Assuming complete dissociation, what is the ph of a 3.24 mg/l ba (oh) 2 solution?

We need to first find the molarity of Ba (OH₂) solution.

A mass of 3.24 mg is dissolved in 1 L solution.

Ba (OH) ₂ moles dissolved - 3.24 x 10⁻³ g/171.3 g/mol = 1.90 x 10⁻⁵ mol

dissociaton of Ba (OH) ₂ is as follows;

Ba (OH) ₂ - - > Ba²⁺ + 2OH⁻

1 mol of Ba (OH) ₂ dissociates to form 2OH⁻ ions.

Therefore [OH⁻] = (1.90 x 10⁻⁵) x2 = 3.8 x 10⁻⁵ M

pOH = - log[OH⁻]

pOH = - log (3.8 x 10⁻⁵)

pOH = 4.42

pH + pOH = 14

therefore pH = 14 - 4.42

pH = 9.58