PH Math

[peace and love]

I found a math formula to calculate PH of a sodium hydroxide solution.

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 : How to Calculate the PH of NaOH

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?

 

[peace and love]

I was able to check my math using the following formula. I confirmed that my calculations above are correct.

Calculate pH of 100 ml solution containing 0.012 g sodium hydroxide (NaOH).
General information:
NaOH is a strong base that accepts a proton from water to produce sodium (Na+) and hydroxyl (OH-) ions in solution. The following equation can be used to accurately calculate pH of a strong base solution with concentrations above 10-5 M:

pH = 14 + log [OH-]

Molecular weight (molar mass) of NaOH = 40.00 g/ mole

Calculations:
Number of moles of NaOH in solution = 0.012 g ÷ 40.00 (g/ mole) = 0.0003 mole
Volume of NaOH solution in liters = 100 ml × 0.001 (liter/ ml) = 0.1 liter

Molar concentration of NaOH solution = 0.0003 mole ÷ 0.1 liter = 0.003 mole/ liter
pH = 14 + log (0.003) = 14 - 2.52 = 11.48

 

[SnozzleBerry]

Yes, it only takes a minimal amount of lye to raise the pH of water. But these calculations don't really help much for extractions, as the bark buffers the pH of the solution, requiring more lye than your calculations indicate.

This is why it takes more lye for STB than A/B approaches.