20‌mL of 0.50M‌HCl=20×0.050m‌mol =1.0m‌mol=1.0‌meq of HCl 30‌mL of 0.10M‌Ba(OH)2 =30×0.1m‌mol =3m‌mol=3×2‌meq =6‌meq‌Ba(OH)2 1 meq of HCl will neutralize 1 meq of Ba(OH)2 Ba(OH)2 left =5‌meq. Tatal volume =50‌mL Ba(OH)2 conc. in final solution =‌