`Ba(OH)_2` is a strong base.
`Ba(OH)_2 -> Ba^(2+)+2OH^-`
`Na_2CrO_4 ` is a basic salt formed from the reaction of `NaOH` and `H_2CrO_4`
`Na_2CrO_4 -> 2Na^++CrO_4^(2-)`
Once these two are mixed it will form the yellow precipitate of `BaCrO_4` .
`Ba(OH)_2 +Na_2CrO_4 -> BaCrO_4+2NaOH`
So the answer is;
`Ba(OH)_2_(aq) +Na_2CrO_4_(aq) -> BaCrO_4_((s))+2NaOH_(aq)`
No comments:
Post a Comment