Calculate the pH of 0.02 m `Ba(OH)_2` aqueous solution assuming `Ba(OH)_2` as a strong electrolyte.

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`