berapakah pH larutan Ba (OH)2 0,05 M

[tex] {oh}^{ - } = b \times mb \\ {oh}^{ - } = 2 \times (5 \times {10}^{ - 2} ) \\ {oh}^{ - } = 10 \times {10}^{ - 2} \\ {oh}^{ - } = {10}^{ - 1} \\ \\ poh = - log( {oh}^{ - } ) \\ poh = - log( {10}^{ - 1}) \\ poh = - log \: 10 \\ \\ ph = 14 - (poh) \\ ph = 14 - ( - log(10) ) \\ ph = 14 + log(10) [/tex]