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)
