Barium hydroxide, #Ba(OH)_2#, is a strong base, so it will react with hydrochloric acid, #HCl#, a strong acid, to produce water and barium chloride, a soluble salt.
Keep in mind that barium hydroxide is not very soluble in aqueous solution, but what does dissolve will dissociate completely to form barium cations, #Ba^(2+)#, and hydroxide anions, #OH^(-)#.
#Ba(OH)_(2(aq)) -> Ba_text((aq])^(2+) + 2OH_((aq))^(-)#
Hydrochloric acid will dissociate completely in aqueous solution to give #H^(+)# cations and #Cl^(-)# anions.
#HCl_((aq)) -> H_((aq))^(+) + Cl_((aq))^(-)#
The chemical reaction between these two compounds will be
#Ba(OH)_(2(aq)) + 2HCl_text((aq]) -> BaCl_(2(aq)) + 2H_2O_((l))#
The complete ionic equation will be
#Ba_((aq))^(2+) + 2OH_((aq))^(-) + 2H_((aq))^(+) + 2Cl_((aq))^(-) -> Ba_((aq))^(2+) + 2Cl_((aq))^(-) + 2H_2O_((l))#
The net ionic equation, which you get if you remove spectator ions, will be
#cancel(Ba_((aq))^(2+)) + 2OH_((aq))^(-) + 2H_((aq))^(+) + cancel(2Cl_((aq))^(-)) -> cancel(Ba_((aq))^(2+)) + cancel(2Cl_((aq))^(-)) + 2H_2O_((l))#
#2OH_((aq))^(-) + 2H_((aq))^(+) -> 2H_2O_((l))#
which is equivalent to
#OH_((aq))^(-) + H_((aq))^(+) -> H_2O_((l))#
