutral with pH=7.
ONLY in pure water at 25 0C :pH + pOH = 14.00
10.1
pH + pOH =14
3
pH + pOH = 14. So pOH = 14 - 1.12 = 12.88 pOH = -log[OH-] [OH-] = 1.31 x 10-13 M
I will assume you are asking about the pH of pure water if pKw is 14.26. The relationship between pH, pOH, and pKw is as follows: pH + pOH = pKw. If it is pure, neutral water (no acids or bases present), then pH = pOH, so: pH + pOH = 14.26 2(pH) = 14.26 pH = pOH = 7.13
ONLY in pure water at 25 0C :pH + pOH = 14.00
The pOH is 6,4.
Average temperature is 60C
its 7
10.1
5-60c = -55
5-60c in faruenhit = -55
5-60c in farnhiet = -55
60C
It can be anywhere in between 60C and 100C (boiling point).
The LCM is 60c.