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
ONLY in pure water at 25 0C :pH + pOH = 14.00
10.1
pH + pOH =14
3
To find the pOH of a solution, you can use the formula pOH = -log[OH⁻]. Given that [OH⁻] = 1.41 × 10⁻¹³, calculate the pOH: pOH = -log(1.41 × 10⁻¹³) ≈ 12.85. Therefore, the pOH of the solution is approximately 12.85.
In pure water, the hydrogen ion (H+) concentration is 1x10^-7 M and the hydroxide ion (OH-) concentration is also 1x10^-7 M. The pH of pure water is 7 (neutral) and the pOH is also 7.
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
pOH = 14 - pH 5 = 14 - pH 9 = pH
The pOH is 6,4.
5-60c = -55
Using the terms "pOH and pH" is not a way to express the hydroxide and hydronium ion concentrations in pure water at 25 degrees Celsius. At 25 degrees Celsius, the concentrations of hydroxide and hydronium ions in pure water are equal: 1.0 x 10^-7 M.
The temperature at which bacteria in water are typically killed is 140F (60C) or higher.