I am trying to figure out the amount of 12.8M NaOH required to bring water from pH 7.2 to pH 10. Below is what I have done so far, but I am not sure if this is the right direction. I appreciate any help.
@ pH 7.2,
[H+] = 6.31*10-8
[OH-] = 1.6*10-4
@ pH 10
[H+] = 10-10
[OH-] = 10-4
I subtracted the [OH-] at pH 10 from pH 7.2
[OH-]10 - [OH-]7.2 = 9.98*10-5
Took that number and divided by concentration of NaOH to get the amount of 12.8M NaOH needed to change the pH of one liter of water.
9.98*10-5/12.8M = 7.8*10-6 liters needed.