I found a math formula to calculate PH of a sodium hydroxide solution.
Quote: 1
Divide the mass of NaOH in the solution by the molar weight (40) of the compound to calculate the number of moles. For example, if you have 2 g of NaOH dissolved in 500 ml of water, then the formula is 2 / 40 = 0.05 moles.
2
Divide the number of moles of NaOH by the volume of the solution (in liters) to calculate the molar concentration of NaOH. In this example, the volume of the solution is 500 ml, or 0.5 L; the molar concentration is 0.05 / 0.5 = 0.1
3
Divide the number 10^(-14) by the molar concentration of NaOH to calculate the concentration of hydrogen ions (H+) in the solution. In this example, the concentration of H+ is 10^(-14) / 0.1 = 10^(-13).
4
Take the logarithm of the concentration of hydrogen ions, and then multiply it by -1 to calculate the pH. In this example, pH is (-1) x log(10^-13) = -1 x (-13) = 13.
Read more :
http://www.ehow.com/how_...4_calculate-ph-naoh.html In the example, 2g of sodium hydroxide dissolved in 0.5L of water is PH 13.
According to my calculations using this formula, it would only take 0.2mg of sodium hydroxide to bring 0.5L of water up to PH 9.
Am I doing this right? Is such a tiny amount of sodium hydroxide really enough to make the water PH 9?