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