SInce #HCl# is a strong acid and #Ba(OH)_2# is a strong base, you are dealing with a neutralization reaction.
Let's start by writing the balanced chemical equation
#2HCl_((aq)) + Ba(OH)_(2(aq)) -> BaCl_2(aq) + 2H_2O_((l))#
Strong acids and strong bases dissociate completely in aqueous solution, so the reaction's complete ionic equation is
#2H_((aq))^(+) + 2Cl_((aq))^(-) + Ba_((aq))^(2+) + 2OH_((aq))^(-) -> Ba_((aq))^(2+) + 2Cl_((aq))^(-) + 2H_2O_((l))#
After eliminating the spectator ions - the ions that can be found both on the reactants, and on the products' side - the net ionic equation is
#H_((aq))^(+) + OH_((aq))^(-) -> H_2O_((l))#
Let's determine the number of moles for each reactant using their respective concentration, #C = n/V#
#n_(HCl) = C_(HCl) * V_(HCl) = 0.10M * 50.0 * 10^(-3)L = 5.0 * 10^(-3)# moles
#n_(Ba(OH)_2) = C * V = 0.20M * 150.0 * 10^(-3)L = 30 * 10^(-3)# moles
We've determined from the balanced chemical equation that 2 moles of #HCl# need 1 mole of #Ba(OH)_2#; however, notice that we have more moles of #Ba(OH)_2# than of #HCl#, which means that the mixture will contain an excess of #Ba^(2+)# and #OH^(-)# ions.
The final concentrations will be
#C_(HCl) = n/V_(TOTAL) = (5.0 * 10^(-3) mol es)/((50.0 + 150.0) * 10^(-3)L) = 0.025M#
#C_(Ba(OH)_2) = (30.0 * 10^(-3) mol es)/((50.0 + 150.0) * 10^(-3)L) = 0.15M#
Therefore,
#[H^+] = 0.025M#
#[Cl^-] = 0.025M#
#[Ba^(2+)] = 0.15M#
#[OH^-] = 0.15M#
