Rabigh's answer is very good, I just want to add an alternative approach to using the dilution calculations equation, C_1V_1 = C_2V_2.
You're dealing with "100 cm"^3 of a 0.1-M NaOH solution. Because sodium hydroxide is a strong base, it will dissociate completely in aqueous solution to give Na^(+) cations and OH^(-) anions.
NaOH_((aq)) -> Na_((aq))^(+) + OH_((aq))^(-)
You can use this solution's molarity to determine the number of moles of sodium hydroxide, which is equal to the number of moles of OH^(-) ions, you have in solution before adding water
C = n/V => n = C * V
n_("NaOH") = "0.1 M" * 100 * 10^(-3)"L" = "0.01 moles NaOH"
When you add the "100 cm"^3 of water, the number of moles of sodium hydroxide remains unchanged; the only thing that changes is the volume of the solution, which implies that the concentration of OH^(-) will change as well.
Think of it like this: same number of moles, twice the volume -> half the initial concentration
C_("new") = n/V_("total") = "0.01 moles"/((100 + 100) * 10^(-3)"L") = "0.05 M"
Use this concentration to determine pOH
pOH = -log([OH^(-)]) = -log(0.05) = 1.3
Therefore, pH is equal to
pH_("sol") = 14 - pOH = 14 - 1.3 = color(green)(12.7)
