The answer is #5.5 * 10^(-14)#.
#NaOH# is a strong base, which means it dissasociates completely into #Na^+# and #OH^-#.
#NaOH(aq) + H_2O(l) <=> Na^+ (aq) + OH^(-) (aq)#
Starting from the given mass of #8.6# grams of #NaOH# and knowing that its molar mass is #40 g/(m o l e)#, we get
#n_(NaOH) = (8.6 g)/(40 g/(m o l e)) = 0.215# moles.
The molar concentration of #NaOH# is
#C = n/V = (0.215 mo l e s)/(1.2 L) = 0.18 M#
Now, a strong base (and a strong acid for that matter) dissasociates completely, which means that the concentrations of #Na^+# and #OH^-# in aqueous solution are equal to the initial concentration of #NaOH#, #0.18 M#.
Therefore, we can calculate the #pOH# using
#pOH = -log([OH^-]) = -log(0.18) = 0.74#
We can determine the #pH# by using
pH #= 14 - pOH = 14 - 0.74 = 13.26#
This means that the #[H^+]# is equal to
#[H^+] = 10^(-pH) = 10^(-13.26) = 5.5 * 10^(-14)#
