The answer is pH = #13.26#.

#NaOH# is a strong base, which means that it dissociates completely to #Na^+# and #OH^-#when placed in water. This also gives us an estimate of its pH, since strong basic solutions usually have a very high #pH#.

#NaOH(aq) <=> Na^+(aq) + OH^(-)(aq) #

Complete dissociation means that the concentrations of#Na^+# cations and #OH^-# anions will be equal to the initial concentration of #NaOH#. We can determine that by calculating the number of #NaOH# moles found in #8.6g#.

#n_(NaOH) = m/(molar mass) = (8.6g)/(40 g/(mol)) = 022 # moles

Therefore, #NaOH# molarity is

#C = n/V_(solution) = (0.22 mol es)/(1.2L) = 0.18M#

We know that #[OH^-]# (the concentration of #OH^-#) is equal to this value, so #[OH^-]# = #0.18M#.

We can now calculate #pOH# by

#pOH = -log([OH^-]) = -log(0.18) = 0.74#

Therefore, the solution's #pH# is

#pH = 14 - pOH = 14 - 0.74 = 13.26#, which matches our initial estimate of a very high #pH#.