(1) what is the ph of a 0.015 m aqueous solution of barium hydroxide (ba(oh)2)? a.12.25 b.1.82 c.12.48 d.1.52 e.10.41

According to the reaction equation:

            Ba(OH)2    ↔     Ba2+     +    2OH-
when   0.015 M            0.015M         2*0.015M

∴ [OH-] = 0.03 m 

so we can get the POH = - ㏒ [OH-] 

by substitution:

∴ POH = -㏒ 0.03

            = 1.52

and then we can get the PH from this formula:

PH + POH = 14 

so by substitution:

∴PH = 14 - 1.52
 
        ≈ 12.48

So3 that’s what i think