Assuming complete dissociation, what is the pH of a 3.09 mg/L Ba(OH)2 solution?

Answer 1

Ba(OH)₂ ==> Ba²⁺ + 2OH^- .. complete dissociation

molar mass Ba(OH)₂ = 171.3 g / mol

moles Ba(OH)₂ = 3.09 mg x 1 g / 1000 mg x 1 mol / 171.3 g = 1.804x10^-5 mols Ba(OH)₂

moles OH^- = 1.804x10^-5 mols Ba(OH)₂ x 2 mol OH^- / mol Ba(OH)₂ = 3.608x10^-5 mols OH^-

Since we have 1 L of solution, [OH^-] = 3.608x10^-5 M

pOH = -log [OH^-] = -log 3.608x10^-5 = 4.44

pH = 14 - 4.44

pH = 9.56