pH of Solution. Help Chemistry?

This is a very interesting question.
Refer to you previous question with 4 dilutions.
The [HCl] was calculated as 1*10^-5M
pH was calculated :
pH = -log [H+]
pH = 5.00

You are now dealing with a basic NaOH solution.
Now with 6 dilutions , [NaOH ] = 1*10^-7M
Do you think that it is correct to say:
[OH-] = 1*10^-7
And you can calculate [H+]
[H+] *[OH-] = 10^-14
[H+] = 10^-14 / 10^-7
pH = -log 1*10^-7
pH = 7.00
This is obviously not correct because if the solution has pH = 7.00 it must be pure water. But this is definitely a dilute solution of NaOH It must have pH > 7.00 due to the presence of the OH- ions in solution.
How then do you calculate the pH of the solution.
At these extremely low concentrations, it is necessary to take into account the contribution of [H+] and [OH-] from the auto-ionisation of the water. These OH- ions add to the OH- ions from the NaOH resulting in a [OH-] > 1*10^-7
The calculation of the exact [OH-] in the solution is complicated. But you get a close approximation if you say
[OH-] from NaOH = 1*10^-7M
[OH-] from H2O = 1*10^-7M
Total [OH-] = 2*10^-7M
To calculate pH , calculate [H+]
[H+] = 10^-14 / ( 2*10^-7)
[H+] = 5*10^-8M
pH = -log 5*10^-8
pH = 7.30

In each time of dilution, 10 mL of the solution is used to make 100 mL of diluted solution.
Dilution factor in each time = 10/100 = 1/10
Concentration of NaOH for the final diluted solution = (0.1 M) × (1/10)⁴ = 1 × 10⁻⁵ M

Each mole of NaOH ionizes in aqueous solution to give 1 mole of OH⁻ ions.
[OH⁻] in the final diluted solution = 1 × 10⁻⁵ M
pOH = -log[OH⁻] = -log(1 × 10⁻⁵) = 5
pH = pKw - pOH = 14 - 5 = 9