[AL]

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.

[Borek]

Looks OK to me.

[Pradeep]

Still you don't have considered about the medium. you have assumed the system as a solution of strong acid. If there is any un dissociated weak acid, or dissociated weak base ( I mean the buffering reaction) your solution will not be correct. The only thing you have to confirm is that, your assumption.