The Ionization of Ba(OH)₂ is as follow,
Ba (OH)₂ → Ba²⁺ + (OH⁻)₂
Or,
Ba (OH)₂ → Ba²⁺ + OH⁻ + OH⁻
As,
Concentration of Ba(OH)₂ is 1.0 × 10⁻³, and each Barium Hydroxde contains two OH⁻ then concenteration of OH⁻ is calculated as,
[OH⁻] = 2 × 1.0 × 10⁻³
[OH⁻] = 0.002 M
Or,
[OH⁻] = 1.0 × 10⁻³
