Question

Calculate the pH of 0.02M Ba(OH)2​

Answer 1

Answer:

12.6

Explanation:

Ba(OH_2) to Ba^(2+)+2OH^-

therefore [OH^-]=2[Ba(OH)^2]

=2xx0.02 =0.04 M

therefore pOH=-log [OH^-]

=1.398 - 1.40

therefore pH=14-1.4=12.6