Assuming complete dissociation, what is the pH of a 4.87 mg/L Ba(OH)2 solution

Assuming complete dissociation, what is the pH of a 4.87 mg/L Ba(OH)2 solution?

convert mg/L to g/L (multiplying by "1g/1000mg")
then convert g/L to mol/L (by dividing by the molas mass of Ba(OH)2)

Ba(OH)2 -> Ba^2+ + 2OH-
Multiply the molarity by two (since when it dissociate, you get 2 OH- ions per Ba(OH)2)

pH= 14- pOH
pOH= -log [OH-]

find pOH by doing -log of the concentration of OH (the molarity found above)
then subtract that from 14.

As a check, make sure the pH is greater that 7 (since it's a base).