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 ofNa^+ 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.
