Rabigh's answer is very good, I just want to add an alternative approach to using the dilution calculations equation, C1V1=C2V2.
You're dealing with 100 cm3 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=nV⇒n=C⋅V
nNaOH=0.1 M⋅100⋅10−3L=0.01 moles NaOH
When you add the 100 cm3 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
Cnew=nVtotal=0.01 moles(100+100)⋅10−3L=0.05 M
Use this concentration to determine pOH
pOH=−log([OH−])=−log(0.05)=1.3
Therefore, pH is equal to
pHsol=14−pOH=14−1.3=12.7
