Answer to Question #125821 in General Chemistry for Drishika

Question

Determine the pH of an aqueous solution of 0.05M Ba(OH)2

Answer 1

Ba(OH)₂ will completely dissociate into Ba²⁺ and 2OH^- ions

OH^- concentration will be double of concentration of Ba(OH)₂

so [OH^-] = 2*0.05 M = 0.1M

pOH= -log([OH^-])

pOH= -log(0.1)

pOH = 1

pH= 14- pOH

pH = 14-1 = 13