Calculate the pH of a 0.206 M Ba(OH)2 solution.

Answer 1

Ba(OH)₂ is a strong base, so it will ionize completely in an aqueous solution to form Barium and hydroxide ions.

Ba(OH)₂ → Ba²⁺ + 2OH^-

For 1 mole of Ba(OH)₂ that ionizes, 2 moles of OH^- are formed.

[OH^-] = 2 x 0.206 M = 0.412 M

pOH = -log [OH^-] = -log 0.412 = 0.39

pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 0.39 = 13.61