1.

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

Answer»

SOLUTION :`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`


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