A 0.0133mol sample of Ba(OH)2 is dissolved in water to make up 1.5L of solution. What is the pH of the solution at 25.0∘C? ?

Answer 1

Ba(OH)2 = Ba2+ + 2OH-

moles of Ba(OH)2 = 0.0133mol 

moles of OH- = 2* 0.0133mol  = 0.0266 moles 

in 1.5 L [OH-] = 0.0266 moles / 1.5L = 0.0177 M

pOH = 1.75

pH = 12.2