Assuming complete dissociation, what is the ph of a 4.82 mg/l ba(oh)2 solution?

Ba(OH)₂ is a strong base and completely ionises into its corresponding ions
Ba(OH)₂ ---> Ba²⁺ + 2OH⁻
1 mol of Ba(OH)₂gives 2 mol of OH⁻ ions. 
molarity of Ba(OH)₂ is 4.82 x 10⁻³ g/L / 171.3 g/mol - 2.81 x 10⁻⁵ mol/L
since 1 mol of Ba(OH)₂gives 2 mol of OH⁻ ions. 
therefore [OH⁻] = 2[Ba(OH)₂]
[OH⁻]  = 2 x 2.81 x 10⁻⁵ mol/L = 5.62 x 10⁻⁵ M
pOH can be calculated using the OH⁻ concentration
pOH = -log [OH⁻]
pOH = - log (5.62 x 10⁻⁵ M)
pOH = 4.25
pH can be calculated as follows
pH + pOH = 14
pH = 14 - 4.25 
pH = 9.75
pH of solution is 9.75